My man, I need to speak some words of gratitude, admiration and support. I just recently discovered your channel and I am in love with your style of videos. You present your journey of having cracked the problem yourself, you give a structured approach to make everything understandable and your never fail to keep the content interesting. Keep up the wonderful work brother!
Set p = k-1. Then, n 2^n = p(p+2). Clearly p is even as n is positive. Set p = 2q. Gives n 2^(n-2) = q(q+1). This leads to n and 2^(n-2) being two consecutive integers (due to odd-even combination). This in turn means that there are only two solutions: n-1 = 2^(n-2) and n+1 = 2^(n-2) leading to n = 2 and 3.
Very smart thinking. I didn't see that n*2^(n-2) =q(q+1) initially but we cannot remove other factors than 2 from the 2^(n-2) part of the product and if we do remove factors then the difference between n*f and 2^(n-2)/f will be 0 or a multiple of 2 leading to the consecutive statement. Really nice and elegant
@@KiLLJoYTH-camI really like the simplicity of this proof but I think that you are right the last step isn't quite right. The conclusion is only roght if n is odd or 2^(n-2) is odd. I think the best way to finish the proof is to note that the even part of n*2^(n-2) must be at most 1 unit larger than the odd part. This even part is greater than or equal to 2^(n-2) and the odd part is smaller than or equal to n. Since 2^(n-2)>n+1 for n>4 (2^3>5+1), you just need to check n=1,2,3,4. And, of course we get the right result then
Sorry for not clarifying the last step, which I put in paranthesis for the sake of brevity. However, my thought process was nicely summarized in the comments by @alucs6362. The only thing I would add is this. * If one of them, n or 2^(n-2), is odd, we are good. * Consider n is even. Since 2^(n-2) has no odd factors, n has to be a multiple of an odd number greater than or equal to 3, i.e. n > 5. Now, n is greater than the odd number and so 2^(n-2) lesser than the even number, leads to 2^(n-2) < (n+1). This inturn gives n < 5, which is a contradiction. So, n can't be even and a multiple of an odd number. Brings back to n and 2^(n-2) being consecutive.
Sorry I didn't clarify the last step fully. I put it in the paranthesis for the sake of brevity. However, it was nicely deciphered in the comments. Thanks all. My reasoning was pretty much along the same lines but with logic by contradiction. Clearly n and 2^(n-2) cannot be both odd. Suppose n and 2^(n-2) are both even. The odd part, which should be at least 3 has to be contained in n. So, n > 5. On the other hand, n is greater than the odd part, while 2^(n-2) is less than the even part. Since the odd and even parts differ by one, 2^(n-2) > (n+1). This leads to n < 5, which is a contradiction. So, one of the two, n and 2^(n-2), is odd and the other is even. And by construction they have to be adjacent numbers.
you can say n2^n + 1 is a square so by (a+b)^2 we can say a^2 + 2a = n2^n let a be 2^x n2^n = 2^2x + 2*2^x n2^n = 2^2x + 2^(x+1) n2^n = 2^(x+1) [ 2^(x-1) + 1] n2^n = [ 2^(x-1) + 1] * 2 ^ (x+1) so here if we assume x + 1 = n then 2^(x-1) +1 should also be equal to n x+1 = 2^(x-1) + 1 x = 2 ^ (x - 1) 2x = 2 ^ x how here you can either make an educated guess that x = 1 and x = 2 satisfies solution or you can solve for x with some rearrangement and lambert w function so if x = 1 then n = x +1 aka n = 2 and for x = 2 ; n = 3 or you can use other equation also for finding out n If any one is reading this please let me know is there any problem with my solution as i am not from maths background
@@pedrogarcia8706 No, they aren't. Similar, but not the same; as I wrote above. Anyway, forgoting is not the way to go. It would be half the scores in a test.
Unfortunetly, this is generally true! Number theory is almost totally neglected. Most math majors have to study it on their own. Thankfully, there are many wonderful books available in this subject!
So we know that k^2 is a perfect square and that k must be odd, then we can write k = 2m + 1 and we have n * 2^n + 1= (2m+1)^2 = 4m^2 + 4m + 1 so that n * n^2 = 4m(m+1) but this m(m+1) must be even so can be written as 2*m (reuse same variable and m is any integer). We then have that n*2^n = 8 * m = 2^3 m or 2^2 * 2m, as m is either odd or even. For 2^1 * 4m and 2^0 * 8m, these are just same as 2^2 * 2m and 2^3 * m. Only solution is n = 2 or 3. Nice little problem! Good work!
for future videos, can you give us a harder problem at the end that we can solve using the same principle with the answer in a pinned comment or in the description or something? Would really love that
For a positive integer n we need to find a positive integer k such that n * 2^n = k^2 - 1 = (k - 1) * (k + 1). It is obvious that this only works if k is an odd integer >= 3 because the RHS must be even. So with k - 1 = 2l we can write n * 2^n = 2l * (2l + 2) = 4I(l + 1) for an integer l >= 1. For n < 5 the only solutions are n = 2 (for k = 3) and n = 3 (for k = 5) which can be easily verified. For n >= 5 there are no further solutions: On the one hand we have n * 2^(n - 2) = l(l + 1), so 2^(n - 2) must divide either l or l + 1 which implies l >= 2^(n - 2) - 1. Therefore n * 2^(n - 2) = l(l +1) >= 2^(n - 2) * (2^(n - 2) - 1), so necessarily n >= 2^(n - 2) - 1 or equivalently 2^(n - 2) n + 1 for all n >= 5, e.g. by mathematical induction. This yields a contradiction. Thus, n = 2, 3 are the only positive integers solving the problem.
I think it is simpler to consider from the line n*2^n=(k-1)(k+1) and well demonstrated that both (k-1) and (k+1) are even numbers, just write k-1=2a and k+1=2a+2 (as even numbers can be always written in that manner for consecutives) Then n*2^n=2a(2a+2)=4a(a+1) and for n>3 we will have n2^(n-2)=a(a+1) Knowing that a and a+1 are consecutive numbers, one on them is odd and their product can never be double multiple of 2 ie can not be multiple of 4. But when we look at the left side of the equation, 2^(n-2) is multiple of 4 conclusion n2^(n-2)=a(a+1) does not have any solution for n>3 By this we have restricted solutions to be less than 3 and becomes easier to solve.
The expression is odd, hence if it was a square, it would be a square of an odd number. So let n2^(n)+1 = (2k+1)^(2) = 4k^(2) + 4k + 1 Hence assuming n >= 2 n2^(n-2) = k(k+1) Exactly one of k and (k+1) is even, so all powers of 2 must be part of it, so if 2^(n-2) > n+1, we cannot factor n2^(n-2) this way. Therefore n+1 >= 2^(n-2) Which means n
Nice work! One point I noticed, however: at 11m45s, you have a chain of 3 inequalities, the 2nd of which is a strict inequality. You then conclude that LHS ≥ RHS, when you could have made that inequality strict, with LHS > RHS. About a minute later, this would have helped you cut the last possibility you were checking. Fred
@@PrimeNewtons Well, it doesn't detract from the correctness of your solution, so it's all to the good. And after looking down the other comments, I see that many others beat me to this observation.
When Michael Penn does a number theory problem, I know I’d never solve it by myself. When you do a number theory problem, I have hope that I might be able to do it or be able to follow you. Thank you. I did 4 years of pure math at university. It was all algebra, calculus, differential equations, analysis, complex numbers, but never any number theory.
Solved it in a similar fashion. Check n=0 as special case then after accepting that RHS is even, put k = 2q+1. Drag out 4 from RHS and get n*2^(n-2) = q*(q+1) for some q integer. n is clearly =>2. In fact n=2 and n=3 are solutions. For n>3, n has to be uneven to match RHS. But clearly 2^(n-2) grows much quicker than n so no more solutions are possible.
I did n=1 to n=6 in my head and my intuition made me feel that there were no other solutions other than 2 and 3. But intuition is not proof and your proof is clear and concise.
n2^n + 1 is odd and assume it is a perfect square call it m^2, then n2^n is even and is a difference of squares so m^2 - 1 = (m + 1)(m - 1) = n2^n and both m + 1 and m - 1 are even, so m + 1 = x2^k so m - 1 = x2^k - 2 so (x^2)2^(2k) - 2x2^k = n2^n However, 2^n | n2^n so 2^n | 2x2^k so 2^(n - k - 1) | x So (C^2)2^(2n - 2) - C2^n = n2^n So (C^2)2^(n - 2) - C = n And (C^2)2^(n - 2) = n + C 2^n = 4n/C^2 + 4/C So for this to be true C | 4 so C = 1, 2, or 4 If C = 1: 2^n = 4n + 4 and 2^n grows faster than 4n + 4, so once 2^n is greater than 4n + 4 we don’t need to check any other n’s, if n = 1 we get 2 = 8 which is not true, if n = 2 we get 4 = 12 which is not true, if n = 3 we get 8 = 16 which is not true, if n = 4 we get 16 = 20 which is not true, if n = 5 we get 32 = 24 which is not true, and the last case, so If C = 2 we get 2^n = n + 2 which if n = 1 we get 2 = 3 which is not true, if n = 2 we get 4 = 4 which is true. So box that, and if n = 3 we get 8 = 5 which is not true and the last subcase for that. If C = 4 we get 2^n = n/4 + 1, which is only true for n = 4z so 2^(4z) = (4z)/4 + 1 or 16^z = z + 1 which for any z > 0, 16^z > z + 1, so we know this case doesn’t work. Now finally we have n = 2, let’s test to make sure it is not an extraneous solution: 2(2^2) + 1 = 2(4) + 1 = 8 + 1 = 9 = 3^2, so yes, this does work, and is our only solution.
It is very simple to prove the theorem let us take a 2^n and its consecutive even terms are 2^n+-2 so lets take 2^n+2=2(2^n-1+1) ,here inside bracket is odd so its obvious that its factorization contains only one 2 and its same for 2^n-1 also!!
I went at this a completely different way but I think it works. Observing that n 2^n + 1 is odd means that the perfect square must be odd and therefore a square of an odd number. Thus 2^n + 1 = (2 m + 1)^2. A bit of simplification and you get n 2^(n-2) = m (m+1). If m is even then m +1 is odd and since 2^(n-2) is even then n = m +1. Plug that in and we get 2^(m-1) = m which has solutions m = 1, 2 and thus n = 2, 3. If m is odd then n = m and 2^(m -2) = (m +1)/m which has no solutions. QED. Anyone see a flaw in the reasoning?
n*2^n+1 is an odd number, which it must be the square of an odd number, therefore we can write n*2^n+1=(2*k+1)^2. This leads to n*2^(n-2) = k*(k+1). We observe that k and k+1 are consecutive integers, which means they are coprime. 2^(n-2) - n > 1 for n≥5. This means that n*2^(n-2) can't possibly be the product of two consecutive integers (for n≥5) since one of them must be at least 2^(n-2) (the other number has to be odd) and the other one can be at most n. That leaves only n ∈ {1,2,3,4} as possible candidates. n=1 doesn't work because 1*2^(1-2) is not an integer and thus can't be a product of integers. n=2 works because 2*(2^(2-2)) = 2 can be factorized into consecutive integers 1*2 (k=1). Therefore 2*2^2+1 = 9 = (2*1+1)^2 n=3 works because 3*(2^(3-2)) = 6 can be factorized into consecutive integers 2*3 (k=2). Therefore 3*2^3+1 = 25 = (2*2+1)^2 n=4 doesn't work because 4*2^(4-2) = 16 can only be factorized into coprime factors as 1*16, which are not consecutive numbers. More values of n just for fun: n=5 -> 5*2^(5-2) = 40 --> coprime factorizations: 5*8, 1*40 n=6 -> 6*2^(6-2) = 96 --> coprime factorizations: 3*32, 1*96 n=7 -> 7*2^(7-2) = 224 --> coprime factorizations: 7*32, 1*224 n=8 -> 8*2^(8-2) = 512 --> coprime factorizations: 1*512 n=9 -> 9*2^(9-2) = 1152 --> coprime factorizations: 9*128, 1*1152 n=10 -> 10*2^(10-2) = 2560 -> coprime factorizations: 5*512, 1*2560 n=11 -> 11*2^(11-2) = 5632 -> coprime factorizations: 11*512, 1*5632 ... It's easy to see that the powers of two completely dominate the coprime factorization for all larger values of n with the remaining factor being much smaller.
Find all positive integers n such that n2^n+1 is a square. n2^n+1=k^2 2(2)^2+1=9=3^2 3(2)^3+1=25=5^2 k = -1, n = 0 k = 1, n = 0 k = ± 5, n = 3 k = ± 3, n = 2 Positive integers of n=2,3
I just started looking at this, but I attacked it differently. I'm looking at trends when the statement is true. It is true when n=2 and n=3. This is as far as I've gotten in a few minutes.
hahaha, I paused the video at the start and tried solving it on my own and went down a completely different (and incorrect) path. I started with: n*2^n+1=x² since I wanted x to be an integer in order for it to be a perfect square, I tried to make it only true for integers. i figured that in order to make f(n) = g(x) only true for x being an integer, I make the following equation: (f(n)-g(x)) + cos(2*pi*x)-1 = 0 if I plug in n2^n+1 in for f(n), and x² for g(x), I get: n * 2^n + 1 - x² + cos(2*pi*x) - 1 = 0 | since +1 and -1 cancel out, simplify n * 2^n - x² + cos(2*pi*x) = 0 ...and then I realized that I can't really get anywhere from there lol.
The problem is to find all positive integers n such that n * 2^(n+1) is a perfect square. To solve this, the expression can be represented as k^2, where k is a square root. The key insight is that k-1 and k+1 must be consecutive even numbers, which leads to a restriction on the possible values of n. By considering the relationship between the linear function 2n and the exponential function 2^(n-1), the valid range of n can be determined to be from 1 to 4. The solutions are then tested by checking if n * 2^(n+1) results in a perfect square for each valid n.
Explanation of the Euclidean algorithm: If b = c (mod a), then gcd(a, b) = gcd(a, c). The reason for this is b = c (mod a) means b = c + ak for some k. Now, let d = gcd(a, b) and g = gcd(a, c). Since g divides a and c, it divides b. Thus, g divides a and b, and thus is a common factor of them. Thus, it divides the gcd of a and b, which is d. Now, rearranging b = c + ak to c = b - ak, it is clear to see that since d divides a and b, it also divides c. Thus, d divides a and c, and is thus a common factor of them, and thus divides their gcd, which is g. Thus, d and g divide each other, and so they are equal. So, in the problem, the reason gcd(x - 1, x + 1) = 2 is because first, we know x = 1 (mod x - 1), so that means x + 1 = 2 (mod x - 1) (just replace x with 1). Thus, gcd(x - 1, x + 1) = gcd(x - 1, 2). gcd(x - 1, 2) = 2 because we know x is odd, and so x - 1 is even, aka it has at least one factor of 2. In general, the Euclidean algorithm is nice for finding the gcd of large integers. Let's say we want the gcd of 78 and 96, we can take 96 mod 78 and get that we just need the gcd of 78 and 18, then we can take 78 mod 18 to get that we just need the gcd of 18 and 6, which is 6. Thus, the gcd of 78 and 96 is 6. In fact, you can even apply the Euclidean algorithm to polynomials. The gcd of f(x) and g(x) is the gcd of f(x) and g(x) (mod f(x)). You can obtain g(x) (mod f(x)) by polynomial long dividing g(x) by f(x) and taking the remainder. Make sure you long divide by the one of f(x) or g(x) with smaller (or equal) degree, because then the remainder will have reduced degree, which is what you want. After all, the purpose of an algorithm is to simplify a problem.
Here's another approach: Find all n such that: n*2^n+1 = m^2. Since m is odd let m = 2*k + 1 for k greater than or equal to 0. This results in: n*2^n+1 = 4*k^2 + 4*k + 1. This simplifies to: n*2^(n - 2) = k*(k + 1). Therefore: n and 2^(n - 2) are consecutive numbers, one odd and one even. 2^a is an integer only when a greater than or equal to 0, so n is greater then or equal to 2. Therefore: n = 0 for 2^(n - 2) odd. n = an odd number for all other solutions. If n = 2 then 2^(n - 2) = 1, k = 1 and m = 3. A good solution. If n = 3 then 2^(n - 2) = 2, k = 2 and m = 5. Also, a good solution. If n = 5 or greater, then: 2^(n - 2) - n is greater than 1, no solution are possible.
this may be wrong. please don't yell at me, just show me my mistake I thought about it using modulo for (k+1) and (k-1) one has to be ≡ 0 mod n and the other ≡ 0 mod 2ⁿ to me that only works when n+1 = k with n = 4, k is greater than 5 and k increases faster than n your explanation is much better/simpler than mine
n*2^n = X^2 = (X-1)(X+1) = 2k*2(k+1) = 4k(k+1) n*2^(n-2) = k(k+1) We have 2 consecutive numbers, one odd and one even. So 2^(n-2) | k or 2^(n-2) | (k+1), and n divides the other number. Therefore |2^(n-2) - n| = |(k+1) - k| = 1. The only 2 numbers where this equality holds are k = 1 and k = 2. With n = 2 and n = 3. When n = 4, the absolute value is 0 and when n> 4, |2^(n-2) - n| > 1. So n=2 and n=3 are the only 2 solutions.
k+1 > k-1, so 2n > 2^(n-1). One question, is there no need to examine n2^n = (k-1)(a2^(n-1)) (the right side part)? We only tried the left part ( n2^n = (a2^(n-1))(k+1) ). 2nd question: @5:39 I am not convinced with this: any number that has many power of 2, the other number is just 2 times odd number. You assume that n is no more divisible by 2 in n*2^n, but suppose n=15. Then n*2^n=3*5*2^(3*5)=(3*2^5)*(5*2^3). Both are odd number times power of 2. Of course, in this example, these numbers are not consecutive numbers, but still we cannot assume one part of odd number times 2 and the other is just power of 2, right?
@@PrimeNewtonsYes, if we choose any number of power of 2 as one of consecutive numbers, the claim is true. However, the only condition we have is n2^n (n is any combination of odd numbers). So, how can we say one of consecutive numbers is just a power of 2? Since n2^n can be broken down to l * m * 2^(n-a+a), it can be one of a pair of ( l*2^(n-a), m*2^a) (l, m: odd numbers, l*m=n). I understand you point if we choose, say 32, 64, 128, or 256... But when not? Remember, we have to "find all" possible answers. Simply put, we need to prove that if the number n2^n can be expressed as the product of two integers, and the difference between those two numbers is 2, then the only way to divide the number is into the two factors 2n and 2^(n-1). And this is not so easy task. Becasue... suppose n2^2=a * b and a - b =2. Then b = -1 +/- SQRT(1 + n2^n). b must be an integer, so (1 + n2^n) must be a squared number, which is exactly the problem itself!!! So, in proving the lemma, we have to solve the problem. We are not allowed to assume the lemma is true unless we solve it without using the lemma. Otherwise, this is a tautology.
Nice problem. If n2^n + 1 is a square then there is a positive integer x such that n2^n = x^2 - 1 = (x - 1)(x + 1). Taking both sides mod 2, it is clear that x must be odd, and thus x - 1 and x + 1 must both be even. Furthermore, the Euclidean algorithm gives us that gcd(x - 1, x + 1) = gcd(x - 1, 2) = 2 because x + 1 = 2 (mod x - 1). Thus, either x - 1 or x + 1 has a factor of 2^(n - 1). Note that for n >= 5, 2^(n - 1), which is the minimal size of the factor that does have 2^(n - 1) as a factor, is bigger than 2n, which is the maximal size of the factor that does not have 2^(n - 1) as a factor, and thus the larger of the two factors, x + 1, must be the one with 2^(n - 1) as a factor. Now notice that 2 = (x + 1) - (x - 1) >= 2^(n - 1) - (x - 1) >= 2^(n - 1) - 2n > 2. This is a contradiction, so we must have n < 5. n = 1 does not yield a solution because 3 is not a square n = 2 does yield a solution because 9 is a square n = 3 does yield a solution because 25 is a square n = 4 does not yield a solution because 65 is not a square Thus, the solutions are n = 2, 3
Method 1) (- x= 3) equation is given Multiplying both sides by (-1) -1*-x=-1*3 Then x=-3 or Method 2) Let the equation be (- x= 3) If we multiply both sides with "MINUS" sign -(- x)= -(3) Then x= -3. Which one is correct or both methods are correct . Please help 🙏🙏
shouldn't one of the numbers be 2 power something while the other is 2 into some odd no. so why did we take' a' into 2^n-1 wouldn't that just mean the other number is 2 ? what exactly is a ?
a is necessary as two consecutive numbers can be i.e: 48 and 50. 48*50 can be written as 75*2^5 or 150*2^4. None of them holds for n*2^n, but we do not know whether there are such two numbers that hold when we start solving.
We can write any square number(>1) in the form 4k or 4k+1 As n2^n+1 is always odd We can say n2^n +1= 4k+1 n2^n=4k Here we can easily say that n=k and 2^n=4 Thus N=2
Does 0 not count as a positive integer or does 1 not count as a square number? Because if both of those are what I think they are you may have missed one. I know technically 0 is neither positive nor negative, but it makes me question if it counts ((0)(2^0) +1 == 1 as far as I can understand
I solved n×2^n=k^2-1 using Lambert W function then used the properties of the W function to find Integer solutions. I could prove that 2 is a solution (without numerical testing) but couldn't use the method to get to n=3. I guess the W function has a property that I am not aware of.
I appreriate the hardwork But i expecteda little more elegant soln This one has many loopholes where your assumption contradicts Either they give a ans or not u should have covered them
6:56 I’m not sure about that method. The left side has n powers of 2 only if n is odd. If n is a power of 2, the term with the multipower of 2 will be a • 2^(n + the number of powers of 2 in a)
3:01 you forget to mention that n is an natural number because else it dont have to be the case! (nevermind i haven´t read the task :D integer over 1) 4:45 you repeat yourself we got it :D (just info! i have no problem with it) 6:04 ok therefore the prove would be nice to knowe 1:05 no please prove that :D :D ^^ but without term reforming :D 11:50 i think it is less and not equal because it can´t be > k-1 and that is less than 2n right? 12:24 i thought you forget it :D 14:15 haven´t we allready faild because equal is not working? 16:26 ok i was right it don´t work 16:35 the old but good advice :D good bye K.Furry ps: 16:42 confused me
n*2^n= k^2-1 n*2^n divisible by k+1 and k-1 n*2^n, n=k+1 or n=k-1, or n is combinable with 2^n. n is not combinable with 2^n, as no perfect square is adjacent to an integer exponent. 2^9+-1 cannot be a perfect square. or k+1,k-1 is divisible by something else. for n=k+1,n=k-1, 2 options: n-1 =k, or n+1=k 2^n+1=k or 2^n-1 =k n+1=2^n-1 n=2, 3=4-1, therefore 2 is a solution, no other solutions of this form, same reasoning. n-1=2^n+1 n=1, 0=3, therefore no solutions. for k+1 divislbe by something else: n= (k-1)/u, 2^n=(k+1)*u, and vice versa. 4 options, Most plausable case is the one stated: n*u+1=2^n/u-1 n=3, u+1=2^3/u-1 u=2, 3=3 n=3,u=2 is a solution. Check the other 3 for any other solutions, but i have finished the video so it is not neccesary. Wonderful as always, harder problem than i realized.
For any n, the expression will be odd, we have an even term plus one. Then if it is a square, then it will be by the form: n * 2^n + 1 = (2k + 1)^2, k is a natural number. We don't use (2k-1) because n * 2^n will be at least 3 for n = 1. Expand the right side and get n * 2^n + 1 = 4k^2 + 4k + 1 n * 2^n = 4k^2 + 4k n * 2^n = 4(k^2 + k) n * 2^(n-2) = k^2 + k n * 2^(n-2) = k(k+1) From here we can say two things: first, n cannot be less than 2 because it would make 2^(n-2)to be less than 1. Second, the right side says that n * 2^(n-2) is the product of two consecutive integers. Since 2^n grows obviously faster than n, there will be a point where 2^n will be so far from n. For example, for n = 6, you get 6 * 2^(6-2) = 6 * 16. Already a big difference, so we don't have to check n >= 6. For n = 2, you get 2 * 2^(2-2) = 2 * 1. Correct For n = 3, you get 3 * 2^(3-2) = 3 * 2. Correct For n = 4, you get 4 * 4. For n = 5, you get 5 * 8. n = 2 and n = 3 are the only solutions.
By observation, it is evident that out of two consecutive even integers, one will be of the form 2*some even number and the other will be of the form 2*some odd number eg. 60 = 2*30 while 62 = 2*31. Thus it is evident that the product of two consecutive even integers must be such that out of the total powers of the 2, one must be from the integer of the form 2*odd number while all the other n-1 must come from the integer of the form 2*even number.
I have 2 questions : 1) why did we ignore n*2^n = (k-1)a2^(n-1) ? 2) n*2^n = (k-1)a2^(n-1) (and the other) It looks like it works but how to we prove it ?
@@glorrin Indeed! But you can get the table of values from n>2^(n-2) as in the video. You get a similar table (same values for n) from n>2^(n-2)-1. These values for n are 1,2,3,4 in both cases, and only 2 and 3 are valid.
If i want to be pedantic: 0 is a solution of the equation , but it is not a valid answer to the question, as the question asked for all positive integers, which by convention excludes 0.
I think we can say that 2n > 2^(n-1) rather than ≥, but all this does is save us from checking n=4. Thanks for the video!
Yaah
I'm going to comment the same 😅
I was thinking of this exactly and was about to comment until I saw this
You’re simply amazing. Excellent presentation.
I think that the most insightful comment of the video was that of building an inequality to bound the number of choices that you would need to try.
He is just fantastic in explaining his methodology before actually resolving. Elfantastico 🎉
Love the process. So clean, clear, and eloquent.
Love from India
Thank you sir
Great learning with you
My man, I need to speak some words of gratitude, admiration and support. I just recently discovered your channel and I am in love with your style of videos. You present your journey of having cracked the problem yourself, you give a structured approach to make everything understandable and your never fail to keep the content interesting. Keep up the wonderful work brother!
Beautiful, outstanding!!!
sir you are the best, your videos makes all us feel math is beautiful
Set p = k-1. Then, n 2^n = p(p+2). Clearly p is even as n is positive. Set p = 2q. Gives n 2^(n-2) = q(q+1). This leads to n and 2^(n-2) being two consecutive integers (due to odd-even combination). This in turn means that there are only two solutions: n-1 = 2^(n-2) and n+1 = 2^(n-2) leading to n = 2 and 3.
Very smart thinking.
I didn't see that n*2^(n-2) =q(q+1) initially but we cannot remove other factors than 2 from the 2^(n-2) part of the product and if we do remove factors then the difference between n*f and 2^(n-2)/f will be 0 or a multiple of 2 leading to the consecutive statement.
Really nice and elegant
Could you expand on why n and 2^(n-2) are proven to be consecutive integers?. For q(q+1) it may be the case, but how does it translate to n* 2^(n-2) ?
@@KiLLJoYTH-camI really like the simplicity of this proof but I think that you are right the last step isn't quite right. The conclusion is only roght if n is odd or 2^(n-2) is odd.
I think the best way to finish the proof is to note that the even part of n*2^(n-2) must be at most 1 unit larger than the odd part. This even part is greater than or equal to 2^(n-2) and the odd part is smaller than or equal to n. Since 2^(n-2)>n+1 for n>4 (2^3>5+1), you just need to check n=1,2,3,4. And, of course we get the right result then
Sorry for not clarifying the last step, which I put in paranthesis for the sake of brevity.
However, my thought process was nicely summarized in the comments by @alucs6362.
The only thing I would add is this.
* If one of them, n or 2^(n-2), is odd, we are good.
* Consider n is even. Since 2^(n-2) has no odd factors, n has to be a multiple of an odd number greater than or equal to 3, i.e. n > 5. Now, n is greater than the odd number and so 2^(n-2) lesser than the even number, leads to 2^(n-2) < (n+1). This inturn gives n < 5, which is a contradiction. So, n can't be even and a multiple of an odd number.
Brings back to n and 2^(n-2) being consecutive.
Sorry I didn't clarify the last step fully. I put it in the paranthesis for the sake of brevity. However, it was nicely deciphered in the comments. Thanks all.
My reasoning was pretty much along the same lines but with logic by contradiction.
Clearly n and 2^(n-2) cannot be both odd.
Suppose n and 2^(n-2) are both even. The odd part, which should be at least 3 has to be contained in n. So, n > 5. On the other hand, n is greater than the odd part, while 2^(n-2) is less than the even part. Since the odd and even parts differ by one, 2^(n-2) > (n+1). This leads to n < 5, which is a contradiction.
So, one of the two, n and 2^(n-2), is odd and the other is even. And by construction they have to be adjacent numbers.
Great vídeo, absolutely amused by your enthusiasm!
"Never stop learning", such a great lesson!
you can say n2^n + 1 is a square so by (a+b)^2 we can say a^2 + 2a = n2^n
let a be 2^x
n2^n = 2^2x + 2*2^x
n2^n = 2^2x + 2^(x+1)
n2^n = 2^(x+1) [ 2^(x-1) + 1]
n2^n = [ 2^(x-1) + 1] * 2 ^ (x+1)
so here if we assume x + 1 = n then 2^(x-1) +1 should also be equal to n
x+1 = 2^(x-1) + 1
x = 2 ^ (x - 1)
2x = 2 ^ x
how here you can either make an educated guess that x = 1 and x = 2 satisfies solution or you can solve for x with some rearrangement and lambert w function
so if x = 1 then n = x +1 aka n = 2 and for x = 2 ; n = 3 or you can use other equation also for finding out n
If any one is reading this please let me know is there any problem with my solution as i am not from maths background
Great Great!!!!❤️❤️👏👏
Thanks brother ❤
You simply forgot about the 2nd case, which leads to 2n >= 2^{n-1}-2, which gives finally the same constraint n
Aren't both cases the same?
@@pedrogarcia8706 No, they aren't. Similar, but not the same; as I wrote above. Anyway, forgoting is not the way to go. It would be half the scores in a test.
Thanks Sir 👍
Unfortunetly, this is generally
true! Number theory is almost
totally neglected. Most math
majors have to study it on
their own. Thankfully, there
are many wonderful books
available in this subject!
Awesome logic!!
Well dine
So we know that k^2 is a perfect square and that k must be odd, then we can write k = 2m + 1 and we have n * 2^n + 1= (2m+1)^2 = 4m^2 + 4m + 1 so that n * n^2 = 4m(m+1) but this m(m+1) must be even so can be written as 2*m (reuse same variable and m is any integer). We then have that n*2^n = 8 * m = 2^3 m or 2^2 * 2m, as m is either odd or even. For 2^1 * 4m and 2^0 * 8m, these are just same as 2^2 * 2m and 2^3 * m. Only solution is n = 2 or 3.
Nice little problem! Good work!
for future videos, can you give us a harder problem at the end that we can solve using the same principle with the answer in a pinned comment or in the description or something? Would really love that
For a positive integer n we need to find a positive integer k such that n * 2^n = k^2 - 1 = (k - 1) * (k + 1). It is obvious that this only works if k is an odd integer >= 3 because the RHS must be even. So with k - 1 = 2l we can write n * 2^n = 2l * (2l + 2) = 4I(l + 1) for an integer l >= 1.
For n < 5 the only solutions are n = 2 (for k = 3) and n = 3 (for k = 5) which can be easily verified.
For n >= 5 there are no further solutions: On the one hand we have n * 2^(n - 2) = l(l + 1), so 2^(n - 2) must divide either l or l + 1 which implies l >= 2^(n - 2) - 1. Therefore n * 2^(n - 2) = l(l +1) >= 2^(n - 2) * (2^(n - 2) - 1), so necessarily n >= 2^(n - 2) - 1 or equivalently 2^(n - 2) n + 1 for all n >= 5, e.g. by mathematical induction. This yields a contradiction.
Thus, n = 2, 3 are the only positive integers solving the problem.
I think it is simpler to consider from the line n*2^n=(k-1)(k+1) and well demonstrated that both (k-1) and (k+1) are even numbers, just write k-1=2a and k+1=2a+2 (as even numbers can be always written in that manner for consecutives)
Then n*2^n=2a(2a+2)=4a(a+1)
and for n>3 we will have n2^(n-2)=a(a+1)
Knowing that a and a+1 are consecutive numbers, one on them is odd and their product can never be double multiple of 2 ie can not be multiple of 4.
But when we look at the left side of the equation, 2^(n-2) is multiple of 4
conclusion n2^(n-2)=a(a+1) does not have any solution for n>3
By this we have restricted solutions to be less than 3 and becomes easier to solve.
class was fun yesterday
The expression is odd, hence if it was a square, it would be a square of an odd number. So let
n2^(n)+1 = (2k+1)^(2) = 4k^(2) + 4k + 1
Hence assuming n >= 2
n2^(n-2) = k(k+1)
Exactly one of k and (k+1) is even, so all powers of 2 must be part of it, so if 2^(n-2) > n+1, we cannot factor n2^(n-2) this way. Therefore
n+1 >= 2^(n-2)
Which means
n
Nice work!
One point I noticed, however: at 11m45s, you have a chain of 3 inequalities, the 2nd of which is a strict inequality.
You then conclude that LHS ≥ RHS, when you could have made that inequality strict, with LHS > RHS.
About a minute later, this would have helped you cut the last possibility you were checking.
Fred
I realized that while editing the video. Another lesson learned. Thanks.
@@PrimeNewtons Well, it doesn't detract from the correctness of your solution, so it's all to the good.
And after looking down the other comments, I see that many others beat me to this observation.
When Michael Penn does a number theory problem, I know I’d never solve it by myself.
When you do a number theory problem, I have hope that I might be able to do it or be able to follow you. Thank you.
I did 4 years of pure math at university. It was all algebra, calculus, differential equations, analysis, complex numbers, but never any number theory.
Michael Penn solutions are Olympiad type. Like if you're studying for math Olympiads his solutions are most of the time really natural.
And only not having contact with NT is another reason to why you think that.
Love the channel. Here's a problem suggestion...
4^5^9 and 5^6^8 both have just over 1 million digits.
Which is bigger?
Solved it in a similar fashion. Check n=0 as special case then after accepting that RHS is even, put k = 2q+1. Drag out 4 from RHS and get n*2^(n-2) = q*(q+1) for some q integer. n is clearly =>2. In fact n=2 and n=3 are solutions. For n>3, n has to be uneven to match RHS. But clearly 2^(n-2) grows much quicker than n so no more solutions are possible.
I did n=1 to n=6 in my head and my intuition made me feel that there were no other solutions other than 2 and 3. But intuition is not proof and your proof is clear and concise.
n2^n + 1 is odd and assume it is a perfect square call it m^2, then n2^n is even and is a difference of squares so m^2 - 1 = (m + 1)(m - 1) = n2^n and both m + 1 and m - 1 are even, so m + 1 = x2^k
so m - 1 = x2^k - 2
so (x^2)2^(2k) - 2x2^k = n2^n
However, 2^n | n2^n so 2^n | 2x2^k so 2^(n - k - 1) | x
So (C^2)2^(2n - 2) - C2^n = n2^n
So (C^2)2^(n - 2) - C = n
And (C^2)2^(n - 2) = n + C
2^n = 4n/C^2 + 4/C
So for this to be true C | 4 so C = 1, 2, or 4
If C = 1: 2^n = 4n + 4 and 2^n grows faster than 4n + 4, so once 2^n is greater than 4n + 4 we don’t need to check any other n’s, if n = 1 we get 2 = 8 which is not true, if n = 2 we get 4 = 12 which is not true, if n = 3 we get 8 = 16 which is not true, if n = 4 we get 16 = 20 which is not true, if n = 5 we get 32 = 24 which is not true, and the last case, so
If C = 2 we get 2^n = n + 2 which if n = 1 we get 2 = 3 which is not true, if n = 2 we get 4 = 4 which is true. So box that, and if n = 3 we get 8 = 5 which is not true and the last subcase for that.
If C = 4 we get 2^n = n/4 + 1, which is only true for n = 4z so 2^(4z) = (4z)/4 + 1 or 16^z = z + 1 which for any z > 0, 16^z > z + 1, so we know this case doesn’t work.
Now finally we have n = 2, let’s test to make sure it is not an extraneous solution: 2(2^2) + 1 = 2(4) + 1 = 8 + 1 = 9 = 3^2, so yes, this does work, and is our only solution.
❤
About the power of two, you forgot the power of love...
i sat the paper that had this question and i got a medal (47 points)!!!
Nice 🎉😂
Omg this is 2023 BMO and I scored 1/10 for this question 😂 what a clean solution!
Next year you'll do better :).
It is very simple to prove the theorem
let us take a 2^n and its consecutive even terms are 2^n+-2
so lets take 2^n+2=2(2^n-1+1) ,here inside bracket is odd so its obvious that its factorization contains only one 2 and its same for 2^n-1 also!!
I went at this a completely different way but I think it works. Observing that n 2^n + 1 is odd means that the perfect square must be odd and therefore a square of an odd number. Thus 2^n + 1 = (2 m + 1)^2. A bit of simplification and you get n 2^(n-2) = m (m+1). If m is even then m +1 is odd and since 2^(n-2) is even then n = m +1. Plug that in and we get 2^(m-1) = m which has solutions m = 1, 2 and thus n = 2, 3. If m is odd then n = m and 2^(m -2) = (m +1)/m which has no solutions. QED. Anyone see a flaw in the reasoning?
n*2^n+1 is an odd number, which it must be the square of an odd number, therefore we can write n*2^n+1=(2*k+1)^2. This leads to n*2^(n-2) = k*(k+1). We observe that k and k+1 are consecutive integers, which means they are coprime.
2^(n-2) - n > 1 for n≥5. This means that n*2^(n-2) can't possibly be the product of two consecutive integers (for n≥5) since one of them must be at least 2^(n-2) (the other number has to be odd) and the other one can be at most n. That leaves only n ∈ {1,2,3,4} as possible candidates.
n=1 doesn't work because 1*2^(1-2) is not an integer and thus can't be a product of integers.
n=2 works because 2*(2^(2-2)) = 2 can be factorized into consecutive integers 1*2 (k=1). Therefore 2*2^2+1 = 9 = (2*1+1)^2
n=3 works because 3*(2^(3-2)) = 6 can be factorized into consecutive integers 2*3 (k=2). Therefore 3*2^3+1 = 25 = (2*2+1)^2
n=4 doesn't work because 4*2^(4-2) = 16 can only be factorized into coprime factors as 1*16, which are not consecutive numbers.
More values of n just for fun:
n=5 -> 5*2^(5-2) = 40 --> coprime factorizations: 5*8, 1*40
n=6 -> 6*2^(6-2) = 96 --> coprime factorizations: 3*32, 1*96
n=7 -> 7*2^(7-2) = 224 --> coprime factorizations: 7*32, 1*224
n=8 -> 8*2^(8-2) = 512 --> coprime factorizations: 1*512
n=9 -> 9*2^(9-2) = 1152 --> coprime factorizations: 9*128, 1*1152
n=10 -> 10*2^(10-2) = 2560 -> coprime factorizations: 5*512, 1*2560
n=11 -> 11*2^(11-2) = 5632 -> coprime factorizations: 11*512, 1*5632
...
It's easy to see that the powers of two completely dominate the coprime factorization for all larger values of n with the remaining factor being much smaller.
Find all positive integers n such that n2^n+1 is a square. n2^n+1=k^2 2(2)^2+1=9=3^2 3(2)^3+1=25=5^2
k = -1, n = 0
k = 1, n = 0
k = ± 5, n = 3
k = ± 3, n = 2
Positive integers of n=2,3
Yeah but you didn't prove those are the only ones.
I just started looking at this, but I attacked it differently. I'm looking at trends when the statement is true. It is true when n=2 and n=3. This is as far as I've gotten in a few minutes.
Watching later into the video, it is clear that these are the only solutions.
2 and 3 are the solution
Can it be proven a parabola only intersects a linear-exponential at most twice?
hahaha, I paused the video at the start and tried solving it on my own and went down a completely different (and incorrect) path.
I started with: n*2^n+1=x²
since I wanted x to be an integer in order for it to be a perfect square, I tried to make it only true for integers.
i figured that in order to make f(n) = g(x) only true for x being an integer, I make the following equation:
(f(n)-g(x)) + cos(2*pi*x)-1 = 0
if I plug in n2^n+1 in for f(n), and x² for g(x), I get:
n * 2^n + 1 - x² + cos(2*pi*x) - 1 = 0 | since +1 and -1 cancel out, simplify
n * 2^n - x² + cos(2*pi*x) = 0
...and then I realized that I can't really get anywhere from there lol.
my math teacher!!
The problem is to find all positive integers n such that n * 2^(n+1) is a perfect square.
To solve this, the expression can be represented as k^2, where k is a square root.
The key insight is that k-1 and k+1 must be consecutive even numbers, which leads to a restriction on the possible values of n.
By considering the relationship between the linear function 2n and the exponential function 2^(n-1), the valid range of n can be determined to be from 1 to 4.
The solutions are then tested by checking if n * 2^(n+1) results in a perfect square for each valid n.
Your proof is technically incomplete. You forgot to check the second case (k-1), but there are no new n to check
Explanation of the Euclidean algorithm:
If b = c (mod a), then gcd(a, b) = gcd(a, c).
The reason for this is b = c (mod a) means b = c + ak for some k. Now, let d = gcd(a, b) and g = gcd(a, c). Since g divides a and c, it divides b. Thus, g divides a and b, and thus is a common factor of them. Thus, it divides the gcd of a and b, which is d. Now, rearranging b = c + ak to c = b - ak, it is clear to see that since d divides a and b, it also divides c. Thus, d divides a and c, and is thus a common factor of them, and thus divides their gcd, which is g. Thus, d and g divide each other, and so they are equal.
So, in the problem, the reason gcd(x - 1, x + 1) = 2 is because first, we know x = 1 (mod x - 1), so that means x + 1 = 2 (mod x - 1) (just replace x with 1). Thus, gcd(x - 1, x + 1) = gcd(x - 1, 2). gcd(x - 1, 2) = 2 because we know x is odd, and so x - 1 is even, aka it has at least one factor of 2.
In general, the Euclidean algorithm is nice for finding the gcd of large integers. Let's say we want the gcd of 78 and 96, we can take 96 mod 78 and get that we just need the gcd of 78 and 18, then we can take 78 mod 18 to get that we just need the gcd of 18 and 6, which is 6. Thus, the gcd of 78 and 96 is 6.
In fact, you can even apply the Euclidean algorithm to polynomials. The gcd of f(x) and g(x) is the gcd of f(x) and g(x) (mod f(x)). You can obtain g(x) (mod f(x)) by polynomial long dividing g(x) by f(x) and taking the remainder. Make sure you long divide by the one of f(x) or g(x) with smaller (or equal) degree, because then the remainder will have reduced degree, which is what you want. After all, the purpose of an algorithm is to simplify a problem.
Well, k is odd, obviously.
But I didn't get it, why
k-1 = a 2^(n-1) etc.
Got to think about it more.
Because a natural number can be 4k, 4k+1, 4k+2, 4k+3. So, two consecutive even numbers cannot be multiple of 4.
You can exclude from the beginning n = a positive power of 2 because n*2^n would be a perfect square, "ruined" by the +1
Here's another approach:
Find all n such that: n*2^n+1 = m^2.
Since m is odd let m = 2*k + 1 for k greater than or equal to 0.
This results in: n*2^n+1 = 4*k^2 + 4*k + 1.
This simplifies to: n*2^(n - 2) = k*(k + 1).
Therefore: n and 2^(n - 2) are consecutive numbers, one odd and one even.
2^a is an integer only when a greater than or equal to 0, so n is greater then or equal to 2.
Therefore: n = 0 for 2^(n - 2) odd. n = an odd number for all other solutions.
If n = 2 then 2^(n - 2) = 1, k = 1 and m = 3. A good solution.
If n = 3 then 2^(n - 2) = 2, k = 2 and m = 5. Also, a good solution.
If n = 5 or greater, then: 2^(n - 2) - n is greater than 1, no solution are possible.
i don't know why you put constant a in front of the power of 2 as it clearly is 1 only. Without a how can you set up the inequality?
this may be wrong. please don't yell at me, just show me my mistake
I thought about it using modulo
for (k+1) and (k-1) one has to be ≡ 0 mod n and the other ≡ 0 mod 2ⁿ
to me that only works when n+1 = k
with n = 4, k is greater than 5 and k increases faster than n
your explanation is much better/simpler than mine
n*2^n = X^2 = (X-1)(X+1) = 2k*2(k+1) = 4k(k+1)
n*2^(n-2) = k(k+1)
We have 2 consecutive numbers, one odd and one even. So 2^(n-2) | k or 2^(n-2) | (k+1), and n divides the other number.
Therefore |2^(n-2) - n| = |(k+1) - k| = 1. The only 2 numbers where this equality holds are k = 1 and k = 2.
With n = 2 and n = 3. When n = 4, the absolute value is 0 and when n> 4, |2^(n-2) - n| > 1. So n=2 and n=3 are the only 2 solutions.
k+1 > k-1, so 2n > 2^(n-1).
One question, is there no need to examine n2^n = (k-1)(a2^(n-1)) (the right side part)? We only tried the left part ( n2^n = (a2^(n-1))(k+1) ).
2nd question: @5:39 I am not convinced with this: any number that has many power of 2, the other number is just 2 times odd number. You assume that n is no more divisible by 2 in n*2^n, but suppose n=15. Then n*2^n=3*5*2^(3*5)=(3*2^5)*(5*2^3). Both are odd number times power of 2. Of course, in this example, these numbers are not consecutive numbers, but still we cannot assume one part of odd number times 2 and the other is just power of 2, right?
The answer is given at 10:56
Not right. The consecutive nature of the numbers is why the claim is true.
@@PrimeNewtonsYes, if we choose any number of power of 2 as one of consecutive numbers, the claim is true. However, the only condition we have is n2^n (n is any combination of odd numbers). So, how can we say one of consecutive numbers is just a power of 2? Since n2^n can be broken down to l * m * 2^(n-a+a), it can be one of a pair of ( l*2^(n-a), m*2^a) (l, m: odd numbers, l*m=n). I understand you point if we choose, say 32, 64, 128, or 256... But when not? Remember, we have to "find all" possible answers.
Simply put, we need to prove that if the number n2^n can be expressed as the product of two integers, and the difference between those two numbers is 2, then the only way to divide the number is into the two factors 2n and 2^(n-1). And this is not so easy task. Becasue... suppose n2^2=a * b and a - b =2. Then b = -1 +/- SQRT(1 + n2^n). b must be an integer, so (1 + n2^n) must be a squared number, which is exactly the problem itself!!! So, in proving the lemma, we have to solve the problem. We are not allowed to assume the lemma is true unless we solve it without using the lemma. Otherwise, this is a tautology.
@@dirklutz2818 I meant, 2n > 2^(n-1), not 2n >= 2^(n-1).
Nice problem. If n2^n + 1 is a square then there is a positive integer x such that n2^n = x^2 - 1 = (x - 1)(x + 1). Taking both sides mod 2, it is clear that x must be odd, and thus x - 1 and x + 1 must both be even. Furthermore, the Euclidean algorithm gives us that gcd(x - 1, x + 1) = gcd(x - 1, 2) = 2 because x + 1 = 2 (mod x - 1). Thus, either x - 1 or x + 1 has a factor of 2^(n - 1).
Note that for n >= 5, 2^(n - 1), which is the minimal size of the factor that does have 2^(n - 1) as a factor, is bigger than 2n, which is the maximal size of the factor that does not have 2^(n - 1) as a factor, and thus the larger of the two factors, x + 1, must be the one with 2^(n - 1) as a factor. Now notice that 2 = (x + 1) - (x - 1) >= 2^(n - 1) - (x - 1) >= 2^(n - 1) - 2n > 2. This is a contradiction, so we must have n < 5.
n = 1 does not yield a solution because 3 is not a square
n = 2 does yield a solution because 9 is a square
n = 3 does yield a solution because 25 is a square
n = 4 does not yield a solution because 65 is not a square
Thus, the solutions are n = 2, 3
Easier explanation is that many powers of 2 must be divisible by 4. Moving by 2 means that modulus 4 must be a 2.
In others words, the size of the odd factor grows linearly but size of the even factor grows exponentially.
Method 1)
(- x= 3) equation is given
Multiplying both sides by (-1)
-1*-x=-1*3
Then x=-3
or
Method 2)
Let the equation be (- x= 3)
If we multiply both sides with "MINUS" sign
-(- x)= -(3)
Then x= -3.
Which one is correct or both methods are correct .
Please help 🙏🙏
method two is the same as method one, the "Minus sign" has an implied one so the two are identical.
If k -1 is 2^n (n>1), then k is divisible by 4. k+3, k+7, etc is divisible by 4, hence k+1 is not divisible by 4 and therefore only contains one 2.
shouldn't one of the numbers be 2 power something while the other is 2 into some odd no. so why did we take' a' into 2^n-1 wouldn't that just mean the other number is 2 ? what exactly is a ?
a is necessary as two consecutive numbers can be i.e: 48 and 50. 48*50 can be written as 75*2^5 or 150*2^4. None of them holds for n*2^n, but we do not know whether there are such two numbers that hold when we start solving.
@@nasancak thankyou so much
I plugged in the first 10 numbers, came up with 2 and 3, and conjecture that will be it.
4:41 The right hand side is divisible by 8 so n is at least 2. However, I have not yet found an easier way to eliminate the other cases.
We can write any square number(>1) in the form 4k or 4k+1
As n2^n+1 is always odd
We can say n2^n +1= 4k+1
n2^n=4k
Here we can easily say that n=k and 2^n=4
Thus N=2
Since your k can be any positive integer, I think all you've shown is that n×2ⁿ is a multiple of 4, which is true for any n ≥ 2.
Does 0 not count as a positive integer or does 1 not count as a square number? Because if both of those are what I think they are you may have missed one. I know technically 0 is neither positive nor negative, but it makes me question if it counts
((0)(2^0) +1 == 1 as far as I can understand
I solved n×2^n=k^2-1 using Lambert W function then used the properties of the W function to find Integer solutions. I could prove that 2 is a solution (without numerical testing) but couldn't use the method to get to n=3. I guess the W function has a property that I am not aware of.
2•2^2+1=9 n=2
I appreriate the hardwork
But i expecteda little more elegant soln
This one has many loopholes where your assumption contradicts
Either they give a ans or not u should have covered them
6:56 I’m not sure about that method. The left side has n powers of 2 only if n is odd. If n is a power of 2, the term with the multipower of 2 will be a • 2^(n + the number of powers of 2 in a)
3:01 you forget to mention that n is an natural number because else it dont have to be the case! (nevermind i haven´t read the task :D integer over 1)
4:45 you repeat yourself we got it :D (just info! i have no problem with it)
6:04 ok therefore the prove would be nice to knowe
1:05 no please prove that :D :D ^^ but without term reforming :D
11:50 i think it is less and not equal because it can´t be > k-1 and that is less than 2n right?
12:24 i thought you forget it :D
14:15 haven´t we allready faild because equal is not working?
16:26 ok i was right it don´t work
16:35 the old but good advice :D
good bye
K.Furry
ps:
16:42 confused me
n*2^n= k^2-1
n*2^n divisible by k+1 and k-1
n*2^n, n=k+1 or n=k-1, or n is combinable with 2^n. n is not combinable with 2^n, as no perfect square is adjacent to an integer exponent. 2^9+-1 cannot be a perfect square.
or k+1,k-1 is divisible by something else.
for n=k+1,n=k-1,
2 options:
n-1 =k, or n+1=k
2^n+1=k or 2^n-1 =k
n+1=2^n-1
n=2, 3=4-1, therefore 2 is a solution, no other solutions of this form, same reasoning.
n-1=2^n+1
n=1, 0=3, therefore no solutions.
for k+1 divislbe by something else:
n= (k-1)/u, 2^n=(k+1)*u, and vice versa.
4 options,
Most plausable case is the one stated:
n*u+1=2^n/u-1
n=3, u+1=2^3/u-1
u=2, 3=3
n=3,u=2 is a solution.
Check the other 3 for any other solutions, but i have finished the video so it is not neccesary. Wonderful as always, harder problem than i realized.
You didn't do the 2n = a(k-1) case.
The answer is given at 10:56
For any n, the expression will be odd, we have an even term plus one. Then if it is a square, then it will be by the form:
n * 2^n + 1 = (2k + 1)^2, k is a natural number.
We don't use (2k-1) because n * 2^n will be at least 3 for n = 1.
Expand the right side and get
n * 2^n + 1 = 4k^2 + 4k + 1
n * 2^n = 4k^2 + 4k
n * 2^n = 4(k^2 + k)
n * 2^(n-2) = k^2 + k
n * 2^(n-2) = k(k+1)
From here we can say two things: first, n cannot be less than 2 because it would make 2^(n-2)to be less than 1. Second, the right side says that n * 2^(n-2) is the product of two consecutive integers.
Since 2^n grows obviously faster than n, there will be a point where 2^n will be so far from n.
For example, for n = 6, you get 6 * 2^(6-2) = 6 * 16. Already a big difference, so we don't have to check n >= 6.
For n = 2, you get 2 * 2^(2-2) = 2 * 1. Correct
For n = 3, you get 3 * 2^(3-2) = 3 * 2. Correct
For n = 4, you get 4 * 4.
For n = 5, you get 5 * 8.
n = 2 and n = 3 are the only solutions.
k-1=a.2^(n-1), what is this ?
the logic is not clear why K+1 or k-1 has to be a2^n-1 . K+1 and k-1 are consecutive even numbers but why it has to be a2^n-1 ?
same, don't understand this part
By observation, it is evident that out of two consecutive even integers, one will be of the form 2*some even number and the other will be of the form 2*some odd number eg. 60 = 2*30 while 62 = 2*31. Thus it is evident that the product of two consecutive even integers must be such that out of the total powers of the 2, one must be from the integer of the form 2*odd number while all the other n-1 must come from the integer of the form 2*even number.
I have 2 questions :
1) why did we ignore n*2^n = (k-1)a2^(n-1) ?
2) n*2^n = (k-1)a2^(n-1) (and the other)
It looks like it works
but how to we prove it ?
The answer is given at 10:56
@@dirklutz2818 not realy you can prove that
from k+1 = a 2^(n-1)
you get 2n >= k-1
and k +1 > 2^(n-1)
but it is not trivial to show
2n > k+1
@@glorrin Indeed! But you can get the table of values from n>2^(n-2) as in the video. You get a similar table (same values for n) from n>2^(n-2)-1.
These values for n are 1,2,3,4 in both cases, and only 2 and 3 are valid.
@@dirklutz2818Yup, I did the same thing
N = 0 works
Edit: ok better worded is, a factor of consecutive 2’s.
n2^n = k^2 - 1
n*ln2*e^(n*ln2) = ln2*(k^2 - 1)
n = W(ln2*(k^2-1))/ln2
If you have a computer to check all the branches, then that works I guess. If not, how do we identify natural solutions?
Hey guys who read my comment. I want you to tell me how to solve these questions. AMATYC 2003 fall Q 4,7,9,14,19 and 2004 spring 3,5,12,13,19,20
Nice and smart !! Actually, n = 0 is also a solution... Thanks for your videos 🙂
If i want to be pedantic: 0 is a solution of the equation , but it is not a valid answer to the question, as the question asked for all positive integers, which by convention excludes 0.
@@zpf6288 Oh, it was just an innocent remark...