Turkish Mathematical Olympiad | 2009 Q1
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- เผยแพร่เมื่อ 20 ส.ค. 2020
- We present a solution to a problem from the Turkish Mathematical Olympiad. This problem involves determining all prime numbers p such that a polynomial evaluated at p is the square of a positive integer.
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The funny thing happens if you brute-force in a programming language with insufficient decimal precision. As expected 2,7 and 11 are answers, but 19262791 also comes out. The reason: sqrt(p**3-4p+9) with p=19262791 is really close to an integer(84...x10**9 + 0.9999961...)
That's why human proofs are human proofs. No computer can battle us.
Why use decimals in a program?
@@ETiDeQuenVesSendo
I don't think it was intentional. The computer might have approximated it.
But that's incorrect since it's not an integer
Hi Michael. I am professor in Spain. I enjoy very much all your imos and Putnam problems. I watch all them. Very, very good job. Thanks for your videos.
I think the hardest part is always coming up with the hints giving the right direction. How do you identify what approach/hints to take?
might seem obscure, but i think knowing about elliptic curves would make the "hints" more obvious.
Experience and intuition
@@John-tb2ds stop spamming!
@@John-tb2ds bloody hell. You folks are clutching at straws aren't you
@@myoung1445 no i just want you to be saved
Your channel has made me love math again
I hope that in the future you'll make videos on Fourier Transform because it's the only topic which is missing from your channel and there are so many interesting concepts about Fourier transform
Here is a shortcut slightly from 8:45, aside from p={2,3}, taking the square of both sides of an inequality ((p^2-2p-9)/9) =< n, We will get the desired bound for p, by dividing p^3 as in the lecture, hence directly concluded that p=
I don't understand why we need to consider two cases as in 9:36 and further. I was able to skip that assumption that "was going out from nowhere" and solve more straighforward way.
On that board we had already found that 3n≥p²-2p+3. Squaring that, we have 9n²≥p⁴-4p³-14p²+36p+81. But we search for p's where n²=p³-4p+9 (from the problem statement), so we just substitute that into this inequality and we get a polynomial inequality for p. It only looks quartic and hard, in fact is's easy to solve: the free term 81 cancels out, so after combining all the terms we have p⁴-13p³-14p²+72p≤0. Let's forget for a minute p is a prime, just let's factor it out and leave a cubic poly'al. It happens that 2 is a root of that cubic, so it factorizes to (p-2)(p²-11p-36)¸ with irrational roots, one of which is negative, anf the other is around 13.6. So we have all four roots or our quartic (around -2.6, 0, 2, around 13.6). Solving this inequality by intervals method we find useable intervals are [around -2.6, 0] and [2, around 13.6]. Now, remembering p was a prime, this translates into p≤13.
Easy part, testing each of p = 2, 3, 5, 7, 11, 13 we have left with 2, 7 and 11, whose are the answer (corresponding n's are 3, 18, 36).
I am very happy now,please make more videos about Turkish Mathematical Olympiads.
Also you can try this one:
For all x,y which are real numbers,
(i) f(f(x²)+y+f(y))=x²+2f(y)
(ii) x≤y --› f(x)≤f(y)
Find the all functions which is f:R--›R
(Turkish Mathematical Olympiad 2012 Second Round Day 1 Question 3)
:)
Bir matematikçi olarak mutlu oldum, Olympiad sorulari nereden bulabilirim, yardımcı olabilir misin?
Nice problem! The only function is the most obvious one, but my proof of this is a couple of pages long. :-)
@@Notthatkindofdr my proof is long to,I asked @Michael Penn for seeing another proof which is shorter than my one
@@alperenkoken I have found another solution which does not need condition (ii) in the problem, and I think is maybe simpler than my previous solution.
Always the last constant determine the roots and always less than last remainder theorem and it tells something about how the last constant is distributed along the function. With the highest power as one. 2 7 and 9+2.
If you look at the original expression mod 4, you get p(cubed) + 1 is congruent to zero or 1.
This means that p = 2 (which checks out) or p is of the form 4k + 3. This reduces the
work considerably.
Never occurred to me to use divisors to set up bounds! Thank you for bestowing these techniques upon us for free :)
yeah that's super sneaky!
if we set the equation to x^2 and rearrange we can factor it as p(p^2-4)=(x+3)(x-3). p is prime so must divide either (x+3) or (x-3). The difference between these two terms =6 therefore let ap =x+-6 and p^2-4= a^2p+-6a. Now gather all terms on the left-hand side: p^2-a^2p-4+-6a=0 and use the quadratic equation to solve for p: p=(a^2+-sqrt(a^4+16+-24a))/2. Now we look at the discriminant. To be a whole number a^4+16+-24a must be a perfect square but a^4 and 16 are both already perfect squares. This means that 24a must be the cross term and it must equal to 8a^2. Therefore a=3. We can now rewrite the equation as p=(9+-sqrt((9+-4)^2)/2 and this simplifies to p=(9+-9+-4)/2 =(18+-4)/2 or +-4/2 which means p=2, 7, 11 after we discount the answer of negative 2
For everyone asking about comparing n against p^2/4 at 8:48 and the early bound of looking at cases p>11 at 1:53 the comparisons aren't required at all and can be proved without it. They are for simplifying the argument a little.
The bound of p>11: For any prime p, n^2 = 9 (mod p) implies (n-3)(n+3) = 0 (mod p) hence p divides either n-3 or n+3. Dropping the assumption p>11 doesn't affect any of the later calculations.
The comparison with p^2/4: Combining the bound (p^2-2p-9)/3 < n with the identity n^2 = p^3-4p+9 we obtain the inequality (p^2-2p-9)^2 < 9(p^3-4p+9). After expanding, simplifying and factorising we obtain p(p-2)(p^2-11p-36)
So when you get to pm^2 +- 6m = p^2 -4, why can't we solve as a quadratic in p, and try to find m values that make p prime?
When I try it doesn't work btw, cause I end up with the sqrt of m^4 +- 24m + 16 which can't be factored into squares afaik. Thx
This was actually the direction I took and I found it to be a little more direct.
Even though m^4±24m+16 can't be factored into polynomials in m, the result can still be a perfect square for specific values of m. The basic observation is that the square root of m^4±24m+16 can't be "too far" from m^2 for large enough m, but it still needs to be an integer; and, actually, you're pretty much restricted to checking when m^4±24m+16 = (m^2+2)^2 or (m^2+4)^2. Only m = 3 ends up working, which leads to the 3 solutions p in {2, 7, 11}.
Another way to solve it: p^3-4p=(n-3)(n+3) either n+3 or n-3 must be divisible by p so p^2-4=a(n+-3)=a(pa+-3)=a^2+-3a. Using the quadratic equation p=(a^2+-sqrt(a^4+16+-24a)/2 however the square root term must also be a perfect square otherwise the answer cannot be an integer therefore 8a=+-24 a^2. Therefore a=+-3. Feeding this back into the original equation (unnecessary because we could have used the previous equation but it is what I did) we get two separate quadratic. Nothing else is possible p^2-9p-22=0 or p^2-9p+14=0 one solution is impossible (-2), the others are 2,7,11
So when he got to P|+-6m-4, why did he say that, because P≧11, P divides half of +-6m-4? Does it have something to do with the -4?
Problem proposition - From Polish edition of Championnat des Jeux Mathématiques et Logiques
final (2003, Question 18):
Find all natural n such that : 3^n+4^n+5^n...+(n+2)^n = (n+3)^n.
Nice
I enjoy watching these proofs, but they punctuate how weak I am this area (number theory, analysis). Can anyone recommend entry level book(s). (I have some math background - engineer, calculus tutor).
but... what was the rank of the elliptic curve?
Instantly thought that primes must end in 1, 3, 7 or 9; and putting those digits through the formula needs to end up with a units digit of 1, 4, 5, 6, 9 or 0. (ie not 2, 3, 7 or 8).
1 minus 4 plus 9 is 6
(3*3*3) = 7 - 2 + 9 is 4
(7*7*7) = 3 - 8 + 9 is 4
(9*9*9) = 9 - 6 + 9 is 2 - so that's the only thing I've ruled out. The prime can't end in 9.
Now let's see what I should have been doing...
Huh, I'm wondering about the +/- portions being added together. Can we do that? I suppose it's because they're both "+/-" rather than one of them being "-/+" ? I was thinking they may partially (or completely) cancel, otherwise.
Thanks
Not sure what part he did that, you'll have to help me out. But in general you just have to be careful with +/-. If two +/- signs are *linked*, as in the choice of one forces the choice of the other, then it would be find to combine in some way. The whole reason -/+ even exists is to be linked with a +/-.
If you've made two independent choices of +/-, you probably shouldn't combine them into 1 equation and instead split into two cases. Otherwise you'd have to index them or something so you remember to split into 4 cases in the end.
@@toast_recon sorry I don't rmember the time-stamp but he went from +/-3m (or something similar to +/-6m
But I think I wrapped my head around it.
Cheers from Turkey! Good problem anyway. Turkey's questions are really good compared with most of the countries -especially the term 2005-2015 the brightest of Turkey National Team-. I hope you can offer more questions from Turkey:)
Another possible answer is to prove that n is divisible by 3. To do this, transpose 9 to the LHS and factor the RHS. Now assume for a moment that p>3. This means that 3 does not divide p (since p is a prime) but 3 must divide at least one of (p-2) or (p+2). Now, this means that p^3-4p is divisible by 3, and so n^2 is also divisible by 3. But this means that 3|n. Now, let n=3m. We have m^2=(p)(p-2)(p+2)/9+1. Rearranging, we have (m+1)(m-1)=p(p+2)(p-2)/9. But since p is odd, so are (p-2) and (p+2). This means that (m-1)(m+1) are both odd.But this implies that gcf(m+1,m-1)=1. Notice also that the two factors differ by 2. This is an important hint
Also, since p is odd prime, gcf(p,p-2)=1, gcf(p+2,p-2)=1, gcf(p,p-2)=1. This also implies that any of p-2 or p+2 is divisible by 9.
Assume 9|(p-2). This means that the product of any two of the three factors p, (p+2), and (p-2)/9 must be either 2 more or 2 less than the third. We have six cases, all of which results in a quadratic equation with one variable. Then solve for p.
Now assume 9|(p+2). This means the product of any two of the three factors p, (p-2) and (p+2)/9 must be 2 more or 2 less than the third. We have six cases, all of which results in a quadratic equation with one variable. Then solve for p.
Now, p=2 will be a solution. Since we assume that p>3, what remains is to test at p=3.
I might have missed something but I think the part about checking 6 cases only works if you know each of the three factors p, (p+2), (p-2)/9 are prime numbers. For e.g. 7 x 11 x 15 = 33 x 35. Each of the numbers on the left are co-prime to each other but if you take the difference of multiplying any two of them together with the third you wouldn't get 2.
@@elephantdinosaur2284 Thanks for noticing the error. I completely missed that part. And also the part where all (p), (p-2) and (p+2) differ by just 2, so no need for quadratic equation in the 6 cases I mentioned.
I think I have to make more analysis to correct the error. Thanks!
15:25
How?
🙌🙌🙌
Always so helpful!
Integarahl boi How? Magic, I guess 🤷♂️
איתן גרינזייד 👍
there should have a resolution more short and more clear. i m sure of that
You can check the comments below nylmoulin Rui of using mod (3).
I found another one from youtube
th-cam.com/video/ctA6JF22ypI/w-d-xo.html
Let p^3-4p+9=k^2 -> (p-2)p(p+2)=(k-3)(k+3), the product of three consecutive numbers with the difference 3 should be described into the product of two numbers with a difference of 6. If p-2 or p+2 is 9m (m=0, 1) (p-2)p(p+2) = m*3p*(3p+/-6), so satisfies conditions. Of course, I should strictly prove that m bigger than 1 cannot satisfies the condition, but ...
Eqn factorises as
(p-2) p (p+2) = (m-3)(m+3)
For p>2 we have three consecutive odd numbers so one must be a multiple of 3. I'm fact a multiple of 9 as RHS is a multiple of 9 if it's a multiple of 3. RHS can be rewritten 9(k-1)(k+1). k=3m.
Obvs p can't be 9 so either (p-2)=9 or (p+2)=9. In fact both give solutions with three consecutive odd numbers (k-1),(k+1),9 and 9,)k-1),(k+1). p=7 or 11.
please answer this. you said on 5:03 that if it is not even it must divide the odd part of the right side but that doesn't hold for every divisor. For instance, 3 divides 21, therefore, it divides 19 + 2. Similarly, we could say that it should divide 19 something that is false.
He is referring to p|(±6m-4)
p|(±6m-4) --> p|2(±3m-2) --> p|2 or p|(±3m-2)
Trivially, 2 is only even prime giving p|2.
So, if p is not even (p>2): p|(±3m-2) which is equivalent to p|(3m±2)
Now, consider m=odd: (3m±2)=odd, hence p divides the odd part of (±6m-4)=±2(3m±2).
Consider m=even: (3m±2)=even, hence p|(3m±2) --> p|2(1½m±1) --> p|(1½m±1).
This process can be repeated until you have removed all factors of 2, leaving p dividing some odd number that is equal to (3m±2)/2ª.
That is what he meant by p divides the odd part of (±6m-4).
The reduction to (3m±2) is sufficient for this problem. No real need to consider cases for (1½m±1) or (¾m±½) etc.
@@Hiltok I found another one from youtube
th-cam.com/video/ctA6JF22ypI/w-d-xo.html
There is a lot ways to solve, but try to think of this that way: p^3-4p=k^2-9
p^2(p-4)=(k+3)(k-3)
Then we can think 'bout which of multipliers from k+3 and k-3 devides by p. For one reason it can' t be only k-3, because then it devides by p^2 and there is unequalety k-3>=p^2. Analogicly it can't be only k+3, so both of k+3 and k-3 devides by p, so 6 devides by p. And then we check p=2 or 3 by hands. And get that only 2 reaches
You did a mistake p^2(p-4) is not p^3-4p, it is p^3-4p^2.
p^2(p-4) = p^3-4p^2
Can someone explain the simplification with the signs at 5:14? I can't quite follow that
+- operator is a built in OR statement.
p | 6m-4 -> p | 2(3m-2) -> p | 3m-2
p | -6m-4 -> p | -2(3m+2) -> p | 3m+2
So in one statement p | 3m +- 2
@@hybmnzz2658 Ok that's much clearer! Thanks
How about your mega favorite number??
I m morrocan students and Next years l will study in first years in university can participate international mathematical competition IMC ? if I can how to inscrire in this competition
You have to participate in the national math Olympics first. But in my time the IMO werw for pre-universitary students.
2:20 Why he can take square on both sides? (n^2 ≡ 9 (mod p) ⇒ n ≡ ±3 (mod p)). Is this a theorem?
Maybe
n^2 = 9 [p]
n^2 - 9 = 0 [p]
n^2 - 3^2 =0 [p]
(n+3)(n-3) = 0 [p]
So p|n+3 and p|n-3
Meaning n+3 = 0 [p] and n-3 = 0 [p]
Or n = +-3 [p]
@@pollitojb3731 Got it! Thanks a lot!
@@pollitojb3731 for p must be greater than 3
Yes it's a theorem... P should be prime but...
Greetings from Turkey
Let = represent congruence , that is a = b mod p means a is congruent to b mod p, p is prime.
64 = 100 mod 3 then 8 = - 10 mod 3 but not 8 = 10 mod 3
100 = 9 mod 13 then 10 = -3 mod 13 but not 10 = 3 mod 13
144 = 64 mod 5 then 12 = - 8 mod 5 but not 12 = 8 mod 5
So n^2 = 9 (mod p) we only have (n = 3 mod p) OR n = -3 (mod p ) but not both.
Many books also use + AND - instead of + OR - , wondering if this is trivial in the video.
I frequently look for materials in your videos to train students for math contests. Thank you.
Who in the world would even think of doing this?? Why not move the 9 to tbe other side and factor it as p(p-2)(p+2)= (n-3)(n+3) and then plug in values to see what works??
Great shirt 👍🤗
why did u take n≤p²/4... why not p²/5 or something else? we would have got a quadratic equation in both cases..
I'm wondering about this as well
Wondering this as well, it looks like black magic to me.
@Hellow so what ?
@Jakub Frei There has to be some other reason no? If that's it then we could just switch out 4 for any perfect square n?
Just as he said, it comes from nowhere... prolly tried with a·p² and found that with 1/4 both inequalities had a useful solution.
Nice.
CAN SOMEONE TELL ME WHAT WE NOTICED THEN WE CHOOSED P^2/4 AS A BOUND FOR N IT STILL SEEMS ABSURD FOR ME
I tried making the ff. Soln, and got the same answers as you, but I have no idea how to justify one of the steps I made...:
Express the RHS as a DOTS. Then, we have
(n-3)(n+3) = (p-2)(p)(p+2).
Obsv: p even, then p 2, but RHS 0, so n 9. That's ok. So we need p odd. But p odd means n even.
So we have the solutio: p = 2
Let n be 2g, then the lhs is ( 2g+3 ) (2g - 3), so p divides one of these terms, let one be Ap, the other is Ap + 6 or Ap - 6. We have either,
Ap(Ap + 6) = (p-2)(p)(p+2).
Ap (Ap - 6) = (p-2)(p)(p+2).
Then transpose around to get
A = ((p^2)-4)/(Ap + 6) or ((p^2)-4)/(Ap - 6)
Treat the denominator and numerator as polynomials of p and use synthetic division, to get
A = (1/A)p + (6/A^2)(p^0) + ((36/(A^2))-4)/(Ap+6) or
A = (1/A)p + (6/A^2)(p^0) + ((36/(A^2))-4)/(Ap-6)
Here is the step I have trouble with. There is a weak argument here, since A is an integer, so the term ((36/(A^2))-4)/(Ap-6) must be 0. I don't know how to connect the two but this is what solvds the problem, anyways if this term is nonzero we have an ugly-looking fraction for all three terms, and... I just don't know how to justify it. Even if the (1/A)p + (6/A^2)(p^0) terms were non integers, it's still more nice and likely to be that the sum of those two is an integer (?) Without the " interference" from the third term.
But, if it is zero, then A = +3 or -3. It's not really important in this solution if we chose to use 2A + 6 or -6 earlier... isn't that strange? Now we solve the equation for p:
The -3 solution:
A = (1/A)p + (6/A^2) + ((36/(A^2))-4)/(Ap-6)
-3 = -p/3 + 2/3; 9+2 = 11 = p.
The 3 solution:
3 = p/3 + 2/3; 9-2 = 7 =p
In addition to the 2 from earlier we now have all solutions of p.
for the part you have trouble with:
If |A| > 6 (possibly there's a bound lower than this that works, but this was the first one I tried), then
|1/Ap| < 1
|6/A^2| < 1
|(36/A^2 - 4)/(Ap +- 6)| |RHS| = |LHS| = |A| > 6, hence 3 > 6, absurd.
And then you check what happens for |A|
That's a hard one. As usual I lasted till about half way.
2:12 => assumes p >= 11
14:50 => gives all prime numbers under 13 as solutions to try, and then so 2 and 7 are.
Excuse me? Where did I get this wrong please
well, assuming p>=11, u have 11 and 13. So u need to check them. But you still need to check primes under 11
@@myxail0
(I understood it juste after sending the comment 😂 but thanks 👍)
Unless I'm missing something, saying p|(3m+2) means p≤3m+2, isn't valid because of negative numbers. For example, p=13 and 3m+2=-52. Obviously this particular m and p doesn't satisfy the equation, but it still is a counter example to p|(3m+2)→p≤3m+2
英語と数学の勉強が同時にできる一石二鳥
That's funny. I've done the same thing for Japanese and math before. I watched 鈴木貫太郎's channel quite a lot in the past.
Respect from one math teacher to another math teacher
Surf Arrakis 😄
Miller's World would be another prime location.
I reached the equation (p-2)*p*(p+2)=(n-3)*(n+3), which means that p divides either (n-3) or (n+3), but couldn't proceed any further without your help. Thanks!
I went here at first as well, but since p-2 and p+2 aren’t necessarily prime I couldn’t see how to go any further.
@@jgcoulth They aren't prime. One must be a multiple of 3 in fact because there are 3 consecutive odd numbers there. Which means 3 divides the RHS. Which means 9 divides the RHS as 3 divides both n-3 and n+3. So we know p-2=9 or p+2=9. These give our solutions.
if we set the equation to x^2 and rearrange we can factor it as p(p^2-4)=(x+3)(x-3). p is prime so must divide either (x+3) or (x-3). The difference between these two terms =6 therefore let ap =x+-6 and p^2-4= a^2p+-6a. Now gather all terms on the left-hand side: p^2-a^2p-4+-6a=0 and use the quadratic equation to solve for p: p=(a^2+-sqrt(a^4+16+-24a))/2. Now we look at the discriminant. To be a whole number a^4+16+-24a must be a perfect square but a^4 and 16 are both already perfect squares. This means that 24a must be the cross term and it must equal to 8a^2. Therefore a=3. We can now rewrite the equation as p=(9+-sqrt((9+-4)^2)/2 and this simplifies to p=(9+-9+-4)/2 =(18+-4)/2 or +-4/2 which means p=2, 7, 11 after we discount the answer of negative 2.
The assumption early on that p>=11 seemed unmotivated. Hardest part of proving something difficult is to not use a truth that follows from already knowing the answer.
The technique provided gives us new insights, but the solution is way too complicated. Just consider 2, get it out of the way and for odd primes, consider primes to be 3k+/-1. For both cases lhs is 0 mod 3. This implies it is 0 mod 9 too as rhs is a perfect square. Hence p(p^2-4)=9n^2-9=9n(n^2-1) .case wise bashing-p+2=9,p-2=9 yields p=7,11. Hence p=2,7,11. Other cases don't yield solutions. Very easy :-)
I would never be able to do that. Not with that method anyway. Maybe there's another....?
It’s a solution but not a very elegant one; could it be that the setters in the olympiad had in mind something simpler? E.g. transform the equation into (n-3)(n+3)=p(p-2)(p+2), or, because obviously both parts are divisible by 9, into 9(m-1)(m+1)=p(p-2)(p+2), and then prove that either p=2, m=1, or both sides represent a product of three consecutive odd numbers, one of which is 9, i.e. p=7 or p=11.
But why wouldn't it be the product of three odd numbers, one of which is A MULTIPLE of 9, rather than simply 9?
@@swenji9113 because the left part should be the product of the same three consecutive odd numbers as the right part (I don’t think it can work otherwise). Two of those numbers are m-1 and m+1 and the third one is 9.
@@swenji9113 @@swenji9113 because the left part should be the product of the same three consecutive odd numbers as the right part (I don’t think it can work otherwise, which I think should be easy to prove). Two of those numbers are m-1 and m+1 and the third one is 9.
Why could we make the assumption that n is bigger than or equal to 11?
It is the first prime after 9.
@@Pacuvio25 so? Sorry, I still don't quite get it
he just made this assumption to make the modulo part a bit easier. but in the end he checked all the smaller cases anyway
He's trying to find some M such that p < M for all solutions, and he doesn't care about M being the least upper bound. He can just pick M to be at least 12. Then, if p 11. That's where he finds he can prove it for M = 16.
Oh ! Yet another transatlantic misunderstanding ! I read Q instead of 9 : a bit more complicated, isn't it 😅
Can you solve this problem?
Find all primes of the form p^p+2 such that p is also a prime. I’ve asked lots of people and none of them could help.
interesting problem for sure.
This looks great - I suggest u post this problem when Michael posts his daily video which he uploads around 5:30 PM india time - most interested viewers usually check his latest video. I have seen a somewhat similar problem solved recently
türkler için söylüyorum bu soru 2009 tübitak lise matematik olimpiyatları 2.aşama 1.gün 1.soru
Zannetmezdim ki Türkiye'den izleyen biri olsun sâhiden sfdfdf.
Merâk ediyordum ikinci aşama mı birinci mi diye ama bir için olmaz gibi geliyordu. Sağ ol.
Hey maşallah
@Hellow valla ben açıyorum 5. dakikadan sonra kafam almıyor kopup kapatıyorum
Az Tr Az Tr sjbdhshdhd (beli burdada)
Baba her yerdeyiz
Valla helal
5:12 5:17 ...wow, I've never seen that in my life. Still not convinced it is true.
If a|b*c and gcd(a,b)=1. Then a|c.
Demonstration:
Since a|b*c then there is an integer k such that b*c=a*k. Also, because gcd(a,b)=1 then there are integers u,v such that au + bv =1. Multiplying both sides of the last ecuation by c we get that acu + bcv = c. Grouping terms, we get a(cu) + (bc)v = c. But because b*c = ak, then we have a(cu) + (ak)v = c, and factorizing a from the left handside of the ecuation we get a(cu + kv) = c. This implies that a|c because there exists an integer m (where m= cu+kv) such that c=am.
Michael used this property, with a =p, b= 2, and c=3m +-2
"notice how i've changed the sign but that's kind of ok as well"
Yet he had to explain why p divides something written as p*(...). If you're gonna explain all the obvious parts then explain the not so obvious ones as well
nerdesiniz beyler
!!
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I am here
I got three different prime number which satisfied the conditions. Those primes are 3,7 and 11.
I got these answer through brute force 😅😅
I meant 2,7, and 11😅😅
Very hard
Very very very
TH-cam has changed
That was quite long....
think i solved in a similar but (for me) easier way. After reasoning mod p we can say p|a+3 or p|a-3.
We also know for FLT p³=p (mod 3) so 3|p³-4p+9=a² and 3|a because 3 it's prime.
I write the first case (p|a+3), the second is similar.
a²=k²p²-6kp+9=p³-4p+9 so p²-k²p+(6k-4)=0
delta=k^4-24k+16, p is an integer so delta must be a perfect square. It's easy to see that delta=(k²-1)²+2k²-24k+15. Because it's a square it can't be (k²-2)
Subbing for the Dune shirt lol
😩😩😩
Not a prime, but 646³ - 4∙646 + 9 = 16419²
how have you found? ahahahah
plzz try this cool problem : find all functions that satisfy the following condition , for all x,y real numbers , f(x^2)-f(y^2)《(f(x)+y)(x-f(y))
XaXuser What does the symbol “《 “ mean?
@@williamadams137 less than or equal
@@williamadams137
Maybe less or equal?
Try his main approaches.
x=0, y=a
x=a, y=0
x=a, y=a
x=a, y=-a
Reply in the comments with any significant findings from those four approaches & then let’s work from there.
The solution is f(x)=x and f(x)=-x, am I wrong???
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🖕
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@@sumukhhegde6677 why did u do that?
:=)
@@________6295 🤫🤦♂️
Okay, and that's a good place to stop....
...you know.... at the end...
Your hairs in every video, are even like this, like that! 😀😉 ...
can you please try this question:
find all natural numbers a, b and c such that
a+2020=b+c
2020a+1=bc
*if someone here in the comments know how to do pls help me too :)
I suggest looking at the general case by replacing 2020 by an integer, n.
Let there be a quadratic equation with roots b and c. Put values of b+c and bc into it in terms of a,
Then calculate discriminant part and solve values of a for which discriminant is perfect square. Once a is known, find b and c using that quadratic equation.
This can be generalized to any n
We get b=2020.111 .. not a natural number.. so no solution
I've edited this comment because i made a calculation mistake ;-; I believe a solution exists.
you can use AM GM inequality to show that (a+2020)/2>=sq root(2020*a+1). sqaure both sides and you have a quadratic equation,the detereminant of which is sq(2020)-4*(sq(2020)-1) which is 16,thus a real a exists satisfying these relations.
as many others have already pointed out,forming a quadratic with b and c as roots is the most intuitive way to go about this.
@@suyashmisra7406 from where that +1 inside of the square come from if you're doing the inequality with a and 2020?
shouldn't be (a+2020)/2≥sqrt(2020a) ?
Noice
Definetly a really hard problem, I tried this, couldn't solve it and then I watched the solution: impossible!
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No offense, but if that's how math is taught in the US, I can see how people hate it