Surely it is easier than this? Just note that (n²+1)² = n⁴ + 2n² + 1 < n⁴ + 3n² + 2 < n⁴ + 4n² + 4 = (n² + 2)². The given expression is strictly sandwiched between two consecutive squares, so cannot be one itself. Also the fourth (respectively second) powers of n are unnecessary: the same thing is true of all n² + 3n + 2 for the same reason.
Those powers are actually necessary. For m^2 + 3m + 2, it is true of all *positive* integers m that it cannot be a square. But if m = -1 or m = -2, then you get 0, which is a perfect square. But there is no real number (let alone integer) n such that n^2 = -1 or -2, which makes the original claim true.
Nicely done. Here is my proof: n⁴ + 3n² + 2 factors as (n² +1)(n² +2). Let n² +1=x. Then it becomes x(x+1), or the product of 2 consecutive integers. But what is x(x+1)? The formula for triangular numbers is well known, it is [x(x+1)]/2. So x(x+1) is simply twice a triangular number. Now consider the squares. Every square is the sum of 2 consecutive triangular numbers. List the triangular numbers: 0,1,3,6,10,15,21, etc. So 0+1 is the first square, or 1². The next square is 1+3, or 2². The next square is 3+6, or 3². Etc. Now consider again x(x+1) Since it is twice a triangular number, that means it is 0+0, 1+1, 3+3, 6+6, etc. But notice this pattern. We have already established that every square is the sum of 2 consecutive triangulars. With the exception of 0, any triangular number added to itself falls short of the very NEXT square. In other words, 1+1 cannot be a square since AT LEAST 3 has to be added to 1 to reach the next square, 2². 3+3 cannot be a square since AT LEAST 6 has to be added to 3 to reach the next square, 3². 6+6 cannot be a square since AT LEAST 10 has to be added to 6 to reach the next square, 4². So the pattern continues to infinity. One other case: 0. In this case, 0+0 IS a perfect square, but plugging it back into the original quartic, n⁴ + 3n² +2 yields 2, which is not a square. QED, proof complete.
Very good proof, nice and easy ! I found one which is longer (and less smart), but it works… there it is : Let X = n^4 + 3 n^2 + 2. Then 4X = 4 n^4 + 12 n^2 + 8, and 4X = (2 n^2 + 4) (2 n^2 + 2) = (2 n^2 + 3)^2 - 1. Then 4X can’t be a square (because it is a square minus 1), and X can’t be a square (because 4 is a square). Thanks for your videos 🙂
Without watching the video the original expression is always even. So if we drop the + 2 we have the reduced expression n^4 + 3n^2 ≡ 2 (mod 4). All even perfect squares are divisible by 4 (because they are q*q where q has at least one 2 in its prime factorization). So test n^4 + 3n^2 for the cases n= 0, 1, 2, 3. All give n^4 + 3n^2 ≡ 0 (mod 4). No cases satisfy thecondition of congruence with 2 (mod 4). So the orginal expression can never be a perfect square.
You don't even need to evaluate n=0,1,2,3. It is a well known fact that every square is 0 or 1 (mod 4) If n² = 0 (mod 4) then n⁴ + 3n² + 2 = 2 (mod 4) and therefore can't be a square. If n² = 1 (mod 4) then n⁴ + 3n² + 2 = 2 (mod 4) and can't be a square QED
My solution: n^4+3n^2+2=k^2 (n^2+1)(n^2+2)=k^2 (n^2+1) and (n^2+2) don't share any factors, since they differ only by 1, so they both have to be squares for (k) to be an integer so: (n^2+1)=a^2 (n^2+2)=b^2 then: b^2-a^2=1 (b-a)(a+b)=1 either: b-a=1 a+b=1 or: b-a=-1 a+b=-1 which makes a=0 and b=+-1 resulting in: n=sqrt(-1) which is not an integer.
ATTEMPT: All square numbers are either 0 or 1 mod 4. Consider when n is even: n^2 is 0 mod 4. n^4 + 3n^2 + 2 = 2 mod 4, so the expression cannot be square. Now consider when n is odd: n^2 = 1 mod 4. n^4 + 3n^2 + 2 = 2 mod 4, so the expression cannot be square. By case analysis we are done.
Very easy visual proof as well, I shall attempt to explain in words: n⁴ + 3n² + 2 can be expressed as n²(n²+3) +2 You can now construct a perfect square that has lengths n². You have three n² by 1 lengths and 2unit lenths to add. Adding one [1*n²]lengths to the top/bottom and the other on the left/right. You nearly have a perfect square but are missing one piece. So you can add 1 to complete the square. At this point our equation would be n²(n²+2) + 1 is always a perfect square. But we still have n²+1 to add. At this point it should become visually apparent there is no way to add n²+1 to n²(n²+2)+1 to make a perfect square. You would always be missing at a minimum n²+2 more. (That is [n²+1]+[n²+2])
n^4 + 2n^2 + 1 = (n^2+1)^2 < n^4 + 3n^2 + 2 < n^4 + 4n^2 + 4 = (n^2 + 2)^2 therefore, we will need a perfect square that is between two consecutive perfect squares.
If n⁴+3n²+2 is a perfect square, it can be written as a²+2ab+b² for every integer in a there will be a possible value of b also integer. For a = n² we have (n²)²+2(n²)b+b² = n⁴+3n²+2 2n²b+b² = 3n²+2 2n²b-3n²=2-b² n²(2b-3) = 2-b² n² = (2-b²)/(2b-3) But n² = a (a+b)² = ((2-b²)/(2b-3)+b)² =(2-b²)²/(2b-3)² + 2(2-b²)/(2b-3)b+b² (4-4b²+b⁴)/(2b-3)² + 2b(2-b²)(2b-3)/(2b-3)²+b²(2b-3)²/(2b-3)² (4-4b²+b⁴+(4b-2b³)(2b-3) +(b(2b-3))²)/(2b-3)² (4-4b²+b⁴+8b²-12b-8b⁴+3b³+(4b⁴-12b³+9b)/(2b-3)² (-3b⁴-9b³+4b²-3b+4)/(2b-3)² This expression can't be simplified in terms of integers factors, what means that a that is n² or b are never both integers at same I was pretty sure I could get a solution by this way, but easier, but that was my best
n^2 and n^4 are always equivalent mod 4 (0^2=0^4=0, 1^2=1^4=1, 2^2=2^4=0, 3^2=3^4=1). Thus the expression is equivalent to n^2+3n^2+2 = 4n^2+2 = 2 mod 4. This also proves the evenness of the expression since any integer that is equivalent to 2 mod 4 is even.
If the product of 2 consecutive numbers is a square, since consecutive numbers are coprime, both numbers must be squares. But the only consecutive squares are 0 and 1. Even without being able to factor it, if k^2 = n^4 + 3n^2 + 2, 4k^2 = 4n^4 + 12n^2 + 8, (2n^2 + 3)^2 - (2k)^2 = 1.
For all integer n, n² = 0 or 1 mod 4, thus n^4+3n²=n²(n²+3)=0 mod 4 thus n^4+3n²+2 = 2 mod 4 for all integer n. However, there is no square integer which is 2 modulo 4.
The first thing that came to my mind is that the expression is also equal to (n^2+1)^2 + (n^2+1), ie it’s a perfect square plus the base of that perfect square , in the form m^2 + m But the next perfect square after m^2 is 2m+1 apart. There is no way m^2+m is a perfect square
I replaced n² with m and to make m² + 3m + 2 = k², and then solved for m² + 3m + 2 - k² = 0 via the quadratic equation. For k² and m are perfect squares and must both be positive integers. But the Q.E. gave me m = (-3 +/- sqrt(k² + 1)) / 2, which is k² is a square, forces m to be irrational or negative.
Rigorously: n⁴+3n²+2 ≡ n⁴-n²+2 = n²(n²-1) + 2 mod 4. Since n² can only be 0 or 1 mod 4, n²(n²-1) = 0 for any n meaning n⁴+3n²+2 ≡ 2 mod 4 and therefore is never a perfect square.
Sir please do a limit question which was came in JEE Advanced 2014 shift-1 question number 57 it's a question of a limit lim as x-->1 [{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4 You have to find the greatest value of a It has 2 possible answers 0 and 2 But I want the reason that why should I reject 2 and accept 0 Because final answer is 0 Please help 😢
For all integers,prove that n^4+3n^2+2 is never a perfect square. First method Suppose there exist some integer m such that m^2=n^4+3n^2+2=(n^2+1)(n^2+2) x^2=4*9 x=±6
Do you think this solution would work? Suppose that n^4 + 3n^2 + 2 = k^2 for some k. Now we can show that because the factorisation of n^4 + 3n^2 + 2 is (n^2 + 1)(n^2 + 2), the expression minimally has a value of 2 at n = 0. [Or you can differentiate the expression, whichever feels more rigorous]. 2 >1^2 and 0^2, so this expression can never equal those squares. Now we consider 2 cases of k^2, where it is a prime square [so k is prime like 2,3,5,7,11], and k is a composite square, [so k is composite like 4,6,8,9,10] If k is a prime square, it's only factors are 1,k and k^2, so we could try to see if any values of n could make a value of k exist. Case 1: n^2 + 1 = 1, n^2 + 2 = k^2 Eq 1 shows that n = 0, so we'd get k^2 = 2 ,but k wouldn't be an integer as 2 isn't a perfect square. So this case is out. Case 2: n^2 + 1 = k, n^2 + 2 = k It is "trivial", but trying to solve the two equations yields 1 = 2, which means these 2 expressions can never equal each other at the same value of n. So the expression can never be a prime square. No prime values of k exist, so no n exists for which the expression is a perfect square. For non prime squares, its factors are 1,k and k^2, alongside any other factors between 1 and k, and k and k^2. We assume that the two factors n^2 + 2 and n^2 + 1 are factors that pair up to make a square that are not equal to each other [as we shown earlier]. So this means that: n^2 + 1 < k < n^2 + 2, where k is the non-prime factor of the non-prime square. This makes sense because factors will always pair up in a number. Since squares have 1 double counted factor, any factor pairs must lie on opposite sides of k. We run into another issue however. n^2 + 1 - (n^2 + 1) = 1, so these numbers are just 1 apart. There is no possible way to sandwich a value of k for any n, because these 2 are consecutive integers. So no value of non-prime k exists as well, so no n exists for which the expression is a perfect square. Since we showed that no value of k can exist for primes and non primes, as well as 0,1 this covers all the positive integers, and since k^2 = (-k^2), this covers the entire number range as well, so no value of k exists for which an integer n can produce a perfect square. As completion, Because the function is even, this also doubles as covering all integers, as f(n) = f(=n), so since we proved it for all positive n, any negative n will have the same result. Hence, no value of n can produce a perfect square. QED.
Surely it is easier than this? Just note that (n²+1)² = n⁴ + 2n² + 1 < n⁴ + 3n² + 2 < n⁴ + 4n² + 4 = (n² + 2)². The given expression is strictly sandwiched between two consecutive squares, so cannot be one itself. Also the fourth (respectively second) powers of n are unnecessary: the same thing is true of all n² + 3n + 2 for the same reason.
Those powers are actually necessary. For m^2 + 3m + 2, it is true of all *positive* integers m that it cannot be a square. But if m = -1 or m = -2, then you get 0, which is a perfect square. But there is no real number (let alone integer) n such that n^2 = -1 or -2, which makes the original claim true.
Nicely done. Here is my proof:
n⁴ + 3n² + 2 factors as (n² +1)(n² +2). Let n² +1=x. Then it becomes x(x+1), or the product of 2 consecutive integers. But what is x(x+1)? The formula for triangular numbers is well known, it is [x(x+1)]/2. So x(x+1) is simply twice a triangular number. Now consider the squares. Every square is the sum of 2 consecutive triangular numbers. List the triangular numbers: 0,1,3,6,10,15,21,
etc. So 0+1 is the first square, or 1². The next square is 1+3, or 2². The next square is 3+6, or 3². Etc. Now consider again x(x+1) Since it is twice a triangular number, that means it is 0+0, 1+1, 3+3, 6+6, etc. But notice this pattern. We have already established that every square is the sum of 2 consecutive triangulars. With the exception of 0, any triangular number added to itself falls short of the very NEXT square. In other words, 1+1 cannot be a square since AT LEAST 3 has to be added to 1 to reach the next square, 2². 3+3 cannot be a square since AT LEAST 6 has to be added to 3 to reach the next square, 3². 6+6 cannot be a square since AT LEAST 10 has to be added to 6 to reach the next square, 4². So the pattern continues to infinity. One other case: 0. In this case, 0+0 IS a perfect square, but plugging it back into the original quartic, n⁴ + 3n² +2 yields 2, which is not a square. QED, proof complete.
Very good proof, nice and easy !
I found one which is longer (and less smart), but it works… there it is :
Let X = n^4 + 3 n^2 + 2. Then 4X = 4 n^4 + 12 n^2 + 8, and
4X = (2 n^2 + 4) (2 n^2 + 2) = (2 n^2 + 3)^2 - 1.
Then 4X can’t be a square (because it is a square minus 1), and X can’t be a square (because 4 is a square).
Thanks for your videos 🙂
Without watching the video the original expression is always even. So if we drop the + 2 we have the reduced expression n^4 + 3n^2 ≡ 2 (mod 4). All even perfect squares are divisible by 4 (because they are q*q where q has at least one 2 in its prime factorization). So test n^4 + 3n^2 for the cases n= 0, 1, 2, 3. All give n^4 + 3n^2 ≡ 0 (mod 4). No cases satisfy thecondition of congruence with 2 (mod 4). So the orginal expression can never be a perfect square.
You don't even need to evaluate n=0,1,2,3. It is a well known fact that every square is 0 or 1 (mod 4)
If n² = 0 (mod 4) then n⁴ + 3n² + 2 = 2 (mod 4) and therefore can't be a square.
If n² = 1 (mod 4) then n⁴ + 3n² + 2 = 2 (mod 4) and can't be a square
QED
@@diegoman8158 Thanks, I didn't think of that (obviously).
My solution:
n^4+3n^2+2=k^2
(n^2+1)(n^2+2)=k^2
(n^2+1) and (n^2+2) don't share any factors, since they differ only by 1, so they both have to be squares for (k) to be an integer so:
(n^2+1)=a^2
(n^2+2)=b^2
then:
b^2-a^2=1
(b-a)(a+b)=1
either:
b-a=1
a+b=1
or:
b-a=-1
a+b=-1
which makes a=0 and b=+-1 resulting in:
n=sqrt(-1) which is not an integer.
My solution
n⁴ + 3n² + 2 = t²
Factoring:
(2n²+3)² = 4t² + 1
(2n²+3+2t)(2n²+3-2t) = 1
This implies t=0 and then:
2n² + 3 = 1
n² = -1, Absurd
I am very happy to find you on this platform! You are a very good teacher and friend. Thank you for your time.
ATTEMPT:
All square numbers are either 0 or 1 mod 4.
Consider when n is even: n^2 is 0 mod 4.
n^4 + 3n^2 + 2 = 2 mod 4, so the expression cannot be square.
Now consider when n is odd: n^2 = 1 mod 4.
n^4 + 3n^2 + 2 = 2 mod 4, so the expression cannot be square.
By case analysis we are done.
Trying n = 0, 1, 2, -1, we find that n⁴ + 3n² + 2 ≡ 2 (mod 4), for all n. Similarly,
n² ≡ 0 or 1 (mod 4), for all n.
Once again an Absolutely beautiful proof. Love number theory.
Very easy visual proof as well, I shall attempt to explain in words:
n⁴ + 3n² + 2 can be expressed as n²(n²+3) +2
You can now construct a perfect square that has lengths n². You have three n² by 1 lengths and 2unit lenths to add. Adding one [1*n²]lengths to the top/bottom and the other on the left/right. You nearly have a perfect square but are missing one piece. So you can add 1 to complete the square.
At this point our equation would be n²(n²+2) + 1 is always a perfect square. But we still have n²+1 to add.
At this point it should become visually apparent there is no way to add n²+1 to n²(n²+2)+1 to make a perfect square. You would always be missing at a minimum n²+2 more. (That is [n²+1]+[n²+2])
n^4 + 2n^2 + 1 = (n^2+1)^2 < n^4 + 3n^2 + 2 < n^4 + 4n^2 + 4 = (n^2 + 2)^2 therefore, we will need a perfect square that is between two consecutive perfect squares.
If n⁴+3n²+2 is a perfect square, it can be written as a²+2ab+b² for every integer in a there will be a possible value of b also integer.
For a = n² we have
(n²)²+2(n²)b+b² = n⁴+3n²+2
2n²b+b² = 3n²+2
2n²b-3n²=2-b²
n²(2b-3) = 2-b²
n² = (2-b²)/(2b-3)
But n² = a
(a+b)² = ((2-b²)/(2b-3)+b)²
=(2-b²)²/(2b-3)² + 2(2-b²)/(2b-3)b+b²
(4-4b²+b⁴)/(2b-3)² + 2b(2-b²)(2b-3)/(2b-3)²+b²(2b-3)²/(2b-3)²
(4-4b²+b⁴+(4b-2b³)(2b-3) +(b(2b-3))²)/(2b-3)²
(4-4b²+b⁴+8b²-12b-8b⁴+3b³+(4b⁴-12b³+9b)/(2b-3)²
(-3b⁴-9b³+4b²-3b+4)/(2b-3)²
This expression can't be simplified in terms of integers factors, what means that a that is n² or b are never both integers at same
I was pretty sure I could get a solution by this way, but easier, but that was my best
n^2 and n^4 are always equivalent mod 4 (0^2=0^4=0, 1^2=1^4=1, 2^2=2^4=0, 3^2=3^4=1).
Thus the expression is equivalent to n^2+3n^2+2 = 4n^2+2 = 2 mod 4.
This also proves the evenness of the expression since any integer that is equivalent to 2 mod 4 is even.
If the product of 2 consecutive numbers is a square, since consecutive numbers are coprime, both numbers must be squares. But the only consecutive squares are 0 and 1. Even without being able to factor it, if k^2 = n^4 + 3n^2 + 2, 4k^2 = 4n^4 + 12n^2 + 8, (2n^2 + 3)^2 - (2k)^2 = 1.
Never stop learning those who stopped learning have stopped living.
Before viewing the video: n⁴ + 3n² + 2 ≡ 2 (mod 4) for all n, but perfect square are either 0 or 1 (mod 4).
n^4+2n^2+n^2+2=(n^2+1)(n^2+2)🎉
Thank you for leaving in the "typos." 😊 It makes you a real person.
For all integer n, n² = 0 or 1 mod 4, thus n^4+3n²=n²(n²+3)=0 mod 4 thus n^4+3n²+2 = 2 mod 4 for all integer n. However, there is no square integer which is 2 modulo 4.
The first thing that came to my mind is that the expression is also equal to (n^2+1)^2 + (n^2+1), ie it’s a perfect square plus the base of that perfect square , in the form m^2 + m
But the next perfect square after m^2 is 2m+1 apart. There is no way m^2+m is a perfect square
I replaced n² with m and to make m² + 3m + 2 = k², and then solved for m² + 3m + 2 - k² = 0 via the quadratic equation. For k² and m are perfect squares and must both be positive integers. But the Q.E. gave me m = (-3 +/- sqrt(k² + 1)) / 2, which is k² is a square, forces m to be irrational or negative.
We could generalize the problem: for what values of k is n^4+3n^2+k a perfect square?
Rigorously: n⁴+3n²+2 ≡ n⁴-n²+2 = n²(n²-1) + 2 mod 4. Since n² can only be 0 or 1 mod 4, n²(n²-1) = 0 for any n meaning n⁴+3n²+2 ≡ 2 mod 4 and therefore is never a perfect square.
Factors (n^2+2)(n^2+1) so is a perfect square if and only if 2 = 1 a contradiction.
Was the first method mathematical contradiction?
Yes . Proof by contradiction
m must be between n^2+1 and n^2+2. n^2+1
14:30 can’t an odd number also be represented by (2k-1)? Does the proof still work by using that?
Thanks Sir 🙏
Perfect square or not, its a perfect equation with four distinct answers. n=i,-i,2^1/2i and -2^1/2i.
Claim: n^2(n^2+3) is divisible by 4 case 1: n is even (2k)^2((2k)^2+3)=4k^2((2k)^2+3) n is odd (2k+1)^2((2k+1)+3)=(2k+1)(4k^2+4k+4)=4(2k+1)^2(k^2+k+1)
n²+2 is never a perfect square because no perfect squares differ by 2, so by extension, n⁴+(never a perfect square) is also never a perfect square.
5:00 this way is kinda wrong, you forgot the absolute value on m (n²+1 < |m| < n²+2), or said that m is a non negative integer
Sir please do a limit question which was came in
JEE Advanced 2014 shift-1 question number 57
it's a question of a limit
lim as x-->1
[{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4
You have to find the greatest value of a
It has 2 possible answers 0 and 2
But I want the reason that why should I reject 2 and accept 0
Because final answer is 0
Please help 😢
sir. 3rd way to integrate 1/(1+x^4)
Coming soon
I love numbers!
For all integers,prove that n^4+3n^2+2 is never a perfect square. First method Suppose there exist some integer m such that m^2=n^4+3n^2+2=(n^2+1)(n^2+2) x^2=4*9 x=±6
Why aren't u using discriminant to prove taking n^2=t
Case 2: n is odd n^2 is odd n^2+3 is even n^2(n^2+3)+2 is even.
X^2=16 then x is not between 4 and 4
Claim: n^4+3n^2+2 is always even. n^4+3n^2+2=n^2(n^2+3)+2 Case 1:n is even n^2 is even n^2(n^2+3)+2 is even.
Since n^4+3n^2 is divisible by 4, n^4+3n^2+2 is never a multiple of 4 since 2
0:47 I miss the smile.....❤❤😅😅
❤❤❤❤
n^4+3n^2+2 is always even.
Sqrt[20736]=144 perfect square root example.
Therefore n^4+3n^2+2 is not a perfect square for all integers.
Love it!!
n^4+3n^2+2=(n^2+1)(n^2+2) It’s in my head.
Do you think this solution would work?
Suppose that n^4 + 3n^2 + 2 = k^2 for some k.
Now we can show that because the factorisation of n^4 + 3n^2 + 2 is (n^2 + 1)(n^2 + 2), the expression minimally has a value of 2 at n = 0. [Or you can differentiate the expression, whichever feels more rigorous].
2 >1^2 and 0^2, so this expression can never equal those squares.
Now we consider 2 cases of k^2, where it is a prime square [so k is prime like 2,3,5,7,11], and k is a composite square, [so k is composite like 4,6,8,9,10]
If k is a prime square, it's only factors are 1,k and k^2, so we could try to see if any values of n could make a value of k exist.
Case 1: n^2 + 1 = 1, n^2 + 2 = k^2
Eq 1 shows that n = 0, so we'd get k^2 = 2 ,but k wouldn't be an integer as 2 isn't a perfect square. So this case is out.
Case 2: n^2 + 1 = k, n^2 + 2 = k
It is "trivial", but trying to solve the two equations yields 1 = 2, which means these 2 expressions can never equal each other at the same value of n.
So the expression can never be a prime square. No prime values of k exist, so no n exists for which the expression is a perfect square.
For non prime squares, its factors are 1,k and k^2, alongside any other factors between 1 and k, and k and k^2.
We assume that the two factors n^2 + 2 and n^2 + 1 are factors that pair up to make a square that are not equal to each other [as we shown earlier]. So this means that:
n^2 + 1 < k < n^2 + 2, where k is the non-prime factor of the non-prime square.
This makes sense because factors will always pair up in a number. Since squares have 1 double counted factor, any factor pairs must lie on opposite sides of k.
We run into another issue however. n^2 + 1 - (n^2 + 1) = 1, so these numbers are just 1 apart. There is no possible way to sandwich a value of k for any n, because these 2 are consecutive integers. So no value of non-prime k exists as well, so no n exists for which the expression is a perfect square.
Since we showed that no value of k can exist for primes and non primes, as well as 0,1 this covers all the positive integers, and since k^2 = (-k^2), this covers the entire number range as well, so no value of k exists for which an integer n can produce a perfect square. As completion, Because the function is even, this also doubles as covering all integers, as f(n) = f(=n), so since we proved it for all positive n, any negative n will have the same result.
Hence, no value of n can produce a perfect square. QED.
n^4+3n^2 is always divisible by 4.
Sqrt[65536]=256 another perfect square root.