# Thread: TA’s Challenge Problem #2

1. ## TA’s Challenge Problem #2

This is the second Challenge Problem I’ve made up.

Prove that $n^{4k}+4m^4$ is not prime for all positive integers $n,\,k,\,m$ with $m>1.$

2. Originally Posted by TheAbstractionist
This is the second Challenge Problem I’ve made up.

Prove that $n^{4k}+4m^4$ is not prime for all positive integers $n,\,k,\,m$ with $m>1.$
$n^{4k}+4m^4=((n^k - m)^2 + m^2)((n^k + m)^2+m^2)$ and both factors are > 1 because m > 1.

3. NCA's identity has a name, its called Sophie Germain's Identity.

4. Wow, that was really quick!

5. Originally Posted by NonCommAlg
$n^{4k}+4m^4=((n^k - m)^2 + m^2)((n^k + m)^2+m^2)$ and both factors are > 1 because m > 1.
For real $m$ and $n$ and integer $k$:

$n^{4k}+4m^4=|n|^{4k}+4|m|^4$

So the restriction to positive integers for $n$ and $m$ is unnecessary.

Also if $m=0$ we have $n^{4k}+4m^4=n^{4k}$ which can never be prime for any integer $n$ (including $0$).

So the problem can be rewordrd to for all positive integers $k$ and integers $n,m$ show that

$n^{4k}+4m^4$

is never prime.

(we may have to be explicit about what we want $0^0$ to mean here or exclude the case where one or both of $n$ and $m$ are zero and $k$ is zero)

CB

6. Originally Posted by CaptainBlack

So the problem can be rewordrd to for all positive integers $k$ and integers $n,m$ show that

$n^{4k}+4m^4$

is never prime.
this is not true. for example if n = m = 1, we'll get the prime number 5. it's true that m, n need not be positive though.

7. Originally Posted by CaptainBlack
For real $m$ and $n$ and integer $k$:

$n^{4k}+4m^4=|n|^{4k}+4|m|^4$

So the restriction to positive integers for $n$ and $m$ is unnecessary.

Also if $m=0$ we have $n^{4k}+4m^4=n^{4k}$ which can never be prime for any integer $n$ (including $0$).

So the problem can be rewordrd to for all positive integers $k$ and integers $n,m$ show that

$n^{4k}+4m^4$

is never prime.

(we may have to be explicit about what we want $0^0$ to mean here or exclude the case where one or both of $n$ and $m$ are zero and $k$ is zero)

CB
That’s a very good point. However, the condition $m>1$ (or $|m|>1$ if you like) is absolutely vital, otherwise you could find a counterexample as NonCommAlg pointed out.

The way I originally did it was to factorize $n^{4k}+4m^4$ as $\left(n^{2k}+2mn^k+2m^2\right)\left(n^{2k}-2mn^k+2m^2\right)$ instead of writing the factors the way NonCommAlg did – and then it isn’t so clear that both factors are actually greater than 1. (One of them certainly is, but the other is iffy.) I then proceeded by considering the expressions $n^{2k}\pm2mn^k+2m^2-1$ as quadratics in $n^k.$ Their determinant is $4m^2-8m^2+4=4(1-m^2)$ and the vital condition $m>1$ ensures that the determinant is negative and therefore that the expressions are positive.

8. Originally Posted by NonCommAlg
this is not true. for example if n = m = 1, we'll get the prime number 5. it's true that m, n need not be positive though.
Aghh.. I had missed that |x|>1 excluded x=+/-1 and so had not considered those cases.

CB