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

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

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- Jun 4th 2009, 08:51 PMTheAbstractionistTA’s Challenge Problem #2
This is the second Challenge Problem I’ve made up. (Nerd)

Prove that $\displaystyle n^{4k}+4m^4$ is not prime for all positive integers $\displaystyle n,\,k,\,m$ with $\displaystyle m>1.$ - Jun 4th 2009, 09:00 PMNonCommAlg
- Jun 4th 2009, 09:04 PMIsomorphism
NCA's identity has a name, its called

*Sophie Germain's Identity.* - Jun 4th 2009, 09:09 PMTheAbstractionist
Wow, that was really quick! (Rofl)

- Jun 4th 2009, 11:04 PMCaptainBlack
For real $\displaystyle m$ and $\displaystyle n$ and integer $\displaystyle k$:

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

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

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

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

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

is never prime.

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

CB - Jun 4th 2009, 11:38 PMNonCommAlg
- Jun 5th 2009, 12:13 PMTheAbstractionist
That’s a very good point. However, the condition $\displaystyle m>1$ (or $\displaystyle |m|>1$ 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 $\displaystyle n^{4k}+4m^4$ as $\displaystyle \left(n^{2k}+2mn^k+2m^2\right)\left(n^{2k}-2mn^k+2m^2\right)$ 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 $\displaystyle n^{2k}\pm2mn^k+2m^2-1$ as quadratics in $\displaystyle n^k.$ Their determinant is $\displaystyle 4m^2-8m^2+4=4(1-m^2)$ and the vital condition $\displaystyle m>1$ ensures that the determinant is negative and therefore that the expressions are positive. - Jun 9th 2009, 02:35 AMCaptainBlack