If $5|n$ then $m=5k$ for $k=1,2,...$. If $n$ is not divisible by $5$ then $n^4 \equiv 1 \mod5$ so we take $m=5k+4$ for $k=1,2,...$.

But the result must be composite for ANY n, not for a GIVEN one. Maybe something like this works: $(m^2+m+mk+\frac{k^2} 2+k+\frac 1 2)(m^2-m-mk+\frac{k^2} 2+k+\frac 1 2)=$ $=m^4+2k^2+k^3+\frac 1 4 k^4+\frac 1 4+k+\frac 3 2 k^2$. All the numbers are integers for k odd, so each $n=2k^2+k^3+\frac 1 4 k^4+\frac 1 4+k$ works.

Kondr, you have the correct idea, which is to express $n^4+m$ as a product of two expressions, but your expressions are a little complicated. Since any product of two expressions without any $n^3, n^2$ or $n$ terms works, $(n^2 + 2kn + 2k^2)(n^2 - 2kn + 2k^2) = n^4 + 4k^4$ would be much easier to write.

"But the result must be composite for ANY n, not for a GIVEN one" Do you mean that it must be one set of integers $m$ for every $n$? It seems strange to me that we can't consider two cases depending on $n$...

Sorry TomciO, I was just skimming through and didn't really look at your post. Your solution is correct and is more elegant!

Tomci0's solution may be elegant, but I still think the choice of m can't depend on n. (if m could depend on n, we can just put m odd for n odd and m even for n even). Because of that only Dblues' nice solution and my ugly solution work.

Right... anyway, the problem statement could be better: "Prove that there exists infinite set of positive integers $m$ such that for every $n$...".

if $ n=4k^4$ and applying sophie germain identity to $ m^4+n$, we have done

For those who don't know: Sophie Germain's problem: When is $ m^4+4n^4$ prime ($ m,n\in \mathbb N$)?

Sorry for the bump, but is this solution correct: If $n$ is odd, let $a$ be an odd integer greater than $1$. Then, $n^4+a$ is an even number greater than $2$ so its not prime. If $n$ is even, let $a$ be an even natural number. Then, $n^4+a$ is an even number greater than $2$ so its not prime.

In this problem, $m$ is fixed, so the solution is wrong.

Raja Oktovin wrote: if $ n=4k^4$ and applying sophie germain identity to $ m^4+n$, we have done now we are finding $m$, not $n$.

Well this is Sophie Germain unmasked. We take $m = 4a^4$ for some positive integer $a.$ Done.

AdithyaBhaskar wrote: Well this is Sophie Germain unmasked. We take $m = 4a^4$ for some positive integer $a.$ Done. I just want to add that a>=2 as for a=1 $z_1$=5 is a prime.

Take $a=4k^4$, for some non-zero integer $k$. We have $$n^4+4k^4=(n^2+2k^2)-4n^2k^2=(n^2+2nk+2k^2)(n^2-2nk-2k^2)$$If $(n^2-2nk-2k^2)=1$, we have that $(n-k)^2+k^2=1$, so since $k$ is positive, we have that $n=k$. Plugging in, we have that $(n^2+2nk-2k^2)=5n^2$, and which is composite due to how we defined $k$. If $(n^2-2nk-2k^2)$ does not equal one, than $n^2+a^4$ is still prime so we win, as desired. $\square$

@above You made a mistake, because $$n^4+4k^4=(n^2+2k^2)-4n^2k^2=(n^2+2nk+2k^2)(n^2-2nk+2k^2)$$

Let $m=p^{2}$ where $p$ is a prime and $p \equiv -1 \pmod 4$. If , $n^{4}+p^{2}=q$ where $q$ is a prime then by Fermat's christmas theorem $ q|n$ and $q|p$ which is clearly contradictory $\square$.

Garou wrote: Let $m=p^{2}$ where $p$ is a prime and $p \equiv -1 \pmod 4$. If , $n^{4}+p^{2}=q$ where $q$ is a prime then by Fermat's christmas theorem $ q|n$ and $q|p$ which is clearly contradictory $\square$. Something is totaly wrong here: Choose $p=11$. Then $p$ is a prime and $p \equiv -1 \pmod 4$. Choose $n=2$. Then $q=n^4+p^2=16+121=137$ is prime, but your conclusion $137|2$ and $137|11$ does not hold and there is no contradiction.

People seem to be missing the fact that $n^4+4m^4$ is a prime for $n=m=1$. So better take $a=4m^4 \forall m \in \{2, 3 \dots \}$