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I have been stuck on this one for months, really simple to state, really giving me trouble.

Show that if $m^4 + 4^n$ is prime, $m>0$, $n>0$, then $m$ is odd and $n$ is even, except when $m=n=1$.

The case $m=n=1$ gives $5$, so that is why it is excluded. Clearly $m$ is odd, for if it was even, the number would be divisible by at least $4$, but I can't seem to get rid of the case where $n$ must be odd.

The only thing I've managed to try is to write $m$ as an odd number, say $m = 2k+1$, and $n = 2\ell + 1$, then get $ m^4 + 4^n = (2k+1)^4 + 4^{2\ell+1} = 16k^4 + 32k^3 + 24k^2 + 8k + 1 + 4^{2 \ell + 1} $ and this is congruent to $0 \mod 5$ unless $k \equiv 2 \mod 5$, and now it gives you an even more ugly polynomial if I write $k = 5j+2$, and it's not working. I've tried other inconclusive approaches, like trying to see if two numbers of some form generate numbers of the same form.. (for instance primes of the form 8k+1 and 8k+7 always generate primes of the same form when multiplied together, stuff like that).

Any ideas? Even just ideas you haven't gave a thought... they're welcome!

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    Tha$n$ks. The reason I proved it this way is because in the book I've found it, it's like in the first chapter and there's no congruences that have been spoken up to where the question has been asked, so I assumed it was possible to do it without them.2011-07-21

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If $m$ even then obviously it is divisible by $2$ and since it is greater than $2$, so not prime. Now if $n$ odd, say $n=2k+1$. Then it can be written as $m^4+4.(2^k)^4$ which, by Sophie Germain Identity is not a prime for $m,n>1$.

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    Good thing I learned this identity then. Asking the question wasn't so stupid after all. Thanks to all of you guys. +1 and checked!2011-07-21
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Oh my god. I've never seen anyone actually do that, but the answer is so cheap that I almost want to delete my own question. I don't know why it took me so long to find the answer. I'll just leave it there and see if anyone has something good to say about it. I'm still open for new original answers.

If $n$ is odd, write it $2k+1$, so that

\begin{align} m^4 + 4^{2k+1} & = m^4 + 2 \cdot 2^{2k+1} \cdot m^2 + 4^{2k+1} - 2 \cdot 2^{2k+1} \cdot m^2 \\\ & = (m^2 + 2^{2k+1})^2 - (2^{k+1}m)^2 \\\ & = (m^2 + 2^{2k+1} - 2^{k+1}m)(m^2 + 2^{2k+1} + 2^{k+1}m) \\\ \end{align} which is composite unless $k=0$ (above says $m$ must be odd), and this case is excluded.

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    Lolll you even mis-spelled my name, you are definitely sleepy. Haha =)2011-07-21