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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    $\begingroup$ Just a remark: a slightly simpler way of showing what you have so far would be, if $n$ is odd and $5\nmid m$, then $$m^4+4^n\equiv m^4+(-1)^n\equiv m^4-1\equiv 1-1\equiv0\bmod 5$$ because $$m^4\equiv 1\bmod 5\text{ for all }m\in\mathbb{Z},5\nmid m$$ by Fermat's Little Theorem. $\endgroup$ – Zev Chonoles Jul 21 '11 at 7:13
  • $\begingroup$ @Zev: Even I was thinking along the same lines. except i forgot to change $4$ to $-1$ in the mod. $\endgroup$ – user9413 Jul 21 '11 at 7:16
  • $\begingroup$ Thanks. 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. $\endgroup$ – Patrick Da Silva Jul 21 '11 at 7:17

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 $m,n>1$.

  • $\begingroup$ I need a more basic proof than that... and I don't know Sophie Germain's identity. Maybe a link to a proof or a proof from yourself, perhaps? If it is a too advanced theorem, I am not interested, since the book I found the question in was asking for an elementary proof, they haven't even mentioned congruences yet. $\endgroup$ – Patrick Da Silva Jul 21 '11 at 7:20
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    $\begingroup$ Proofs don't get much more basic than that -- it's just a factorization: en.wikipedia.org/wiki/Sophie_Germain#Honors_in_number_theory $\endgroup$ – joriki Jul 21 '11 at 7:21
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    $\begingroup$ Here is a link explaining the Sophie Germain identity. It is not advanced at all, it is a simple factorization identity, provable using the distributive property. Note that among the applications listed in the linked page, is a special case of the result you are asking about. $\endgroup$ – Zev Chonoles Jul 21 '11 at 7:22
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    $\begingroup$ $m^4+4q^4=m^4+4m^2q^2+q^4-4m^2q^2$ and "difference of squares" is the magic phrase. $\endgroup$ – Gerry Myerson Jul 21 '11 at 7:23
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    $\begingroup$ 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! $\endgroup$ – Patrick Da Silva Jul 21 '11 at 7:35

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.

  • $\begingroup$ $m^{4}+4^{n} = (m^{2}+2^{n})\cdot (m^{2}-2^{n})$ $\endgroup$ – user9413 Jul 21 '11 at 7:31
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    $\begingroup$ Are you serious? I think you're sleepy... =) you're funny. $\endgroup$ – Patrick Da Silva Jul 21 '11 at 7:36
  • $\begingroup$ @Partick: Yes, I am :( $\endgroup$ – user9413 Jul 21 '11 at 7:55
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    $\begingroup$ Lolll you even mis-spelled my name, you are definitely sleepy. Haha =) $\endgroup$ – Patrick Da Silva Jul 21 '11 at 8:18

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