# If $m^4 + 4^n$ is a prime, for $n,m\geq2$, then $m$ is odd and $n$ is even?

One thing which i got is $m^4 + 4^n$ is congruent to $1 \pmod 8$ when both $n,m$ are odd...
Is it an iff condition?

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Please check whether my edit did not unintentionally change the meaning of your question. –  P.. Jan 15 '13 at 12:47

$1^4+4^1=5$ is prime, but n is not even.

$m=1$ and $n=6$, $1+4^6=17\times 241,$ so is not prime.

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sorry m not equal to 1 –  Mambo Jan 15 '13 at 12:34
@user49236 : what about $n=0$? –  Henry Jan 15 '13 at 12:45
@user49236 I just edited the answer. –  ՃՃՃ Jan 15 '13 at 12:54
$3^4+4^9=5^2\cdot17\cdot617$ –  P.. Jan 15 '13 at 12:55
Sorry, I made a mistake. –  ՃՃՃ Jan 15 '13 at 12:55

It follows easily from $$x^4+4y^4=(x^2-2xy+2y^2)(x^2+2 xy+2 y^2) \ .$$

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If $n=2k+1$ take $x=m$ and $y=2^k$ –  PAD Jan 15 '13 at 13:09

There are primes when $m=1$ or when $n=0$ (indeed all but one of them are the same). You are probably excluding them.

For $n=1$ you have $m^4+4^1 = ((m-1)^2 +1) \times ((m+1)^2 +1)$ which is not prime unless $m=1$.

For $n \gt 1$ you have $m^4+4^n \equiv 0 \text{ or } 1 \mod 8$, which leads to your question

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If you make some arithmetic modulo $\,4\,$:
$$m^4+4^n=m^4=0,1\pmod 4\Longrightarrow \,\,m\,\,\text{must be odd (why?)}$$
$$n=2k+1\Longrightarrow 4^n=4\cdot (4^k)^2\,$$