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

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- Feb 27th 2010, 03:47 PMeykePrimes
Show that if $\displaystyle m^4+4^n$ is prime, then $\displaystyle m$ is odd and $\displaystyle n$ is even, except when $\displaystyle m=n=1$.

- Feb 27th 2010, 05:03 PMNonCommAlg
well, if $\displaystyle m$ is even, then obviously $\displaystyle 4 \mid m^4 + 4^n$ and so $\displaystyle m^4 + 4^n$ cannot be prime. if $\displaystyle n=2k+1, \ k \geq 1,$ then putting $\displaystyle a=2^k$ we have $\displaystyle 4^n=4^{2k+1}=4a^4.$

now $\displaystyle m^4+4^n=m^4+4a^4=(m^2+2a^2+2ma)(m^2+2a^2-2ma)$ and so $\displaystyle m^4+4^n$ cannot be prime because $\displaystyle m^2+2a^2-2ma=(m-a)^2+a^2 \geq a^2 > 1.$