Prove that $n^4+4^n$ is never prime. Here $n$ is any natural number greater than $1$. I have tried by induction hypothesis but to no avail. Can it be done by considering cases when $n$ is odd and when it is even?
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When $n$ is even we already know it is not a prime. Tried binomial expansion with $n=2k+1$? – Zelos Malum Dec 07 '15 at 06:22
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There's an intense feeling I've seen this before, but I can't locate a duplicate. Anyway, the case for odd $n$ can be proven by contradiction I believe. – Corellian Dec 07 '15 at 06:42
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@AndréNicolas: What do you mean by duplicate? The factorization generalizes to $n^4 + 4k^4$, but that doesn't do anything here because we already know the theorem when $n$ is even. – user21820 Dec 07 '15 at 06:42
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The question was answered here and earlier. – André Nicolas Dec 07 '15 at 07:01
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@AndréNicolas: Sorry I didn't realize that what I said in my comment above does settle the problem as $4^{n-1}$ is indeed a fourth power when $n$ is odd... – user21820 Dec 07 '15 at 07:02
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(1) If $n$ is odd, $n^4+4^n=(n^2+2^n-2^{n+1\over2}n)(n^2+2^n+2^{n+1\over2}n)$.
(2) If $n$ is even then $4|n^4+4^n$.
cr001
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The only thing missing is to show that $n^2+2^n-2^{(n+1)/2}n\ne 1$ or equivalently that $n^4+4^n>n^2+2^n+2^{(n+1)/2}n$ which is easily seen by dividing through by $4^n$. – DanielWainfleet Dec 07 '15 at 08:30
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on binomial expansion of (2k+1)^2,all terms are multiples of 4 but one term is 1.so 4 cannot be taken common.