Prove that $\,(n+2)^4 + 16\cdot4^n\,$ is composite number for every $\,n\,$ natural number ?
I end up in a conclusion proving for $n$-odd and $n$-even but, I’m not sure if it's correct, is there any other possible solution for this ?
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Angelo
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Mustafë Ismajli
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For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – Another User Nov 24 '22 at 18:11
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Seems like treating odd and even $n$ separately is best. $n$ even, after all, is quite obvious. – Thomas Andrews Nov 24 '22 at 18:12
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1With $m=n+2$ this is $m^4+4^m$. – Sil Nov 24 '22 at 18:14
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5There's a general factorization $x^4+4y^4=(x^2-2xy+2y^2)(x^2+2xy+2y^2)$ which will work for the odd case. – Thomas Andrews Nov 24 '22 at 18:15