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Conjecture:

Let $m$ be a positive integer. Then $$f(m)=(2m)^{2m+1}+m^{2m+1}\cdot (2m+1)^m+(2m+1)^{2m}$$ is not a prime number.

One can prove it when $m$ is odd number, it is clear that $f(m)$ is an even number in this case.

But for $m$ even, it is not easy to show the conjecture. See some example values: $$f(2) = 4^5 + 2^5 \cdot 5^2 + 5^4 = 2449 = 31\cdot 79$$ $$f(4) = 8^9 + 4^9 \cdot 9^4 + 9^8 = 1897191233 = 7\cdot 53\cdot 73\cdot 70051$$ $$f(6) = 12^{13} + 6^{13} \cdot 13^6 + 13^{12} = 63171766713176497 = 281\cdot 2003681\cdot 112198777$$

Using Maple, one can verify the conjecture for even $m≤1300$.

For $m=1008$, the answer to this question contains a proof, because using Fermat's little theorem it shows $f(1008)\equiv 5 \pmod 5$.

Can we show a factorization, that is, can we look for $g(m),h(m)\in Z[x]$, such that $$f(m)=g(m)\cdot h(m)?$$

Improve this question
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    $\begingroup$ What is the motivation for this question? $\endgroup$
    – Jose Arnaldo Bebita
    Commented May 8, 2018 at 12:27
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    $\begingroup$ Crossposted at MSE. $\endgroup$
    – Dietrich Burde
    Commented May 8, 2018 at 14:43
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    $\begingroup$ It's impossible for there to be a factorization $f(m)=g(m)h(m)$ where $g,h\in\Bbb Z[x]$, because $f(m)$ is not a polynomial. $\endgroup$
    – Greg Martin
    Commented May 8, 2018 at 17:00
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    $\begingroup$ If $m \equiv 8$ mod $20$ then $f(m)$ is divisible by 5. Similarly if $m \equiv 4,16,22,30$ mod $42$ then $f(m)$ is divisible by 7. One can find other examples by noticing that if $p$ is prime then $f(m)$ mod $p$ depends only on $m$ mod $p(p-1)$ by Fermat's little theorem. $\endgroup$
    – François Brunault
    Commented May 8, 2018 at 17:13
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    $\begingroup$ If we use the heuristics that a number $n$ is prime with probability $1 / \log n$, the expected number of $m \le x$ with $f(m)$ prime is $\sim \log \log x$. I don't think checking for $m \le 1300$ is good enough support. $\endgroup$
    – Dongryul Kim
    Commented May 9, 2018 at 3:41

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