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Let $m$ be positive integers. According to this Question conjecture $$f(m)=(2m)^{2m+1}+m^{2m+1}\cdot (2m+1)^m+(2m+1)^{2m},\forall m\in N^{+}$$ not prime number

I have proved when $m$ is an odd number, it is clear $f(m)$ is an even number.

But for $m$ even, it is not easy to show that, because $$f(2)=4^5+2^5*5^2+5^4=2449=31\cdot 79$$ $$f(4)=8^9+4^9*9^4+9^8=1897191233=7\cdot 53\cdot 73\cdot 70051$$ $$f(6)=12^{13}+6^{13}*13^6+13^{12}=63171766713176497=281\cdot 2003681\cdot 112198777$$ $\cdots\cdots\cdots$.

and for large $m$,such $m=1018$ the answer see Question because $f(1008)\equiv 5 \pmod 5$, others case maybe use factorization,can we look for $g(m),h(m)\in Z[x]$,such $$f(m)=g(m)\cdot h(m)?$$

math110
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    Any reason to believe this? Have you tested the first few cases of even $m$? The numbers grow quite quickly... – lulu May 08 '18 at 11:11
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    Checked with Maple for even $m \leq 1300$, seems to hold, so far... – Sil May 08 '18 at 11:13
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    @lulu,so I conjecture – math110 May 08 '18 at 11:41
  • Now have 51 viewed 51 times,no have some thought ? so this problem is hard?In this case, I'm asking the question at MO:https://mathoverflow.net/questions/299705/show-this-number-always-is-composite-number – math110 May 08 '18 at 12:18
  • @inequality Yes, this question is hard, unless someone finds a prime. There are cases with no small prime factor, and I think that there is a prime of that form. I started PFGW, but this led to trouble because I am currently running a test with PARI/GP, which takes ages for primality tests of large numbers. So, be a bit patient. – Peter May 08 '18 at 12:33
  • Just a note, the linked MO question has lots of issues, missing letters, spaces, etc... ("intgers", "number,it",...), sentences are not finished... Just maybe read after yourself... – Sil May 08 '18 at 12:47
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    No prime for $m\le 4\ 000$ – Peter May 09 '18 at 10:45
  • No prime for $m\le 5\ 000$ – Peter May 10 '18 at 08:02
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    $f(m) \sim \sqrt{e} m^{3m+1} 2^m$, and heuristically the probability of a number this size being prime is approximately $1/\log(f(m)) \sim 1/(3 m \log m)$. Since $\sum_m 1/(m \log m)$ diverges, we might expect (in the absence of any elementary reason to rule out primes) that there are infinitely many primes of this form. However, the sum grows very slowly, so if there are no such $m\le 5000$ it is quite likely that the first $m$ is very large indeed: maybe something like $5000^{e^3} \approx 2 \times 10^{74}$. – Robert Israel May 10 '18 at 23:40
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    If that is the case, we may never know: no computer that could fit in the known universe could even store the digits of $f(m)$, let alone check it for primality. – Robert Israel May 10 '18 at 23:48

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