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 $$f(m)=g(m)\cdot h(m)?$$