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Conjecture:

Let $m$ be a positive integer. Then $$f(m)=(2m)^{2m+1}+m^{2m+1}\cdot (2m+1)^m+(2m+1)^{2m},\forall m\in N^{+}$$$$f(m)=(2m)^{2m+1}+m^{2m+1}\cdot (2m+1)^m+(2m+1)^{2m}$$ 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 Useusing 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 $$f(m)=g(m)\cdot h(m)?$$

Conjecture:

Let $m$ be a positive integer. $$f(m)=(2m)^{2m+1}+m^{2m+1}\cdot (2m+1)^m+(2m+1)^{2m},\forall m\in N^{+}$$ 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 Use Fermat's little theorem $f(1008)\equiv 5 \pmod 5$.

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

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

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math110
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Conjecture:

Let $m$ be a positive integer. $$f(m)=(2m)^{2m+1}+m^{2m+1}\cdot (2m+1)^m+(2m+1)^{2m},\forall m\in N^{+}$$ 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 Use Fermat's little theorem $f(1008)\equiv 5 \pmod 5$.

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

Conjecture:

Let $m$ be a positive integer. $$f(m)=(2m)^{2m+1}+m^{2m+1}\cdot (2m+1)^m+(2m+1)^{2m},\forall m\in N^{+}$$ 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 $f(1008)\equiv 5 \pmod 5$.

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

Conjecture:

Let $m$ be a positive integer. $$f(m)=(2m)^{2m+1}+m^{2m+1}\cdot (2m+1)^m+(2m+1)^{2m},\forall m\in N^{+}$$ 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 Use Fermat's little theorem $f(1008)\equiv 5 \pmod 5$.

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

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