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# Need help with a proof.

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Let \$P(x)\$ be a nonconstant polynomial, where all the coefficients are nonnegative integers. Prove that there exist infinitely many positive integers \$n\$ such that \$P(n)\$ is composite. I have been able to get three cases. When a0=0, when a0 is greater than 1, and when a0=1. I have solved all cases except when a0=1. Can someone help me with that case. (a0 refers to the constant in the general equation of a polynomial)

Jun 23, 2020

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I don't know what sort of method that you are using, but here's one that covers all three of your cases.

I've set it out for a quadratic, (to avoid too much LaTex), but it's easy enough to extend to a general polynomial.

Let \(\displaystyle P(x) = ax^{2} + bx + c\) , where a, b and c are positive integers

and then let

\(\displaystyle P(t) = at^{2} + bt + c = T.\)

T will be a positive integer greater than t.

Then, for any positive integer k,

\(\displaystyle P(t+kT)= a(t+kT)^{2}+b(t+kT)+c,\\=a(t^{2}+2tkT+k^{2}T^{2})+b(t+kT)+c,\\=at^{2}+bt+c+2atkT+ak^{2}T^{2}+bkT ,\\= T+T(2atk+ak^{2}T+bk),\\=T(1 + 2atk+ak^{2}T+bk).\)

So, P(t + kT)  will be composite for any positive integers t and k.

Jun 24, 2020