Prove that if $p$ is a some polynomial with integer coefficients, $M$ is an integer such that $p(n)$ is prime for every $n\ge M $. Then $p$ is constant.
Please help , I have no clue how to prove it.
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3Welcome to MathSE! Having "no clue" is no good excuse for not including what you think about the problem and what you have tried. So please edit your question to include what you have tried and we can hint accordingly without spoiling the fun of solving the problem! – Oct 29 '20 at 15:29
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if $,p(x!+!M),$ is prime for all $,x\in\Bbb N,$ then $,\deg p(x!+!M) = 0,$ so $,\deg p(x) = \deg p(x!+!M) = 0\ \ $ – Bill Dubuque Oct 30 '20 at 00:00
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Suppose $p$ is not constant. We know that $p(M)=q$ with $q$ prime. Since $p$ is not constant $\lim_{n \to \infty} |p(M+nq)|=\infty$. That is, for $n$ big enough, $|p(M+nq)|>q$. But notice that $$ p(M+nq) \equiv p(M) \equiv 0 \ mod \ q. $$ So $M+nq \geq M$ and $p(M+nq)$ is a composite number.
PAM1499
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$p=a_kx^k+ \cdots a_1x+a_0$ for $a_i$ integers. Then $p(M+nq)=a_k(M+nq)^k+ \cdots + a_1(M+nq)+a_0$. But we know $M+nq \equiv M \ mod \ q$. So... – PAM1499 Oct 29 '20 at 17:06