$p(x)$ be a polynomial over $\mathbb{Z}$. If $P(a)=P(b)=P(c)=-1$ with integers $a,b,c$.Then $P(x)$ has no integral roots

Question

Let $\mathbb{P}(x)$ be a polynomial over $\mathbb{Z}$.

If $\mathbb{P}(a)$=$\mathbb{P}(b)$=$\mathbb{P}(c)$=$-1$ with integers $a,b,c.$

Then $\mathbb{P}(x)$ has no integral roots

Answer 1

Hint $\ $ By CRT/Lagrange, $\rm\ p(x)\, =\, -1 + (x\!-\!a)(x\!-\!b)(x\!-\!c)\,f(x)\ $ for some $\rm\ f(x)\in \Bbb Z[x].\:$ Therefore, if $\rm\:p(n) = 0\:$ for $\rm\: n\in \Bbb Z\:$ then $\rm\ (n\!-\!a)(n\!-\!b)(n\!-\!c)\, f(n)\, =\, 1\:$ so $\,\ldots$