# How many different integer solutions exist for $p(x)=23$

Let the polynomial $$p(x)=x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n,~n\in\Bbb{N}$$ with all integer coefficients be such that $$p(b_1)=p(b_2)=p(b_3)=p(b_4)=p(b_5)=19$$ for five distinct integers $$b_1,b_2,b_3,b_4,b_5$$. How many different integer solutions exist for $$p(x)=23$$?

The given condition implies that $$b_i$$'s are roots of $$p(x)-19$$. Hence all the $$b_i$$'s divide $$a_n-19$$, from that how do we infer about $$a_n-23$$?. Please help.

Write $$p(x) = (x-b_1)\cdots (x-b_5)q(x)+19$$
Now if $p(x)=23$ then we have $$4 =(x-b_1)\cdots (x-b_5)q(x)$$
so $$|(x-b_1)|\cdots |(x-b_5)|\cdot |q(x)|=4$$ which means that 3 factors among $|x-b_1|,...|x-b_5|$ must be $1$ (else we would get $\geq 8$ on left side). So two values $x-b_1,...x-b_5$ must be the same. A contradiction. So there is no integer $x$ for which $p(x)=23$.