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 $p(x)=23$.

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