MHB Can Polynomial Constraints Determine a Unique Integer Solution?

[anemone]

Here is this week's POTW:

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Let $P(x)$ be any polynomial with integer coefficients such that $P(21)=17,\,P(32)=-247$ and $P(37)=33$.

Prove that if $P(N)=N+51$ for some integer $N$, then $N=26$.

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[anemone]

Congratulations to kaliprasad for his correct solution(Cool)!

Model solution from other:

Spoiler

If $P(N)=N+51$ for some integer $N$, then $P(x)-x-51=(x-N)Q(x)$ for some polynomial $Q(x)$ by the factor theorem.

Note $Q(x)$ has integer coefficients because $P(x)-x-51=P(x)-P(N)-(x-N)$ is a sum of $a_i(x_i-N_i)$ terms with $a_i's$ integer.

Since $Q(21)$ and $Q(37)$ are integers, $P(21)-21-51=-55$ is divisible by $21-N$ and $P(37)-37-51=-55$ is divisible by $37-N$ is 16, we must have $N=26$ or 32.

However, if $N=32$, then we get $-247=P(32)=32+51$, a contradiction. Therefore, $N=26$.