How do you find the value of $N$ given $P(N) = N+51$ and other information about the polynomial $P(x)$? Problem:
Let $P(x)$ be a polynomial with integer coefficients such that $P(21)=17$, $P(32)=-247$, $P(37)=33$. If $P(N) = N + 51$ for some positive integer $N$, then find $N$.

I can't think of anyway to begin this question so any help will be appreciated.

Alternative approach: Polynomial Remainder Theorem (kind of).
$$p(x) = k(x)(x-21)(x-32)(x-37)+r(x)$$
$r(x)$ is the remainder after division by a degree 3 polynomial, so $r(x)$ is at most degree 2:
$$r(x)=ax^2+bx+c$$
$$p(21)=17=r(21) \therefore 17 = 17^2a+17b+c$$
$$p(32)=-247=r(32) \therefore -247 = 32^2a+32b+c$$
$$p(37)=33=r(37) \therefore 33 = 33^2a+33b+c$$
Solving the simultaneous equations gives
$$r(x)=5x^2-289x+3881$$
Now, using our equation linking $p(x)$, $k(x)$ and $r(x)$ we know:
$$(N-21)(N-32)(N-37)\ |\ p(N) - r(N) = N+51-5N^2+289N-3881$$
We know that $a|b$ if $\exists c$ such that $ac=b$. So we are looking for a multiplier of the $LHS$ that gives integer solutions for $N$ when we make the $LHS$ equal to the $RHS$. Let's try a multiplier of $1$ to start with:
$$N^3-90N^2+2633N-24864 = -5N^2+290N-3830$$
$$\therefore N^3-85N^2+2343N-21034=0$$
Let's try $(N-26)$ as a factor, long division gives:
$$(N-26)(N^2-59N+809)=0$$
And $(N^2-59N+809)$ has no integer factors, so $N=26$ is the solution.

By the way, this question is from the British Mathematical Olympiad 1987A and is discussed here: Q4 from 23rd British Mathematical Olympiad 1987A
