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.
 A: 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
A: Since $P$ is with integer coefficients then 
$$\forall x,y \in \mathbb{Z}:(x-y) | \left(p(x) - p(y)\right)$$ then we will get 
$$p(n)-p(21) = n+34 \Rightarrow n-21 |n+34 \Rightarrow  n-21 | 55$$
$$p(n)-p(32) = n+298 \Rightarrow n-32 |n-298 \Rightarrow n+298 |330$$
$$p(n)-p(37) = n+18 \Rightarrow n-37 |n+18 \Rightarrow n-37 |55 $$
So I think if we do all calculations  we are done.
A: $ \overbrace{{\rm Note}\,\ \ p(x)\!=\!x\!-\!4\,\ {\rm at}\ x\!=\!21,37}^{\textstyle{\rm so\ let}\ \ p(x)\!-x\!+\!4\, =:\, f(x)\qquad\!}\,$ so $\,f\,$ has roots $\,\!21,37\,$ $\Rightarrow$ $\,f^{\phantom{|^|}}\!\!\!\! = (x\!-\!21)(x\!-\!37)g,\,$ $g\in\Bbb Z[x]\,$ so evaluating at $\,x\!=\!n\Rightarrow$ $\ \color{#c00}{55}^{\phantom{|^|}}\!\!\! = f(n) = (\color{#90f}{n\!-\!21})(\color{#0a0}{n\!-\!37})g(n),\,$ so test $\rm\color{#90f}{fac}\color{#0a0}{tors}$ of $\color{#c00}{55}$ differing by $16^{\phantom{|^|}}\!$ 
A: The hint.
$$p(n)-p(21)=n+51-17=n-21+55,$$
Thus, $55$ divisible by $n-21$.
Now, make the similar things with $p(n)-p(32)$ and with $p(n)-p(37)$.
Indeed, $$p(n)-p(37)=n+51-33=n-37+55,$$
which gives that $55$ divisible by $n-37$ and
$$p(n)-p(32)=n+51+247=n-32+330,$$ 
which gives that $330$ divisible by $n-32$.
Now, since $55$ divisible by $n-37$, we obtain:
$$n-37\in\{-55,-11,-5,-1,1,5,11,55\}$$ or
$$n\in\{-18,26,32,34,38,42,48,92\}$$
and we see that only $n=26$ is valid.
Done! 
A: Since $P(x)$ is a polynominal with integer coefficients, $a-b$ divides $P(a)-P(b)$
$N-21$ divides $P(N)-P(21)$, which implies that $N+34$ is divisible by $N-21$. 
$\therefore$ $N-21|55$
Proceed in the same manner for $37$ and $32$, which yields $N-32|330, N-37|55$. 
Let $N-21=X$. Then $X|55$ and $X-16|55$. All divisors of $55$ are $-55,-11,-5,-1,1,5,11,55$. This implies that $X=5$ or $11$. If $X=11$, $N=32$. A contradiction. Therforefore $X=5$. 
$\therefore N=26$ 
