integral roots for $f(x) = 41$ if $f(x) = 37$ has 5 distinct integral roots. Given a polynomial $f(x)$ with integral coefficients and $f(x) = 37$ has 5 distinct integral roots, find the number of integral roots of $f(x) = 41$?
My Approach: Say $f(x) = (x-r_1)(x-r_2)(x-r_3)(x-r_4)(x-r_5)g(x) + 37$, where $r_i$ are the distinct integers. 
Now for $f(x) = 41$ we have $(x-r_1)(x-r_2)(x-r_3)(x-r_4)(x-r_5)g(x) = 4$, so the factors can be $\pm 1, \pm2$ or $\pm 4$. Given $r_i$ are distinct at most two of them will give $\pm1$, then there can be both of $\pm 2$ or one of $\pm 4$. This is where I get lost, since even if I use all of $\pm1, \pm2$, I will be still be left with one $x-r_i$ factor. What about that? Does it matter that 37 and 41 are primes, or is it just a coincidence?
Thanks in advance.
 A: You are essentially finished. There cannot be an integer $a$ such that $(a-r_1)(a-r_2)(a-r_3)(a-r_4)(a-r_5)g(a)=4$. For the $a-r_i$ are distinct integers, and no product of $5$ distinct integers can divide $4$.
A: Hint $\ $ Finish by applying the following
Key Idea $\ $ The possible factorizations of a polynomial $\in\Bbb Z[x]$ are constrained by the factorizations of the integer values that the polynomial takes. For a simple example, if some integer value has few factorizations (e.g. a unit $\,\pm1 $ or prime $p$) then the polynomial must also have few factors, asssuming that that the factors are distinct at the evaluation point. More precisely
If $\, f(x) = f_1(x)\cdots f_k(x)\,$ and $\,f_i\in\Bbb Z[x]\,$ satisfy $\color{#0a0}{f_i(n) \ne f_j(n)}\,$ for $\,i\ne j,$ all $\,n\in \Bbb Z$
$\quad \color{}{f(n) =\pm1}\,\Rightarrow\, k\le 2\ $ else $1$ would have $\rm\,3\,\ \color{#0a0}{distinct}$ factors $\,f_1(n),f_2(n),f_3(n)$  
$\quad f(n) = \pm p\,\Rightarrow\, k\le \color{#c0f}4\ $ since a prime $p$ has at most $\,\color{#c0f}4\,$ distinct factors $\,\pm1,\pm p$
Remark $\ $ One can push the key idea to the hilt to obtain a simple algorithm  for polynomial factorization using factorization of its integer values and Lagrange interpolation. The ideas behind this algorithm are due in part to Bernoulli, Schubert, Kronecker. See this answer for references.
