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.

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 $4$.
