Number of possible polynomials 
Let $a,b,c,d$ be four integers (not necessarily distinct) in the set $\{1,2,3,4,5\}$. Find the number of polynomials of the form $x^4+ ax^3 + bx^2 + cx +d$ which is divisible by $x+1$.

My Try:
Let $f(x) = x^4+ ax^3 + bx^2 + cx +d$, then $f(-1) = 0$. Thus $1+ (b+d) = c+a$. On counting cases I got 80 permissible cases. Is there a way to solve the above equation $1+ (b+d) = c+a$?

The number of solutions of $a-b+c-d=1$ for $a,b,c,d\in\{1,2,3,4,5\}$ can be counted as the coefficient of $x$ in the Laurent series
$$ \left(x+x^2+x^3+x^4+x^5\right)^2\left(\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}\right)^2,$$
which is also the coefficient of $x$ in 
$$ \left(\frac{1}{x^4}+\frac{2}{x^3}+\frac{3}{x^2}+\frac{4}{x}+5+4x+3x^2+2x^3+x^4\right)^2, $$
so it is given by 
$$ 2\cdot 1+3\cdot 2+4\cdot 3+5\cdot 4+4\cdot 5+3\cdot 4+2\cdot 3+1\cdot 2=2\sum_{k=1}^{4}k(k+1)=80 $$
as claimed.
