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The zeroes of the function $f(x)=x^{2}-a x+a$ are integers. What is the sum of the possible values of $a$? (Adapted From 2015 AMC 10A Problems, Question \#23) 
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By Vieta\textquotesingle s Formula, $a$ is the sum of the integral zeros of the function, and so $a$ is integral.  Because the zeros are integral, the discriminant of the function, $a^{2}-4a$, is a perfect square, say $k^{2}$. Then adding $4$ to both sides and completing the square yields $$ (a-2)^{2}=k^{2}+4$$.  Therefore, $(a-2)^{2}-k^{2}=4$ and $$((a-2)-k)((a-2)+k)=4$$.  Let $(a-2)-k=u$ and $(a-2)+k=v$; then, $a-2=\frac{u+v}{2}$ and so $a=\frac{u+v}{2}+2$.  Listing all possible $(u, v)$ pairs such that $a$ is integral: $(2,2), (-2,-2)$ (not counting transpositions because this does not affect $u+v$). Then, $a=4,0$.  These $a$ sum to $4$. 