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The zeroes of the function $f(x)=x^{2}-a x+2 a$ are integers. What is the sum of the possible values of $a$? (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, $a^{2}-8 a$, is a perfect square, say $k^{2}$. Then adding $16$ to both sides and completing the square yields $$ (a-4)^{2}=k^{2}+16 $$.  Therefore, $(a-4)^{2}-k^{2}=16$ and $$((a-4)-k)((a-4)+k)=16$$.  Let $(a-4)-k=u$ and $(a-4)+k=v$; then, $a-4=\frac{u+v}{2}$ and so $a=\frac{u+v}{2}+4$.  Listing all possible $(u, v)$ pairs such that $a$ is integral: $(2,8),(4,4),(-2,-8),(-4,-4)$ (not counting transpositions because this does not affect $u+v$). Then, $a=9,8,-1,0$.  These $a$ sum to $16$. 