how many positive integer solutions to the following equation? $a^2 + b^2 + 25 = ab + 5a + 5b$
I have tried looking for a factorisation that could solve this question but couldn't find anything useful - found $(a+b+5)^2$ - don't know if this is useful
The equation does look similar to an equation of a circle - can you use this idea?

The intuition is that the right-hand side is, with a few possible exceptins, smaller than the left-hand side.
Note that $(a-b)^2\ge 0$, so $ab\le \frac{a^2+b^2}{2}$. Thus the right-hand side is $\le \frac{a^2+b^2}{2}+5a+5b$.  
It follows that
$$(a^2+b^2+25)-(ab+5a+5b)\ge \frac{a^2+b^2}{2}+25-5a-5b.$$
The right=hand side above is 
$$\frac{1}{2}\left(a^2+b^2-10a-10b+50\right).$$
This is 
$$\frac{1}{2}\left((a-5)^2+(b-5)^2\right),$$
which is $\gt 0$ unless $a=b=5$.
Remark: We can make the proof more myaterious by writing the magic identity
$$2\left[(a^2+b^2+25)-(ab+5a+5b)\right]=(a-5)^2+(b-5)^2+(a-b)^2,$$
and concluding that the left-hand side is $0$ precisely if $a=b=5$.
