Find the number of ordered pairs of positive integers (x,y) that satisfy $x^{2} - xy + y^{2} = 49$

Question

I tried to start this quest but won't end up with any answer. First I made a complete square on RHS which is like this $x^{2} - xy - xy + xy + y^{2}$ Then converted it into $xy+{\left( x - y\right) }^{2}$ And as LHS is a perfect square Which is $7^2$ So RHS must be a square but I don't know how to proceed further?

Answer 1

If $x=y$, then $x=y=7$. So suppose $x>y$ (the other case is obviously symmetric). Notice that the equation can be rearranged as $x(x-y)=(7-y)(7+y)$, so $y<7$. Hence, you just have to check $6$ cases.

Answer 2

Do this. Multiply both sides by $4$: $$ x^2-xy+y^2=49\Rightarrow (2x-y)^2+3y^2 =196. $$ Now, this immediately yields, $y^2\leqslant \frac{196}{3}<66$, thus, $y\leqslant 8$. The rest is easy.