2019/009) Find all non-negative integers x,y satisfying $(xy-7)^2 = x^2 + y^2$

We have adding 2xy on both sides

$(xy-7)^2 + 2xy = x^2 +y^2 + 2xy = (x+y)^2$
or $(x^2y^2 -14xy + 49) + 2xy = (x+y)^2$
or $(x^2y^2 -12xy + 49)  = (x+y)^2$
$(x^2y^2 -12xy +36 ) + 13 = (x+y)^2$
or $(xy-6)^2 + 13 = (x+y)^2$
or $13 = (x+y)^2 - (xy-6)^2 = (x+y+xy - 6)(x + y -xy + 6)$
or $13 = ((x+1)(y+1) -7)(5 - (x-1)(y-1))$
this is product of 1 and 13 so we have
$(x+1)(y+1) -7 = 13$ and $5 - (x-1)(y-1) = 1$ giving $x+1)(y+1) = 20$ and $(x-1)(y-1) = 6$
giving (x,y) = (3,4) or (4,3)

Or 
$(x+1)(y+1) -7 = 1$ and $5 - (x-1)(y-1) = 13$ giving $x+1)(y+1) = 8$ and $(x-1)(y-1) = -8$
giving (x,y) = (0,7) or (7,0)

So solution set $(x,y) = (0,7)\, or \,(7,0)\, or\, (3,4)\, or\, (4,3)$