How NOT to do an Olympiad Math Problem

[alexarnold]

At 17:02 you got that factoring wrong, that's x^3 - 4x, but the point still stands as 8< 9 = 3^2, it's still >0 when X>2 for positive integers

[juuso]

Yes most definitely this is NOT the way to solve this. If I end up with messy polynomials like you did, I figure I've taken the wrong path and try something else.

Even if I would get a solution via pure stubborness like you did, it would still bother me what is the better less messy way.

My approach was to change variables:

u = x + y
v = x*y.

Then the initial equation becomes:

(v-6)^2 + 13 = u^2

Using x^2-y^2 = (x+y)(x-y) and the fact that 13 is prime ie. 13 = 13 * 1 immediately gives all possible values for u and v and then for x and y.