this question has bothered me a lot.

if you arent familiar: https://www.youtube.com/watch?v=Y30VF3cSIYQ

was wondering if this is a valid solution to this famous problem:

(a^2+b^2)/(ab+1) = n, so a^2+b^2-nab=n.
let the greater or equal integer be a. a ? b
let a new integer c = bn-a, c,a,b,n are all integers
since a^2+b^2-nab=n and bn = a+c, n=a^2+b^2-a(a+c)=n, n = b^2 - ac
since n = b^2 - ac and n > 0 and and a ? b it follows c<b
now we must prove c ? 0. to do this we assume it isnt: c<0
when this is the case a > bn (c = bn-a) and n > ac (n = b^2 - ac and a,b,n are all positive and c is negative)
since |c| > 0 it follows n > a which contradicts a > bn when b ? 0
(in the case b = 0 it can be easily observed that n = a square integer, in this case a^2)
so c ? 0
we are almost done now just substitute a = bn - c (from c = bn-a) into n = b^2 -ac
we get: n = b^2 - c(bn-c), *n = b^2 + c^2 - nbc*
notice we have the same exact equation we began with except a has become b and b has become c
since the same conditions are present we can rewrite n again with c_2 = cn-b
and as we proved b>c and so b>c>c_2>c_3>c_4.......>c_[k-1] > c_k=0 as we keep rewriting with smaller and smaller integers
it follows that eventually at some integer in this sequence (c_n) we will arrive at zero.
and at this point n = (c_[k-1])^2 + 0^2 - 0 and since all numbers in this sequence are integers n is an integer squared