In the last post we conjectured that given fixed integers with , the equation

has only finitely many solutions in integers. Let us prove this conjecture.

If or then we are done. So assume that . If has no solution we are done. Otherwise, let be a solution. Assume towards a contradiction that is the infinite set of solutions of . For any , we have

i.e.

.

This means that

is equal to a constant for infinitely many . Since , it follows that

for infinitely many . In particular,

for infinitely many with and . So by the corollary to Theorem 1 of [1],

for some positive integers . So (exercise). But then divides for all , implying is bounded over all . But then, by , must also be bounded over all . Thus is finite, a contradiction.