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
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 ,
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.