A Fresh Power Diophantine Equation

anbn+2=(a+b)n1 a^n-b^{n+2}=(a+b)^{n-1}

aa and bb are relatively prime positive integers and n(>1)n\, (>1) is an integer. Find all solutions to the equation above and enter your answer as (a+b+n).\sum (a+b+n).


This is a generalization of the problem from 1997 Russian Olympiad:

For prime numbers pp and q,q, solve p3q5=(p+q)2.p^3-q^5=(p+q)^2.

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