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an−bn+2=(a+b)n−1 a^n-b^{n+2}=(a+b)^{n-1}an−bn+2=(a+b)n−1
aaa and bbb are relatively prime positive integers and n (>1)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).∑(a+b+n).
This is a generalization of the problem from 1997 Russian Olympiad:
For prime numbers ppp and q,q,q, solve p3−q5=(p+q)2.p^3-q^5=(p+q)^2.p3−q5=(p+q)2.
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