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No naturals with 4m(m+1) = n(n+1) - [Canada MO 1977]

Source:

September 29, 2011
quadratics

Problem Statement

If f(x)=x2+xf(x) = x^2 + x, prove that the equation 4f(a)=f(b)4f(a) = f(b) has no solutions in positive integers aa and bb.