Olympiad workbookIOQMQuadratic Equation

Let a, b be positive integers satisfying a^3-b^3-a b=25 . Find the largest possible value of a^2+b^3 .

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a^3-b^3-a b=25 for a=4 and b=3 Because for any greater number a^3-b^3-a b>25 To prove this if a>b , then a^3-b^3-a b aligned & =(b+t)^3-b^3-b(b+t), t>0 & =(3 t-1) b^2+ (3 t^2-t ) b+t^3 is always greater than 4 aligned then b 3 So, a^2+b^3=4^2+3^3=43

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