Maths Olympiad Prep

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Number theory Difficulty 5.4 AIME, harder Prove it Slovenia

Find the smallest positive integer of the form 3a2ab22b43a^2 - ab^2 - 2b - 4, where aa and bb are some positive integers.

Solution

The answer is 22. If a=4a = 4 and b=3b = 3, then 3a2ab22b4=23a^2 - ab^2 - 2b - 4 = 2.

It therefore suffices to show that the equation 3a2ab22b4=13a^2 - ab^2 - 2b - 4 = 1 has no solutions in positive integers. We can rewrite the equation as 3a2ab2=2b+53a^2 - ab^2 = 2b + 5. The right-hand side is odd, so the left-hand side must be odd as well. So, aa must be odd and bb must be even, which implies that b2b^2 is divisible by 44, and a2a^2 gives a remainder of 11 when divided by 44. It follows that the left-hand side gives the remainder of 33, and the right-hand side gives the remainder of 11 because 2b2b is divisible by 44. This contradiction shows that the equation has no solution in positive integers.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.