perplexus.info :: Just Math : Some Powers Sum Square II

Some Powers Sum Square II (Posted on 2013-12-18)

Determine all possible pairs (P, Q) of integers such that :
(19P + Q)¹⁸ + (P + Q)¹⁸ + (P + 19Q)¹⁸ is a perfect square.

No Solution Yet Submitted by K Sengupta

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solution and happy new year

| Comment 1 of 2

p=0, q=0 is a solution.

For p<>0, q=0, p^18 * (19^18 + 2) must be a square which implies (19^18 + 2) = (4k + 3) is a square which is impossible.

If the greatest common factor of (p,q) = m>1, we can factor m^18 from both sides of the equation to get an equation of the same form as the original.

So take m=1. Then one of p,q is odd and the other even, making each term of the LHS odd and their sum = (4k + 3), again impossible.

Posted by xdog on 2014-01-01 01:25:20

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