Perfect squares.

For natural n>1 let 3n+1 be a perfect square.

Prove (n+1) can be written as a sum of 3 perfect squares, not all of which may be distinct.

1 Answers

341

Hari Shankar ·2010-04-27 10:45:03

3n+1 = (3kÂ±1)² gives n+1 = 3k²Â±2k+1 = k²+k²+(kÂ±1)²

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