Hello! This is a nice problem from Putnam a few years ago, for fun.

Show that there are infinitely many triples of consecutive integers, each of which is the sum of two squares.

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- Dec 3rd 2009, 11:10 PMBruno J.Sums of squares : putnam problem
Hello! This is a nice problem from Putnam a few years ago, for fun.

Show that there are infinitely many triples of consecutive integers, each of which is the sum of two squares. - Dec 3rd 2009, 11:40 PMNonCommAlg
- Dec 3rd 2009, 11:46 PMBruno J.
Very nice! Your intuition is very good.

My solution is different. I show that given any triple $\displaystyle n-1, n, n+1$, you can construct another triple. Since sums of two squares are closed under product, $\displaystyle (n-1)(n+1)=n^2-1$ is a sum of two squares, and thus we obtain $\displaystyle n^2-1,n^2,n^2+1$. So it suffices to show that we have one such triple, and 8,9,10 does the job.

Yours is better though (Happy) - Dec 3rd 2009, 11:55 PMNonCommAlg