# Sums of squares : putnam problem

• Dec 3rd 2009, 11:10 PM
Bruno 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 PM
NonCommAlg
Quote:

Originally Posted by Bruno J.
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.

let $\displaystyle n=4k^2(k^2+1), \ k \in \mathbb{Z}.$ then $\displaystyle n=(2k^2)^2 + (2k)^2, \ n+1=(2k^2 + 1)^2, \ n+2=(2k^2+1)^2 + 1.$
• Dec 3rd 2009, 11:46 PM
Bruno 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 PM
NonCommAlg
Quote:

Originally Posted by Bruno 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)

your solution is pretty good too! (Clapping)