# Thread: Sums of squares : putnam problem

1. ## 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.

2. 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 $n=4k^2(k^2+1), \ k \in \mathbb{Z}.$ then $n=(2k^2)^2 + (2k)^2, \ n+1=(2k^2 + 1)^2, \ n+2=(2k^2+1)^2 + 1.$

3. Very nice! Your intuition is very good.

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

Yours is better though

4. Originally Posted by Bruno J.
Very nice! Your intuition is very good.

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

Yours is better though
your solution is pretty good too!