MHB Problem of the Week # 260 - Apr 25, 2017

[Ackbach]

Here is this week's POTW:

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Prove that there exist infinitely many integers $n$ such that $n,n+1,n+2$ are each the sum of the squares of two integers. [Example: $0=0^2+0^2$, $1=0^2+1^2$, $2=1^2+1^2$.]

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[Ackbach]

Re: Problem Of The Week # 260 - Apr 25, 2017

This was Problem A-2 in the 2000 William Lowell Putnam Mathematical Competition.

Congratulations to Kiwi for his correct solution, which follows:

Spoiler

Let's find an infinite number of such 'triplets' of the form \(a^2+(n-1)^2=n^2-1,n^2,n^2+1^2\) with a, a non negative integer.

Clearly, there are an infinite number of triplets such that the second and third entries have the required form and are successive integers.

Assume there is only a finite number of fully compliant triplets. There is certainly at least one triplet found by setting n=1 and a=0 to give (0,1,2).

If there is a finite number of these triplets, then there must be a largest such triplet. Let that triplet be defined by n=k.

Now \(a^2+(k-1)^2=k^2-1\) and

\(a^2+k^2-2k+2=k^2\) therefore

\(a=\sqrt{2k-2}\)

The LHS is an integer and the RHS is divisible by square root of 2 so a is divisible by 2.

Now replace a with a+2 and we solve for n.

\(a+2=\sqrt{2n-2}\)

\(a^2+4a+6=2n\)

The LHS is divisible by 2 so n is an integer, but n > k because:

\(k=\frac{a^2+2}2 \) < \(n=\frac{(a+2)^2+2}2 \)

We have found n>k such that
\((a+2)^2+(n-1)^2=n^2-1,n^2,n^2+1^2\) is a triple greater than \(a^2+(k-1)^2=k^2-1,k^2,k^2+1^2\)

A contradiction so there are an infinite number of compliant triplets.