More fun in 2015, Part 35

There is an old and elegant theory on how to write a positive integer as a sum of two squares. Here is a good summary , in just two pages. The proofs, based on Gaussian integers, are pretty straightforward.

Let's apply this theory to our problem. If $n$ is a square-free positive integer with $m$ prime factor of the form $4k+1$ , then there are $\frac{1}{2}(3^m-1)$ ways to write $n^2=a^2+b^2$ for an unordered pair ${a,b}$ of positive integers, each giving us a Pythagorean triangle with hypotenuse $n$ . The smallest number $m$ such that $\frac{1}{2}(3^m-1)>2015$ turns out to be $m=8$ . To construct the smallest $n$ with the required property, let's multiply the 8 smallest primes of the form $4k+1$ , namely $n=5*13*17*29*37*41*53*61$ . The largest prime factor of $n$ is $\boxed{61}$ .