If $p_{i}$ and $p_{j}$ are two primes of the form $4k+1$ , with $p_{j} > p_{i}$,
show that if $p_{j} \neq$ sum of two squares $p_{i}$ is also not equal to sum of two squares.
It is well known that primes of the form $4k+1$ can be written as a sum of two squares. However, my task is to prove this using infinite descent. So my strategy was to assume contrary and take an arbitrary prime which is not the sum of two square (say $p_{j}$) and to prove that if $p_{j}$ is not a sum of two squares, $p_{i}$ is also not a sum of two squares. By this, I will arrive at the number $5$ in the end which is the sum of two squares and hence our assumption will become false.
I took two cases, one was that $p_{j}$ is a sum of three squares and the other was that $p_{j}$ is a sum of four squares. In either case, my strategy was to derive a contradiction by putting $p_{i}$ a sum of two squares. Moreover, my case division exhausted all the cases since by the sum of squares theorem, a number can be either a square, sum of two squares, sum of three squares or sum of four squares, and a prime cannot be a square. However, I cannot think of a way to derive a contradiction from either case.
Any help will be appreciated.
Thanks!