If $a$ and $b$ are relatively prime to each other then each factor of $a^2 + b^2$ is the sum of two squares.

Question

If $a$ and $b$ are relatively prime to each other then each factor of $a^2 + b^2$ is the sum of two squares.

I have tried but couldn't able to find anything that leads to the conclusion.

EDIT : I want to mention that I have read two propositions of Euler from wikipedia.

(1) If a number is the sum of two squares and if it is divisible by a prime which is also a sum of two squares then the quotient thus obtained is also a sum of two squares.

With the help of this proposition the second proposition of Euler is proved which asserts that

(2)If a number is the sum of two squares which is divisible by a prime which is not the sum of two squares then there exists a factor of the quotient thus obtained which is not the sum of two squares.

With the help of (2) it has been proved that

(3) If $a$ and $b$ are relatively prime to each other then each factor of $a^2 + b^2$ is the sum of two squares.

Finally with the help of (3) Fermat's theorem of two squares has been proved.

My problem is I fail to grasp the entire proof of (3) from wikipedia.Can anybody please help me understanding it?

Thank you in advance.

Answer 1

There are a few steps, given the way this is asked. the beginning is that $a^2 + b^2$ is not divisible by $4$ or by any prime $q \equiv 3 \pmod 4,$ as such would force a common factor of $a,b;$ in the first case, $2,$ in the second $q.$ Quadratic reciprocity.