Any product of primes in the form of 4n+1 is the sum of 2 relatively prime squares

Question

How can we prove the theorem that any product of primes in the form $4n+1$ is the sum of 2 relatively prime square numbers?

EDIT: My question is not the same as the responses that were linked. I'm asking specifically about the product of primes in the form $4n+1$ being the sum of $2$ relatively prime squares, not just primes in the form $4n+1$ being the sum of squares.

Answer 1

Since every prime of the form 4n+1 is the sum of two squares, and

$\begin{array}\\ (a^2+b^2)(c^2+d^2) &=a^2c^2+a^2d^2+b^2c^2+b^2d^2\\ &=a^2c^2\pm 2abcd+b^2d^2+a^2d^2\mp 2abcd+b^2c^2\\ &=(ac\pm bd)^2+(ad\mp bc)^2\\ \end{array} $

the product of any number of primes of the form 4n+1 is the sum of two squares.

If the two squares are not relatively prime, some prime divides both of them, and so the square of that prime divides each square and so divides the product of the primes.

But the product of the primes (unless two of them are the same) is only divisible by primes to the first power, a contradiction.

Answer 2

I think more work is needed to show that the squares can be chosen to be coprime. The above answer assumes that the primes are all distinct, but that isn't stated in the question.

For example having got $13 \times 5 = 8^2 + 1^2$, we can extend to $$13 \times 5 \times 5 = (8^2 + 1^2)(1^2 + 2^2) = 10^2 + 15^2 = 17^2 + 6^2,$$ where the first solution is invalid because $10$ and $15$ aren't coprime. We need to show that at least one of the two solutions is valid.

Suppose $a^2 + b^2$ is a product of primes $4n + 1$, and $c^2 + d^2$ is a prime of that form. If the first solution is invalid then some prime $q$ divides both $ac + bd$ and $ad - bc$. Hence $$q \vert (ac + bd)c + (ad - bc)d = a(c^2 + d^2),\quad q \vert (ac + bd)d - (ad - bc)c = b(c^2 + d^2).$$ Since $q$ doesn't divide both $a$ and $b$, we have $q \vert c^2 + d^2$, hence $q = c^2 + d^2$.

If both solutions are invalid, then $c^2 + d^2$ divides $ac - bd$ as well as $ac + bd$, hence divides $2ac$ and $2bd$. Since $c^2 + d^2$ doesn't divide $2c$ or $2d$ it must divide both $a$ and $b$, contrary to hypothesis.