Sum of two squares

Number Theory Level 2

Is it true, that for any two coprime integers ( $x, y$ ), any of the positive divisors of $x^2+y^2$ can be expressed as sum of two perfect squares.

For example: If $x=1, y=3$ , then the statement above holds true (here $x^2+y^2=10$ ), because $\begin{aligned} 1 & = 0^2+1^2 \\ 2 & = 1^2+1^2 \\ 5 & = 1^2+2^2 \\ 10 & = 1^2+3^2 \end{aligned}$

Yes, it is true No, it is false

1 solution

Jesse Nieminen
Aug 17, 2018

Straight forward corollary of the sum of two squares theorem is that, if there is no prime $p = 4k + 3$ dividing $n$ then $n$ can be expressed as sum of two squares.

Now, let's show that all odd prime factors of $x^2 + y^2$ are of form $p = 4k + 1$ .

Assume contrary, i.e. there exists $p = 4k + 3$ such that $p \mid x^2 + y^2$ .

It's well-known that $p$ is a prime in gaussian integers meaning that since $x^2 + y^2 = (x + yi)(x - yi)$ we must have $p \mid x + yi$ or $p \mid x - yi$ .

However, this is well-known to be impossible since $x$ and $y$ are coprime.

Hence, all of the factors of $x^2 + y^2$ can be represented as sums of two squares. $\square$

Sum of two squares

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