Quadratic Congruence modulo square-free integer

Question

If $m$ is a square-free integer, show that $x^{2} + y^{2} \equiv k\pmod{m}$ has a solution $\forall k\in\mathbb{N}$. This means that we need to prove existence of such $m$ for all $k\in\mathbb{N}$.

So far I've got this:

Prime decomposition of each square-free integer has exactly one factor for each prime in it. Therefore $m= p_1...p_v$.

Suppose that $k=a+b$ for some $a,b \in \mathbb{N}$, then $x^2 \equiv a \ \pmod{m}$, $y^2 \equiv b\pmod{m}$ would imply $x^2 + y^2 \equiv k\pmod{m}$.

Using Chinese remainder theorem $x^{2} \equiv a \pmod{m}$ requires

$x^2 \equiv a \pmod{p_1}$

...

$x^2 \equiv a \pmod{p_v}$,

$y^2 \equiv b \pmod{p_1}$

...

$y^2 \equiv b \pmod{p_v}$,

Therefore we nedd to show that $\forall a,b \in \mathbb{N}$ we can find primes $p_{1},...,p_{v}$ which satisfy the above mentioned congruences.

I would appreciate an advice or a hind, because I'm not sure whether this is the right path to the solution.

Answer 1

For $p_i>2$ not every residue is a quadratic residue, so the congruences $$x^2\equiv a\pmod{p_i}\qquad\text{ and }\qquad y^2\equiv b\pmod{p_i},$$ do not admit solutions for all integers $a$ and $b$ with $k=a+b$.

Also, after applying the Chinese remainder theorem you seem to get confused: The question is not to find primes $p_1,\ldots,p_n$ such that the congruences hold. You are given $m=p_1\cdot\ldots\cdot p_n$ and must show that there exist $a$ and $b$ with $k=a+b$ such that the congruences hold. But unfortunately finding such $a$ and $b$ explicitly for all $k$ and $m$ is not practically doable.

Edit: Upon reading again, this confusion already starts on the very first line; you do not need to prove existence of $m$ for all $k$, you need to prove existence of $x$ and $y$ for all $k$ and $m$.

Your idea to use the Chinese remainder theorem is a good start though; you can use this to reduce the problem to showing that for every prime number $p$ the congruence $$x^2+y^2\equiv k\pmod{p},$$ has a solution for all $k\in\Bbb{N}$. This problem has been posed and answered on this site before.