<s>[INST] $(a^2+b^2)\mid (ax+by)$, implies $\gcd(x^2+y^2,a^2+b^2)>1$ Suppose $a,b,x,y$ are natural number such that $(a^2+b^2)\mid (ax+by)$, prove that $\gcd(x^2+y^2,a^2+b^2)>1$. Can anyone give me a hint to solve the problem? I am rather stuck, thanks!
 [/INST] First notice that $$(a^2+b^2)(x^2+y^2)=(ax+by)^2+(bx-ay)^2$$
If $a=da_1$ and $b=db_1$ where $(a_1, b_1) = 1$ the equation can be rewritten as $$(a_1^2+b_1^2)(x^2+y^2)=(a_1x+b_1y)^2+(b_1x-a_1y)^2$$
Suppose now that $p$ is a prime such that $p \mid a_1^2+b_1^2$. Since we have $$d^2(a_1^2+b_1^2)=(a^2+b^2)\mid (ax+by)=d(a_1x+b_1y)\rightarrow d(a_1^2+b_1^2) \mid (a_1x+b_1y)$$ we would get that $p \mid a_1x + b_1y$ and hence from the equation it follows that $p \mid (b_1x-a_1y)$. So we would have that $$p \mid y(a_1x+b_1y) + x(b_1x-a_1y)=b_1(x^2+y^2)$$ Notice that $p \not \mid b_1$ since otherwise we would have that $p \mid b_1$ and from $p \mid a_1^2+b_1^2$ we would get that $p \mid a_1$ and that would effectively mean that $(a_1, b_1) \not = 1$. So we are left with the option of $p \mid x^2+y^2$. Since $p \mid d^2(a_1^2+b_1^2)= a^2+b^2$ we get that $(a^2+b^2, x^2+y^2)$ is at least $p$.
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