If $x+y$ is a factor of $x^2$ prove it's also a factor of $y^2$ If $X+Y$ is a factor of $X^2$ prove $X+Y$ is a factor of $Y^2$.
I have tried the rmainder theorem, attempted factorisation but those don't work.
 A: So $$x+y\mid x^2$$ Since $x+y\mid y^2-x^2$ we have also $$x+y\mid x^2+(y^2-x^2)=y^2$$
A: $(x+y)x-xy=x^2$ so that $x+y|xy$
Hence $$x+y|(x+y)y-xy=y^2$$
A: $x,y$ and $k$ are integers.
$x^2=k(x+y)$;
$((x+y)-y)^2=$
$ (x+y)^2-2y(x+y)+y^2=k(x+y);$
$y^2=k(x+y) -(x+y)^2 +2(x+y)y;$
Hence?
A: Or let $x=dx_1,y=dy_1$ with ${\rm gcd}(x_1,y_1)=1$. Then, $x_1+y_1\mid dx_1^2$ holds. It is not hard to see that ${\rm gcd}(x_1+y_1,x_1)={\rm gcd}(x_1,y_1)=1$, and thus, ${\rm gcd}(x_1+y_1,x_1^2)=1$, and thus, $x_1+y_1\mid d$, that is, $x+y=d(x_1+y_1)\mid d^2 \mid y^2$, as claimed.
A: By the Polynomial Congruence Rule, $ $ if $\,f\,$ any polynomial with integer coef's $ $  (OP is $\,f(z) = z^2)$
$\!\!\bmod\, x\!+\!y\!:\ \ x \equiv -y \ \Rightarrow\ f(x)\, \equiv\, f(-y)\, $ 
hence we conclude that $\  f(x)\equiv 0\!\iff\! f(-y)\equiv 0,\ $  i.e. $\ x\!+\!y\mid f(x)\!\iff\! x\!+\!y\mid f(-y)$
Remark $ $ Negating $\,y\,$ above yields $\, x-y\,\mid\, f(x)-f(y).\,$ This is the Universal Factor Theorem, known since high-school in the special case where the indeterminate $y$ is specialized to a number $a$ and reformulated as $\,x\!-\!a\mid f(x) \iff x\!-\!a\mid f(a)\iff f(a) = 0,\,$ i.e. $\,x\!-\!a\,$ is a factor of $\,f(x)$ $\iff a\,$ is a root of $\,f(x)$
