Show that $\gcd(a+b,a-b)=1\iff\gcd(a^2+b^2,a^2-b^2)=1.$ 
Problem. Let $a$ and $b$ be two integers. Show that
  $$\gcd(a+b,a-b)=1\iff\gcd(a^2+b^2,a^2-b^2)=1.$$

I am really clueless for how to approach this problem. First let's try $(\Rightarrow)$. We can write
$$x(a+b)+y(a-b)=1.$$
Then I tired to use the identity $a^2+b^2=(a+b)^2-2ab$ but it doesn't lead anywhere.
 A: Let $p$ be a prime number distinct of $2$ suppose that $p$ divides $a+b$ and $p$ divides $a-b$, then $p$ divides $(a+b)(a-b)=a^2-b^2$ and $p$ divides $(a-b)^2+(a+b)^2=2(a^2+b^2)$ and thus $p$ divides $a^2+b^2$.
If $a+b$ and $a-b$ are even, $a$ and $b$ have the same parity. This implies that $a^2$ and $b^2$ have the same parity and $a^2+b^2, a^2-b^2$ are even. Thus $gcd(a^2+b^2,a^2-b^2)=1$ implies that $gcd(a+b,a-b)=1$.
Let $p$ be a prime, suppose that $p$ divides $a^2-b^2$ and $p$ divides $a^2+b^2$, $p$ divides $(a^2-b^2)+(a^2+b^2)=2a^2$, it divides also $2b^2$, thus $p$ divides $2a$ and $2b$. 
It divides    $a^2-b^2=(a+b)(a-b)$, if $p$ divides $a+b$, it divides $(a+b) -2b=a-b$ if $p$ divides $a-b$, it divides $(a-b)+2b=a+b$. Thus $p$ divides $a+b$ and $p$ divides $a-b$, we deduce that $gcd(a+b,a-b)=1$ implies that $gcd(a^2+b^2,a^2-b^2)=1$.
A: Take $a-b=x$ and $a+b=y$
We are given $\gcd ( x^2 +y^2 , xy) =1$
Take $x=pm$ and $y= pn$ with $(m,n)=1$
Manipulate the given expression and obtain $p=1$, giving the desired result.
