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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.

Simon
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2 Answers2

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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 $gcd(a^2+b^2,a^2-b^2)=1$.

Tsemo Aristide
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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.

N.S.JOHN
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  • We are actually given $\gcd ( \frac{x^2 +y^2 }{2}, xy) =1$, but the proof proves a stronger inequality – N.S.JOHN Sep 16 '16 at 16:42