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Prove that if $\gcd(x, y) = 1$ then $\gcd(x + y, x - y) = 1$ or $2$.

I know that any linear combination of $x, y$ is multiple of 1 since $\gcd(x, y) = 1$ then the set would be $\{1, 2, 3, 4, 5, \ldots, x - y, \ldots, x + y, \ldots \}$, $x$ and $y$ cannot be both even. But $x$ and $y$ can be both odd primes or even and odd for their GCD to be 1. If $x$ and $y$ are odd prime then $x + y$ and $x - y$ will give even integers, so their GCD could be 2. If $x$ and $y$ are even and odd then their GCD can be either 1, 3 or 5.

This is what I think. Is it in the correct way?

Martin Sleziak
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Ron Ald
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1 Answers1

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Suppose $p$ is 1 or prime and $p$ divides $x-y$ and $x+y$ then $p$ divides $2x$ and $2y,$ since if $p$ divides into two numbers it divides into their sum and difference.

Either $p$ divides $2$ or $x$.

Either $p$ divides $2$ or $y$.

If $p$ divides into $2$ it is $1$ or $2.$ If it divides $x$ and $y,$ it is $1.$ We are done.

Mark Joshi
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