Given that $a$ and $b$ are relatively prime and not both odd, prove that $a^2+b^2$ and $a^2-b^2$ are also relatively prime.
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3False. Eg $a=3,b=5$ – almagest Jan 20 '20 at 17:41
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1in fact parity shows any time both $a,b$ are odd, both expressions are even. – Jan 20 '20 at 17:44
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Truth. a,b even then? – Peter Szilas Jan 20 '20 at 17:45
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$a,b$ even means they aren't coprime. – Jan 20 '20 at 17:47
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Sorry everyone. I forgot to include not both odd. – Truth-seek Jan 20 '20 at 17:52
2 Answers
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The correct result is that $a^2+b^2$ and $a^2-b^2$ are relatively prime unless $a$ and $b$ are both odd.
If $a$ and $b$ are both odd, then $a^2+b^2$ and $a^2-b^2$ have g.c.d. $2$.
N.B. Any common divisor of $a^2+b^2$ and $a^2-b^2$ must divide both their sum, $2a^2$ and their difference, $2b^2$. Since $a$ and $b$ are coprime the only possibilities for the g.c.d. are $1$ and $2$.
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Can you prove the amended statement? It's not obvious and this isn't a complete answer without that. – Matt Samuel Jan 20 '20 at 17:48
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2@MattSamuel gcd must divide $2b^2$ so it's either 2, or a factor of $b^2$ but those can't divide $a^2$ proof complete. – Jan 20 '20 at 17:54
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The statement is invalid.
Let $a=x^2-1$ and $b=2x$. $$\begin{align} a^2+b^2 &=(x^2+1)^2 \\ a^2-b^2 &=(x^2-2x-1)(x^2+2x+1).\end{align}$$ When $x$ is odd, they have a common factor of $2$.
Whoops, didn't realise the question was edited until after.
Mr Pie
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