Let $p$ and $q$ be odd primes. Prove that $\gcd(p + q, p - q) = 2$. I have considered EEA to multiply it out, but I'm unsure where to go from there.
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Martin Sleziak
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user242743
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2you need $p \neq q$ – Will Jagy Dec 02 '13 at 03:40
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See also: Prove $\gcd(a+b, a-b) = 1$ or $\gcd(a+b, a-b) = 2$ if $\gcd(a,b) = 1$ – Martin Sleziak Jul 27 '17 at 13:25
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If $\gcd(p-q,p+q)=d$, then we definitely have $d \mid p-q+p+q=2p$. This implies that either $d=2$ or $d=p$. The last case implies that $p=q$, so if $p$ and $q$ are distinct primes, we must have $d=2$.
primes.against.humanity
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First off, you need $p\not=q$ because $\gcd(3+3,3-3)=\gcd(6,0)=6$.
Since $p,q$ are odd, we know that $p+q,p-q$ are even. Thus, $2|p+q$ and $2|p-q$. Therefore, $\gcd(p-q,p+q)\geq2$. It remains to show that $\gcd(p-q,p+q)\leq2$.
Since $p,q$ are distinct odd primes, we have $\gcd(p,q)=1$. By Bezout's Identity, there exists integers $a,b$ such that $ap+bq=\gcd(p,q)=1$. Let $x=a+b$ and $y=a-b$. Observe that $(p+q)x+(p-q)y=2$. Thus, by Bezout's Identity, $\gcd(p+q,p-q)\leq2$. This is what we desired.
Note: In some places, Bezout's identity is called GCD Characterization Theorem.
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