• February 3rd 2010, 02:10 PM
nikie1o2
Well, for starters, you ought to specify that we're referring to the equations $(a,b,c)=(u^2-v^2,2uv,u^2+v^2)$. Also, this theorem is false: $u$ and $v$ must not only have opposite parities, they must be relatively prime. Example: $6$ and $3$ have opposite parity, yet $(27,36,45)$ is NOT primitive. The correct theorem should read: "If $(u,v)=1$, then $(a,b,c)$ is primitive." Moving on...
To prove by contradiction, we start with the premise " $u$ and $v$ have no common divisors AND there exists a prime $p$ that divides $a,b,c$." We will show this premise to be impossible:
Lemma: if $(u,v)=1$, $u$ and $v$ must have opposite parities, one even one odd. $p|b$ so $p|2uv$, so $p=2$ or $p|u$ or $p|v$. If $p=2$ then since $u$ and $v$ have opposite parities, so $p|u$ or $p|v$. WLOG, suppose $p|u$. Now, $p|a$ so $p|u^2-v^2$. Since $p|u$, $p|v^2$ so $p|v$. Thus, $(u,v)\neq1$, and we have our contradiction. Therefore our premise is false.