Proof by contradiction

**nikie1o2** · February 3rd 2010, 01:10 PM

If u and v have different parities( one is even and one is odd), then ( a,b,c) is primitive.

Fact: If (a, b, c) is not primitive, then there exists a prime number p and natural numbers a`, b`, c` , such that a=pa`, b=pb` , c=pc`

Fact 2: if p is a prime number, and m,n are elements of N such tha p|mn , then p|m or p|n

**Media_Man** · February 5th 2010, 04:17 PM

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.

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