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 Pythagorean triples
09-22-2010, 08:55 AM
Post: #1
 elim Moderator Posts: 544 Joined: Feb 2010 Reputation: 0
Pythagorean triples
(1) Find all $(a,b,c) \in \mathbb{N}^+$ such that $a^2+b^2=c^2$
(2) Find all $(a,b,c) \in \mathbb{N}^+$ such that $a^2+b^2=2 c^2$
09-23-2010, 07:23 PM
Post: #2
 elim Moderator Posts: 544 Joined: Feb 2010 Reputation: 0
RE: Pythagorean triples
A positive integer triple $(a,b,c)$ is called Pythagorean if $a^2+b^2=c^2$
A Pythagorean triple $(a,b,c)$ is prime if $gcd(a,b,c)=1$
(1) Let $(a,b,c)$ be a prime Pythagorean triple, then $c \equiv a-b \equiv 1 \mod {2}$
$\quad\quad$Since $\gcd(a,b)>1 \Rightarrow \gcd(a,b,c)>1$ and the sum of two odd squares is not a square
$\quad\quad$Assume $2\mid b$ we then have $2 \nmid a, \; 2 \mid (c \pm a)$ and $\displaystyle{\frac{b^2}{4} = \frac{c+a}{2}\frac{c-a}{2}}$ a perfect square. Since
$\quad\quad \displaystyle{\gcd(\frac{c+a}{2},\frac{c-a}{2})=\gcd(c,\frac{c-a}{2})}\le \gcd(c,c-a)=\gcd(a,c)=1$, we see that
$\quad\quad \displaystyle{\frac{c-a}{2} = m^2, \; \frac{c+a}{2} = n^2, \; \gcd(m,n)=1}$ Therefore the set of prime Pythagorean triples is
$\quad\quad P_0 =\{(a,b,c) | c = m^2+n^2,\; \{a,b\} = \{n^2-m^2,2mn\}, \; m<n,\; 2 \nmid (n-m), \; \gcd(m,n) = 1\}$
$\quad\quad$ While the set of Pythagorean triples is
$\quad\quad P_* = \{(a,b,c) | c = m^2+n^2, \; \{a,b\} = \{n^2-m^2,2mn\}, \; m,n \in \mathbb{N}^+,\; m<n\}$
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