Pythagorean triples - Printable Version +- OMath! (http://math.elinkage.net) +-- Forum: Math Forums (/forumdisplay.php?fid=4) +--- Forum: Number Theory (/forumdisplay.php?fid=7) +--- Thread: Pythagorean triples (/showthread.php?tid=97) Pythagorean triples - elim - 09-22-2010 08:55 AM (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 $a^2+b^2=2 c^2$ RE: Pythagorean triples - elim - 09-23-2010 07:23 PM 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