피타고라스 세 쌍

 

정의1

$a,b,c \in \mathbb{N}$, a triple $(a, b, c)$ is a Pythagorean triple if $a^{2} + b^{2} =c^{2}$

 

# 줄여서 $PT$라고 쓰기도 한다.

 

 

 

정의2

A $PT$ $a,b,c$ is a primitive Pythagorean triple ($PPT$) if $gcd(a,b,c) = 1$

 

 

 

How to find $PPT$?

 

Suppose 1 : $(a, b) =$ (even, even)

Then $c^{2}$ is even

$\Rightarrow$ $c$ is even

$\Rightarrow$ $gcd(a,b,c) \geq 2$

Contradiction

 

Suppose 2 : $(a, b) =$ (odd, odd)

$a = 2x+1, b=2y+1$ for some $x,y \in \mathbb{Z}_{\geq 0}$  $\Rightarrow$  $c = 2z$  for some $z \in \mathbb{Z}$

 

위를 $a^{2} + b^{2} =c^{2}$에 대입하면

$\Rightarrow$ $(2x+1)^{2} + (2y+1)^{2} =(2z)^{2}$

$\Rightarrow$ $4x^{2} + 4x + 4y^{2} + 4y + 2 =4z^{2}$

$\Rightarrow$ $2x^{2} + 2x + 2y^{2} + 2y + 1 =2z^{2}$

좌변은 odd이고 우변은 even이다.

Contradiction

 

WLOG, WMA $(a,b) =$ (odd, even) (Then c is odd)

$a^{2} + b^{2} =c^{2}$

$\Rightarrow$  $a^{2} = c^{2} - b^{2} = (c-b)(c+b)$

 

Claim 1 : $c-b$ and $c+b$ are squares

 It sufficies to show that
 $gcd(c-b, c+b) = 1$ since $(c-b)(c+b)$ is a square

 Cl 1 - Suppose 1 : $d|c-b$ and $d|c+b$
 Then $d|(c+b) + (c-b)$ $\Rightarrow$ $d|2c$
 and   $d|(c+b) - (c-b)$ $\Rightarrow$ $d|2b$
 $gcd(b,c) = 1$  since  $gcd(a, b,c) = 1$ and $a^{2} + b^{2} =c^{2} $
 $\therefore$  $d=1$ or $d=2$

 $c =$ odd, $b = $ even $\Rightarrow$ $c - b =$ odd 
 $\Rightarrow$ $d \neq 2$
 $\therefore$  $d=1$ 
 $\Rightarrow$  $gcd(c-b, c+b) = 1$

 

Cl 1 - Conclusion : $a^{2} = t^{2}s^{2} \; (t^{2} = c-b, s^{2} = c+b$)

$\Rightarrow$  $ a = st, \; b = \frac{s^{2}-t^{2}}{2}, \; c = \frac{s^{2}+t^{2}}{2}$

$(s, t : odd, \; s > t \geq 1, \; gcd(s,t)=1)$

 

 

 

정리1

Let $a,b,c \in \mathbb{N}$ suppose $a$ is odd, $b$ is even. Then

$(a,b,c)$ is a $PPT$

$\Leftrightarrow $  $a = st, \; b = \frac{s^{2}-t^{2}}{2}, \; c = \frac{s^{2}+t^{2}}{2}$ for some $s,t \in \mathbb{N}$ s.t.

$(s, t : odd, \; s > t \geq 1, \; gcd(s,t)=1)$

 

pf)

$\Rightarrow)$

Done

 

$\Leftarrow)$

  We need to show that $gcd(a, b,c) = 1$ and $a^{2} + b^{2} =c^{2} $

  clearly $(st)^{2} + ( \frac{s^{2}-t^{2}}{2})^{2} = ( \frac{s^{2}+t^{2}}{2})^{2}$

  show that $gcd(st, \frac{s^{2}-t^{2}}{2}, \frac{s^{2}+t^{2}}{2}) = 1$

 

  Suppose 1 : $p|st, \frac{s^{2}-t^{2}}{2}, \frac{s^{2}+t^{2}}{2}$  ($p$ is prime)

  Since $p$ is a prime and $p|st$,

  so $p|s$  or  $p|t$

 

  S1 - Case 1 : $p|s$

  $s=px$ for some $x \in \mathbb{N}$

  Since $ p|\frac{s^{2}-t^{2}}{2}$,

  so $s^{2}-t^{2} = 2py$ for some $y \in \mathbb{N}$

  $\Rightarrow$ $(px)^{2} - t^{2} = 2py$

  $\Rightarrow$ $(px)^{2} -2py = t^{2} $

  Since $ p | (px)^{2} -2py$,

  so $p|t^{2}$ $\Rightarrow$ $p|t$

  Since $p|s, \; p|t$ and $gcd(s,t)=1$,

  so $p=1$

  $\therefore$   $gcd(st, \frac{s^{2}-t^{2}}{2}, \frac{s^{2}+t^{2}}{2}) = 1$

 

  S1 - Case 2 : $p|t$

  Simliar to S1 - Case 1