📂Number Theory

A Primitive Pythagorean Triple Can Be Expressed With Only Two Odd Numbers

Theorem ¹

For three natural numbers $a,b,c$ satisfying $a^2 + b^2 = c^2$, $$ \begin{align*} a =& st \\ b =& {{s^2 - t^2 } \over {2}} \\ c =& {{s^2 + t^2 } \over {2}} \end{align*} $$ there exist two coprime odd numbers $s>t$ satisfying the above.

Explanation

According to this theorem, there is practically no reason to call a Pythagorean triple a ’triple’ anymore. Being able to reduce the number of variables is unconditionally a good thing, no matter what the subject is.

Proof

Since $(a,b,c)$ is a Pythagorean triple, either $a$ or $b$ is even. If both $a,b$ were even, then $c^{2} = a^{2} + b^{2}$ would also be even, so $(a,b,c)$ would have the common divisor $2$, which violates the premise of being a primitive Pythagorean triple. Therefore, only one of $a$ and $b$ must be even. Without loss of generality, let $a$ be odd and $b$ be even.

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Part 1. $\gcd \left( c-b, c+b \right) =1$

Since $(a,b,c)$ is a Pythagorean triple, $a^2 + b^2 = c^2$. Moving $b^2$ to the right-hand side, $$ a^2 = c^2 - b^2 = (c+b)(c-b) $$ If $b$ and $c$ had a common divisor $d_{1} \ne 1$, then $a^{2}$ would be expressed, for some $B_{1},C_{1}$, as $$ a^{2} = c^{2} - b^{2} = d_{1}^{2} \left( C_{1}^{2} - B_{1}^{2} \right) $$ so $a$ would also be a multiple of $d_{1}$, contradicting the assumption of being a primitive Pythagorean triple; hence $b$ and $c$ are coprime. Also, suppose $(c+b)$ and $(c-b)$ had a common divisor $d_{2} \ne 1$. Then $d_{2}$ would also be a divisor of $$ 2b = (c+b) - (c-b) \\ 2c = (c+b) + (c-b) $$ and since $b$ and $c$ are coprime, $d_{2}$ must be $2$. However, likewise, for some $C_{2} , B_{2}$, $$ a^{2} = c^{2} - b^{2} = d_{2}^{2} \left( C_{2}^{2} - B_{2}^{2} \right) $$ so $d_{2} = 2$ would be a divisor of $a$, which violates $a$ being odd.

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Part 2. Definition of $s,t$

For the square $a^{2}$ of $a$ to be expressed as the product of $(c-b)$ and $(c+b)$, $$ a^{2} = (c-b)(c+b) $$ since $(c-b)$ and $(c+b)$ were coprime in Part 1, each must be the square of some number $s,t$. Now, letting $$ s^2 := c + b \\ t^2 := c - b $$ since $b$ is even and $c$ is odd, $s,t$ are coprime odd numbers satisfying the following. $$ st = \sqrt{(c-b)(c+b)} = a \\ {{s^2 - t^2 } \over {2}} = b \\ {{s^2 + t^2 } \over {2}} = c $$

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Part 3. Verification

$$ \begin{align*} \left( st \right)^{2} + \left( {{ s^{2} - t^{2} } \over { 2 }} \right)^{2} &= s^{2} t^{2} + {{ s^{4} - 2 s^{2} t^{2} + t^{4} } \over { 4 }} \\ =& {{ s^{4} + 2 s^{2} t^{2} + t^{4} } \over { 4 }} \\ =& \left( {{s^2 + t^2 } \over {2}} \right)^{2} \end{align*} $$ Also, primitive Pythagorean numbers, when expressed in terms of $s,t$, satisfy the following. $$ \gcd \left( st , {{ s^{2} - t^{2} } \over { 2 }} \right) = 1 \\ \gcd \left( {{ s^{2} - t^{2} } \over { 2 }} , {{ s^{2} + t^{2} } \over { 2 }} \right) = 1 \\ \gcd \left( {{ s^{2} + t^{2} } \over { 2 }} , st \right) = 1 $$ Therefore, $\displaystyle st, {{s^2 - t^2 } \over {2}}, {{s^2 + t^2 } \over {2}}$ form a primitive Pythagorean triple.

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