# Given primitive Pythagorean triple $(a,b,c)$ with $a<b<c$, if $a$ is odd, then $a^2 = b+c$

I've noticed that, for any pythagorean triple $(a,b,c)$ arranged least to greatest, if they have no common divisor (i.e., if the triple is primitive) and $a$ is odd, then following holds:

$$a^2=b+c$$

For example, for $(3,4,5)$, we have $4+5=9$.

Is there a way to prove this? Or an intuitive reason it's true?

• This is by no means true. Take $(20,21,29)$. Then $20^2\neq 21+29$ – Dietrich Burde May 3 '18 at 18:04
• @DietrichBurde Sorry, forgot to mention $a$ is odd. – NMister May 3 '18 at 18:08
• Then take $(33, 56, 65)$. Again $33^2\neq 56+65$. – Dietrich Burde May 3 '18 at 18:09

Suppose $a^2+b^2=c^2$, then
$a^2=c^2-b^2=(c+b)(c-b)$ if we want this to be equal to $c+b$ we need $c-b=1$.
So this works only for triples $(\sqrt{2x+1},x,x+1)$ or equivalently $(2k+1,2k^2+2k,2k^2+2k+1)$.
Note for a pythag triple $a^2+b^2=c^2$ that $$a=x^2-y^2$$ $$b=2xy$$ $$c=x^2+y^2$$ Essentially what youre saying is that if $x^2-y^2$ is odd, that: $$(x^2-y^2)^2=x^2+y^2+2xy$$ I can factorise this: $$(x+y)^2(x-y)^2=(x+y)^2$$ $$(x-y)^2=1$$ $$x-y=\pm1$$ $$x=y\pm1$$
So this only works for a select few pythagorean triples. For the example you gave $(3,4,5)$, $x=2, y=1$ so $x=y+1$ and the criteria is met.