Find two primitive pythagorean triples $(a,b,c)$ with $a$ odd, $b$ even and $c=a+2$

Question

Find two primitive Pythagorean triples $(a,b,c)$ with $a$ odd and $b$ even, such that $c=a+2$.

I know the Conditions for primitive triples are: $ a^2 + b^2 = c^2, \quad (a,b) = 1$.

Answer 1

Define new integer variables $x,y$ such that

$a = 2y+1$ and $b = 2x$

Since $c = a+2$ then $c = 2y+3$ and substituting these into the Pythagorean identity we have

$(2y + 1)^{2} + (2x)^{2} = (2y + 3)^{2} $

we can expand the brackets

$4y^{2}+4y+1 + 4x^{2} = 4y^{2}+12y+9$

cancel like terms and rearrange

$ x^{2} = 2y+2$

This tells us that any even x^2 greater than 2 will give a valid y from which we can find a corresponding Pythagorean triple. (even because $2y+2$ is even and odd x will yield odd x^{2})

x | y | (a,b,c)

---

2 | 1 | (3,4,5)

4 | 7 | (15,8,17)

6 | 17| (35,12,37)

8 | 31| (63,16,65)

etc...

Answer 2

How about $a=3$, $b=4$, $c=5$?

Edit: Sorry, thought you just needed one. Since $c = a+2$, you can write $$ a^2 + b^2 = (a+2)^2$$ which simplifies to $$ b^2 = 4(a+1) $$ Thus pick $a := x^2 - 1$ for some $x \in \mathbb{Z}$, $x \geq 2$. This gives you $b = 2x$ and $c = x^2 + 1$.

$x=2$ gives $(3,4,5)$

$x=3$ yields $(6,8,10)$ which is not primitive, but $x=4$ works, giving $(15,8,17)$.

Answer 3

A simple google search yields multiple results:

a= 99, b = 20, c = 101