Question

Pythagorean triples are given by the formulas x2 - y2, 2xy, and x2 + y2. Use the formulas for the Pythagorean triples to find a right triangle with leg lengths of 16 and an odd number. Show all of your work for full credit.

Answer 1

First of all, let me break down the formula: $x^2-y^2,\ 2xy,\ x^2+y^2$ is always a Pythagorean triple because you have

$(x^2-y^2)^2+(2xy)^2 = x^4-2x^2y^2+y^4+4x^2y^2 = x^4+2x^2y^2+y^4 = (x^2+y^2)^2$

So, for any $x,y>0$ you can choose (let's suppose $x>y$), these three numbers will always fall in the form $a^2+b^2=c^2$, which is also the rule that works for right triangles. So, every time you choose two numbers $x,y>0$, the legs will be $x^2-y^2$ and $2xy$, while the hypotenuse will be $x^2+y^2$.

We have to find a right triangles with legs 16 and an odd number. Well, the legs in the triple we are given are $x^2-y^2$ and $2xy$, so we want one of this to be 16, and the other to be odd. But $2xy$ can't be odd, because it has a 2 factor in it. So, it must be 16: we have

$2xy=16 \iff xy=8$

The only ways we can choose two numbers $x>y>0$ such that their product is 8 are:

$x=8,\ y=1,\quad x=4,\ y=2$

In the first case, the legs are

$x^2-y^2 = 64-1 = 63,\ 2xy = 16$

In the second case, the legs are

$x^2-y^2 = 16-4 = 12,\ 2xy = 16$

In this case both legs are even, so the only good choice is $x=8,\ y=1$

So, the triangle we're working with has legs

$x^2-y^2 = 64-1 = 63,\ 2xy = 16$

and hypotenuse

$x^2+y^2 = 64+1 = 65$