# Thread: In the right triangles with sides...

1. ## In the right triangles with sides...

In the right triangles with sides (3,4,5) and (5,12,13), one side is just 1 less than the hypotenuse. Find all such integer-sided right triangles. [If you cannot do that, at least try to determine whether or not there are infinitely many of them.]

Thanks!

2. Originally Posted by NikoBellic
In the right triangles with sides (3,4,5) and (5,12,13), one side is just 1 less than the hypotenuse. Find all such integer-sided right triangles. [If you cannot do that, at least try to determine whether or not there are infinitely many of them.]

Thanks!
Since $\displaystyle c$ is odd and $\displaystyle c-b=1$, this implies that $\displaystyle b$ is even, so $\displaystyle b=2mn$ and we know $\displaystyle c=m^2+n^2$.

So $\displaystyle c-b = m^2-2mn+n^2 = (m-n)^2=1\implies m=n+1$.

Thus $\displaystyle a=m^2-n^2 = (n+1)^2-n^2 = 2n+1$

$\displaystyle b=2mn = 2(n+1)n = 2n^2+2n$

$\displaystyle c=m^2+n^2 = (n+1)^2+n^2 = 2n^2+2n+1$ where $\displaystyle n$ is arbitrary.

3. Hello, NikoBellic!

In the right triangles with sides (3,4,5) and (5,12,13),
. . one side is just 1 less than the hypotenuse.
Find all such integer-sided right triangles.

Pythagorean triples are generated by: .$\displaystyle \begin{Bmatrix}a &=& m^2-n^2 \\ b &=& 2mn \\ c &=& m^2+n^2 \end{Bmatrix}$ . where: .$\displaystyle m,n \in I^+\:\text{ and }\,m > n.$

Let $\displaystyle m \:=\:n+1 \qquad\text{Then: }\:\begin{Bmatrix}a &=& 2n+1 \\ b &=& 2n^2+2n \\ c &=& 2n^2+2n+1 \end{Bmatrix}$

And we have: .$\displaystyle (3,4,5),\;(5,12,13),\;(7,24,25),\;(9,40,41),\;(11, 60,61),\;\hdots$