1. ## Pythagorean Triples II

Find all the Pythagorean triples that form an arithmetic sequence.

2. Originally Posted by MATNTRNG
Find all the Pythagorean triples that form an arithmetic sequence.
Is this a challenge problems? If not, let's see some work.

3. Hello, MATNTRNG!

Find all the Pythagorean triples that form an arithmetic sequence.

Assume that Pythagorean triples consist of positive integers.

We have: . $(a,\:a+d,\:a+2d)$

Hence: . $a^2 + (a+d)^2 \:=\:(a+2d)^2$

. . $a^2 + a^2 + 2ad + d^2 \:=\:a^2 + 4ad + 4d^2$

. . . . $a^2 -2ad - 3d^2 \:=\:0$

. . . . $(a+d)(a-3d) \:=\:0$

And we have: . $\begin{Bmatrix}a+d &=& 0 & \Rightarrow & d &=& \text{-}a \\
a - 3d &=& 0 & \Rightarrow & d &=& \frac{a}{3} \end{Bmatrix}$

Since $d = \frac{a}{3}$, the Pythagorean triples are: . $\left(a,\:\frac{4}{3}a,\:\frac{5}{3}a\right)$
. . where $\,a$ is any positive multiple of 3.

4. A useful strategy would be to denote the sides $a-d, a,$ and $a+d$, where $a$ and $b$ are integers; these form an arithmetic sequence. We assume that $d>0$.

The following relation holds: $(a-d)^2+a^2=(a+d)^2$. From $a^2-2ad+d^2+a^2=a^2+2ad+d^2$, we find that $a^2=4ad$. But $a>0$ since $a-d$ is one of the triangle's sides, so we can divide by $a$ to get $a=4d$.

Thus Pythagorean triples that form an arithmetic sequence are given by $3d, 4d, 5d$. In particular, this implies that the only Pythagorean triple that consists of consecutive postive integers is $3, 4, 5$.