# Pythagorean Triples II

• Nov 1st 2010, 05:20 PM
MATNTRNG
Pythagorean Triples II
Find all the Pythagorean triples that form an arithmetic sequence.
• Nov 1st 2010, 06:02 PM
Drexel28
Quote:

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.
• Nov 1st 2010, 09:23 PM
Soroban
Hello, MATNTRNG!

Quote:

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.
• Nov 2nd 2010, 07:13 AM
melese
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 $3, 4, 5$.