# 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: .$\displaystyle (a,\:a+d,\:a+2d)$

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

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

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

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

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

Since $\displaystyle d = \frac{a}{3}$, the Pythagorean triples are: .$\displaystyle \left(a,\:\frac{4}{3}a,\:\frac{5}{3}a\right)$
. . where $\displaystyle \,a$ is any positive multiple of 3.
• Nov 2nd 2010, 07:13 AM
melese
A useful strategy would be to denote the sides $\displaystyle a-d, a,$ and $\displaystyle a+d$, where $\displaystyle a$ and $\displaystyle b$ are integers; these form an arithmetic sequence. We assume that $\displaystyle d>0$.

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

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