Pythagorean Triples II

**MATNTRNG** · November 1st 2010, 05:20 PM

Find all the Pythagorean triples that form an arithmetic sequence.

**Drexel28** · November 1st 2010, 06:02 PM

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.

**Soroban** · November 1st 2010, 09:23 PM

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: . $\begin{Bmatrix}a+d &=& 0 & \Rightarrow & d &=& \text{-}a \\<br /> a - 3d &=& 0 & \Rightarrow & d &=& \frac{a}{3} \end{Bmatrix}$

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.

**melese** · November 2nd 2010, 07:13 AM

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 $3, 4, 5$ .

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