An algebra problem by atishay jain

Algebra Level 2

True or False:

Any positive integer divisible by 4 is part of a pythagorean triplet whose values form an arithmetic progression.

True False

2 solutions

Jon Haussmann
Dec 13, 2017

Take $3n$ , $4n$ , $5n$ .

Atishay Jain
Dec 10, 2017

Let say a is any natural number divisible by 4, $a = 4d$

$a^2 = 4ad$ (where a can't be 0)

$a^2 + d^2 - 2ad + a^2 = a^2 + d^2 +2ad$

$(a+d)^2 = (a-d)^2 + a^2$

Now, Here a - d, a & a + d are in A.P. as well as are pythagorean triplet

Your solution is incomplete/wrong. Current version only shows that if $(a,b,c)$ is a pythagorean triple with $a<b<c$ and $a, b$ and $c$ are in arithmetic progression, then $b$ is a multiple of $4$ .

You are supposed to show the converse.

Muhammad Rasel Parvej - 3 years, 6 months ago

We can solve it in a reverse manner like