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#### kaliprasad

##### Well-known member

- Mar 31, 2013

- 1,346

- Thread starter kaliprasad
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- Thread starter
- #1

- Mar 31, 2013

- 1,346

- Thread starter
- #2

- Mar 31, 2013

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No ans yet. The quesion may be boring.

I shall give a hint

if (x,y,z) is a Pythagorean triplet then $(x-y)^2$, $z^2$ and $(x + y)^2$ are in AP

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- Mar 31, 2013

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But does there exist an AP whose 3 consecutive terms are perfect squares

Solution:

As they are in AP we have

$b^2-a^2 = c^2-b^2$

or $a^2+c^2 = 2b^2$

this has a solution and we know that

if $x^2 + y^2 = z^2 $

then $(x+y)^2 + (x-y)^2 = 2(x^2+y^2) = 2z^2$

so if (x,y,z) is a Pythagorean triplet the $(x-y)^2 ,\, z^2, \, (x+y)^2$ are perfect squares and are in AP.

Or

$a= x-y$

$ b = z^2$

$v= x+ y$

for example

(3,4,5) is Pythagorean triplet so $(4-3)^2,\, 5^2,\,(4+3)^2$ or 1,25,49

(5,12,13) Pythagorean triplet so $(12-5)^2,\, 13^2,\,(12+5)^2$ or 49,169,289

Parametric form of Pythagorean triplet is

$(m^2-n^2),\, (2mn),\, m^2 + n^2$

So Parametric form of the required AP is

$ (m^2-n^2-2mn)^2,\,(m^2+n^2)^2,\,(m^2-n^2+2mn)^2 $