3 consecutive terms of AP that are perfect square

Well-known member
find parametetric representation of 3 perfect squares which are successive terms in AP ($x^2$,$y^2$,$z^2$) such that $x^2,y^2,z^2$ are successive terms of AP. find x,y,z

Well-known member
find parametetric representation of 3 perfect squares which are successive terms in AP ($x^2$,$y^2$,$z^2$) such that $x^2,y^2,z^2$ are successive terms of AP. find x,y,z
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

Well-known member
It is proved that( I do not know the proof) we cannot have an AP whose 4 successive terms are in perfect squares
But does there exist an AP whose 3 consecutive terms are perfect squares

Solution:

Let the 3 consecutive term be $a^2,b^2,c^2$

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

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