2011/012) To construct an AP of integers so that 3 successive elements are perfect squares.

It is proved that 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