Squares and Arithmetic Progressions

Squares and Arithmetic Progressions
Observe the Squares 1², 5², 7².. these numbers viz . . 1, 25 and 49 are in an Arithmetic progression. So one wonders if there are more such examples. Consider one more such triple 7², 13² and 17²… looks like there can be infinitely many. We shall try proving such a statement .. and what if we look for 4 such squares?

let a², b²,  and c² be in arithmetic progression so that b²-a²=c²-b²,

or 1-(a/b)²= (c/b)²-1.

For convenience we write x=a/b, y=c/b.

This implies 1-x²= y²-1. Factorising we get

(1+x)/(y+1)=(y-1)/(1-x).........(1)

let t be the common value of (1) which gives us

1+x=t(y+1)

 AND y-1=t(1-x)

OR

x-ty=t-1.................[2]

tx+y=t+1..............[3]

solving for x and y we get the following parametrization

x=(t²+2t-1)/(t²+1)

y=(-t²+2t+1)/(t²+1).
For t=1/2 we get x=1/5, y=7/5 so that we get 1, 1/25 and 49/25. Clearing denominators we get 1, 25 and 49...the triple we started with. 

Similarly taking some other value for 't' we can arrive at other triples of squares that form an A.P. For instance t=2/3 leads us to 49, 169 and 289..the common difference being 120. The parametrization we have arrived at implies that we can get an infinitude of such triples!!.

Next we ask the question if there are 4 such square numbers which form an A.P. This question was asked by Fermat and the answer is NO. One cannt get hold of 4 squares satisfying such a property. The proof for this assertion was first given by Leonhard Euler!!

ADAPTED FROM THE BOOK-Numbers and the beginning of Algebra By Shailesh Shirali [Little Mathematical Treasures]