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What's the Longest Arithmetic Progression of Perfect Squares?

An arithmetic progression is a sequence of numbers in which the difference between any element and its predecessor is constant.

An infinite arithmetic progression of distinct integers cannot consist entirely of perfect squares.


Let element ai of the arithmetic progression be given by bi+c, where b and c are integers.

If b were zero, the elements would not be distinct, and if b were negative, some elements would be negative (and thus not perfect squares) so b>0. This is an increasing arithmetic progression.

Let an be an element of the progression that is greater than or equal to b².

an is a perfect square, so an integer, d, exists such that an=d²

an >= b² so d >= b.

an+1 = d²+b < d²+2d+1 = (d+1)²

So now we have an+1 greater than d² but less than (d+1)², so an+1 could not be a perfect square.

So infinite arithmetic progressions of perfect squares don't exist.  What about finite ones?

{1, 25, 49} is a 3-element arithmetic progression.  Let's look at what makes it tick.

25-1 = 24, and 49-25 = 24.  These are the differences of squares.  So

(5-1)(5+1) = 24, and (7-5)(7+5) = 24

From this we see there are three integers, a, b, and c such that

(b-a)(b+a) = (c-b)(c+b)

I wrote a computer program (an excel spreadsheet, actually) that looks for such triplets.  It found 67 sequences of three elements whose constant difference was 10,000 or less.  The first few are 

{1�, 5�, 7�}, which have a constant difference of 24
{2�, 10�, 14�}, which have a constant difference of 96
{7�, 13�, 17�}, which have a constant difference of 120
{3�, 15�, 21�}, which have a constant difference of 216
{7�, 17�, 23�}, which have a constant difference of 240

Some pairs of sequences have the same constant difference, but they can't be combined to make one 4-element sequence.  Some examples are

{1�, 29�, 41�}, 
{23�, 37�, 47�}, which both have a constant difference of 840

and even a triplet,

{2�, 58�, 82�}, 
{46�, 74�, 94�}, 
{97�, 113�, 127�}, which all have a constant difference of 3360

The last few with constant differences less than or equal to 10,000 are

{142�, 170�, 194�}, which have a constant difference of 8736
{199�, 221�, 241�}, which have a constant difference of 9240
{7�, 97�, 137�}, which have a constant difference of 9360
{20�, 100�, 140�}, which have a constant difference of 9600
{63�, 117�, 153�}, which have a constant difference of 9720

But it found no sequences of four elements.  Not one.  I even enhanced the program to check for differences of up to 100,000,000, and found 17883 3-element sequences, but still no sequences of four elements.

Here's another interesting pattern I notice when glancing at the thousands of 3-element sequences produced by my program:  The constant differences are all multiples of 24.  Why should that be?  I imagine that if I explore all the various ways the constant differences can be expressed, I'll see why they must contain two factors of 2 and one factor of 3.  So let's start looking at facts about the constant difference between these squares.

If the arithmetic progression consists of {a, b, c} then it's true that b�-a� = c�-b�.

This also means that (b-a)(b+a) = (c-b)(c+b), and that equals (c-a)(c+a)/2.  Do I see the factor of 24 here?  No, frankly, I don't.

But the differences of squares are also the sum of arithmetic progressions of odd numbers.  For example, 5�-1� = 3+5+7+9, and 7�-5� = 11+13.

The sum of m consecutive odd numbers starting with n is m(n+m-1).  You can derive that by adding up the series n + n+2 + ... + n+2m-2 to the same series written backwards, then dividing by two.

The sum of the next p consecutive odd numbers starting where the first sequence left off is p(n+2m+p-1).  So we have


May be this is enough to find a factor of 24 on both sides of this equation.  Then after that, I'll add the mystery fourth element of the arithmetic progression, and show that it can't exist.

Continue the Exploration of Arithmetic Sequences of Squares

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