Arithmetic Sequences of Perfect Squares

Question 1: why does a 3-element arithmetic progression of perfect squares have a constant difference that is a multiple of 24?

In other words, if c^2 - b^2 = b^2 - a^2, a, b, c integers, why does 24|(b^2-a^2) ?

There are two approaches to answering this:

First Answer

First, suppose a^2, b^2, c^2 are in an arithmetic sequence whose constant difference isn't a multiple of 8.  Without loss of generality GCD(a,b) = GCD(b,c) = 1. Note that a and c can't both be even since then b^2 = (a^2 + c^2)/2 is even so a and b would share a factor of 2.  This means that at most one of a, b, c is even.

Now if x is an odd integer then it is easily checked that x^2 = 1 (mod 8) and if x is even then either x^2 = 0 (mod 8) or x^2 = 4 (mod 8).  Therefore working mod 8 the triple (a^2, b^2, c^2) must be one of:

(0,1,1), 
(4,1,1), 
(1,0,1), 
(1,4,1), 
(1,1,0), or 
(1,1,4)

It is clear that none of these are in arithmetic sequence, so this is a contradiction establishing that the common difference must be a multiple of 8.

Now if the arithmetic progression has common difference d = 1 (mod 3) then a^2, b^2, c^2 must each have distinct residues (mod 3); in particular one of them must be equal to -1 (mod 3) which is impossible.  Likewise if the common difference d = -1 (mod 3) then a^2, b^2, c^2 have distinct residues (mod 3) which leads to the same contradiction.  So we must have d = 0 (mod 3).

Therefore the common difference is a multiple of 8 and 3, so is a multiple of 24.

Thanks to Michael Doré for this answer.

Second Answer

This question of arithmetic sequences of squares is related to Pythagorean Triples.

Let {a^2, b^2, c^2} be in arithmetic sequence.

The difference c^2 - a^2 must be an even number because it's twice the difference b^2 - a^2.

So either c and a are both odd, or they're both even.  In either case, c-a is even and c+a is even.

Let n = (c-a)/2, and
let m = (c+a)/2.

Then a=m-n, so a^2 = m^2 - 2mn + n^2, and
c=m+n, so c^2 = m^2 + 2mn + n^2

b^2 = (a^2+c^2)/2, because a^2, b^2, c^2 are in arithmetic sequence.
b^2 = m^2 + n^2
So m, n, b are a Pythagorean Triplet.

The constant difference of the original arithmetic sequence, b^2-a^2, is equal to 2mn

In a Pythagorean Triplet it is always seen that either m or n is divisible by 3, and either m or n is divisible by 4.  So 2mn is divisible by 3 and it is divisible by 8, so it is divisible by 24.  This is proved in the "Interesting Facts" section of Pythagorean Triples

So here my second question remains: Are there any 4-element arithmetic progressions of perfect squares (I don't think so).  Some effort to understand this question is given on the next page.

Related pages in this website

Number Theory

Theorems Involving Perfect Squares -- answers questions such as "why is the square root of x irrational unless x is a perfect square?" and other fundamental questions about perfect squares.

Pythagorean Theorem

Pythagorean Triples -- derives the formula for finding them, and gives some interesting facts about them, such as this: of the two legs of the triangle, at least one is a multiple of three, and at least one is a multiple of four.

Prove that the area of a right triangle with integer sides is not a perfect square.

The webmaster and author of the Math Help site is Graeme McRae.
     [home]  [email]  [search]  [Links to Math Sites]  [Whiteboard]