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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, 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.
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.
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 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 this Math Help site is Graeme McRae.