This page gives you the background you will need to understand the proofs that there aren't 4 squares in arithmetic sequence. You could skip this, and go directly to Proof 1 or Proof 2, but then you would miss out on some really fun facts about Pythagorean Triples, their "generators", and arithmetic sequences of three squares, each of which can generate the others. (That is, for each Pythagorean Triple, there is a corresponding arithmetic sequence of three squares, and a pair of relatively small numbers r,s that I call the "generator" of both the triple and the sequence.)
If you decide to skip this preamble, and then get stuck reading one of the proofs, you can always come on back here for a quick refresher.
GCD is the Greatest Common Divisor (sometimes called the GCF, where F stands for Factor). Normally, you think of the greatest common divisor as being the biggest divisor. But in Number Theory, the term "greatest" has a special meaning, which I'll illustrate with the official definition of GCD:
c=GCD(a,b) iff c divides a and c divides b and any number, d, that divides both a and b also divides c
I should point out that "divides", too, has a special meaning: "a divides b", usually written a|b means
there exists an integer, k, such that ak=b. It's counterintuitive that 11 divides 0, but it does because 11k=0 has an integer solution. It's even more outrageous that 0 divides 0, but, again, 0k=0 has an integer solution (lots, in fact).
Since GCD is such a basic building block of Number Theory, sometimes it ("GCD", that is) goes without saying -- that is, the expression
(a,b) is sometimes written to mean GCD(a,b). Don't worry though -- I won't do that to you. I'm just telling you in case you go surfing the 'net and find this somewhere.
I should point out that negative numbers are fair game, although their GCD is always taken to be nonnegative. GCD(-6,15)=3 because 3|-6, and 3|15, and any number, d, that divides both -6 and 15 also divides 3. You see, all the other common divisors of -6 and 15 are "smaller" than 3 in the sense that they divide 3.
One more crazy example: It's not common to consider GCD of some number and zero, but
GCD(11,0)=11 because 11|11, and 11|0, and any number, d, that divides both 11 and 0 also divides 11.
There's a special word for the relationship between two numbers a and b such that GCD(a,b)=1. These numbers are said to be "coprime", which means sharing no common factors other than 1. Note that 1 is coprime to every number including itself!
The Pythagorean Triple Generators
Whenever you have a Primitive Pythagorean Triple (a,b,c), where, WLOG, c is biggest and b is even, it's always possible to find two numbers r,s of opposite parity (one odd, the other even) and coprime such that
a = r² - s²
b = 2rs
c = r² + s²
I call these numbers r,s the "generators" of the Pythagorean Triple. It's interesting to note that there is a bijection (an exact pairing) from generators (r,s) to primitive triples. That is, for any pair of generators there is exactly one primitive triple, and for any primitive triple, there is exactly one pair of generators.
Maybe that's a lot to throw out all at once "Primitive" means GCD(a,b,c)=1, and it turns out that a,b,c are also pairwise coprime. WLOG means "without loss of generality". We're saying that (3,4,5) and (4,3,5) are really just one triple, and we'll write it with the even number in the middle, and the biggest number last. The three numbers can't all be even because the triple is primitive. So at least one of them has to be odd. Parity tells you that a or b is odd, but not both -- to see this, carry out the calculation a²+b²=c² mod 4, and since each square has to be equivalent to 1 or 0, you know that c must also be odd. WLOG, we always put the even element of the triple in the middle.
Note 1: This convention of putting the even number in the middle is just my approach; it's not necessary to list Pythagorean Triples this way, and most people don't bother. It's just me.
Note 2: My way doesn't always list the three numbers in ascending order. For example,
(15,8,17) is a primitive triple listed with the even number in the middle, and its elements aren't in ascending order.
At this point, if you're lost, you must do this: start over from the top of this page, and understand everything up to this point. Or else give up. These are your only two choices. There's an email link at the bottom of the page, where you can ask me for help or, as an alternative, you can tell me what a pompous ass I am. I enjoy both kinds of email. So I guess there are more than two choices after all.
Good, you're reading on. That means you understand the fact that a Primitive Pythagorean Triple is composed of three elements, and my way of listing them gives the first as the difference of two squares, the second as twice the product of two numbers, and the third as the the sum of the same two squares. For example, (15,8,17) is (4²-1², 2*4*1, 4²+1²), so r and s are 4 and 1.
It's easy to find r and s, given the primitive triple (a,b,c)
r² = (c + a)/2,
s² = (c - a)/2,
rs = b/2.
Often, you'll see that a proof glosses over the method of finding r and s, and just says that given the triple (a,b,c) there exist r,s such that blah blah blah... -- just so you know, it's not hard to find r and s.
The last thing I need to say in this section is that wonderful things happen when a, b, and c are themselves perfect squares. If a,b,c are perfect squares, then let a=A², b=B², and c=C². Then r and s can be found such that
A² = r² - s²
B² = 2rs
C² = r² + s²
which sets up all sorts of new triples: (A,s,r), and (r,s,C) which are both smaller than the original triple (A²,B²,C²).
This kind of thing (finding smaller numbers with a given property) sets up the "Infinite Descent" proof, in which you first assume that a number with a certain property is minimal, and then you find another number with the same property that's smaller than the first number.
Relationship between 3-element arithmetic progressions of squares and Pythagorean Triples
You may be excited to learn that it's easy to find a 3-element arithmetic progression of perfect squares. In fact, there's a bijection from the set of such 3-element progressions to Pythagorean Triples. In this section I'll show you how to create either one from the other.
Let (a,b,c) be a triple. Since a and b uniquely determine c, there is only one triple with a given value of a and a given value of b. Consider these three values:
These three values are in arithmetic progression because their constant difference is 2ab. Also, since (a,b,c) is a triple, all three are squares: The first number is (a-b)²; the second is c²; and the third is (a+b)². So every triple determines a unique arithmetic sequence of three squares.
Now, let's do the reverse.
Let x², y², z² be in arithmetic sequence.
((z+x)/2)² + ((z-x)/2)² =
(z²+2zx+x²)/4 + (z²-2zx+x²)/4 =
z²/2 + x²/2 =
Since ((z+x)/2)² + ((z-x)/2)² = y²,
( (z-x)/2, (z+x)/2, y )
is a Pythagorean Triple, with the first two numbers in some order. (I say "some order" because the sum of the first two numbers in the triple is z, an odd number. That means one of them is odd and the other is even, but I don't know which is which.) The numbers are integers, because x and z are either both odd or both even.
Moreover, it's easy to check that if the triple is primitive, then the arithmetic sequence will be primitive (pairwise coprime) and odd; and if the arithmetic sequence is primitive, then the triple is primitive.
The Generators of an Arithmetic Sequence of Three Squares
Finally, given any arithmetic sequence of three squares, x²,y²,z², you can get all the way back to the generators, r and s. This is interesting, to me, anyway, because when you go from the sequence of squares to the triple, the numbers get smaller, and when you go from the triple to the generators, the numbers get smaller still. You can also go in the other direction: given any pair of coprime opposite-parity generators, you can get a triple, and from there, you can get the sequence of three squares.
Let x², y², z² be a primitive arithmetic sequence.
( (z-x)/2, (z+x)/2, y )
is a primitive triple, with the first two numbers in some order. The generators can be found this way:
r² = (y+(z-x)/2)/2
s² = (y-(z-x)/2)/2
-- OR --
r² = (y+(z+x)/2)/2
s² = (y-(z+x)/2)/2
(one set of equations will give perfect squares)
Going the other way is a lot easier: Given coprime, opposite parity, integers r,s (with r>s), the arithmetic sequence is
It is instructive to multiply these expressions out:
r4 - 4r3s + 2r2s2 + 4rs3 + s4,
r4 + 2r2s2 + s4,
r4 + 4r3s + 2r2s2 - 4rs3 + s4
By expressing the arithmetic sequence of squares this way, you can clearly
see that the common difference in the arithmetic sequence is
4r3s-4rs3. This factors as 4rs(r+s)(r-s). It is easy to see, now, that this common difference is a multiple of 24, because either r or s is even, so 4rs is a multiple of 8, and at least one (exactly one, it turns out) of r, s, r+s, r-s is a multiple of 3.
Well, now that you've read this preamble, I'm sure you're just itching to get on with the proofs. Here they are:
Arithmetic Sequence of Squares -- a series of rambling discussions about the possible existence of an arithmetic sequence of four squares
The webmaster and author of this Math Help site is Graeme McRae.