Arithmetic Progression Proof

Date: 07/19/2006 at 04:59:21
From: sheryl
Subject: can you show me the proof please..

Can you please show me the proof that the product of four positive
integers in an arithmetic progression cannot be the square of an
integer?  Thanks!

Date: 07/19/2006 at 14:46:20
From: Doctor Vogler
Subject: Re: can you show me the proof please..

Hi Sheryl,

Thanks for writing to Dr. Math.  I don't know of an easy way to prove 
this, so I will sketch the proof that I can manage, and you are 
welcome to ask for more details of any part that is confusing.  But I 
will deliberately leave out many details so that you can work through 
them yourself.

Let's suppose that you have four positive integers in an arithmetic 
progression,

  x, x+y, x+2y, x+3y,

and their product is a square.  You'd like to prove that this is 
impossible.  First of all, explain why you can assume that x and y 
are relatively prime (i.e. gcd(x,y) = 1).

Next, we'd like to say that if x and y are relatively prime and

  x(x+y)(x+2y)(x+3y)

is a square, then each of the factors x, x+y, x+2y, x+3y, are 
squares.  This is not true.  However, you can prove that

  gcd(x, x+y) = 1
  gcd(x+y, x+2y) = 1
  gcd(x+2y, x+3y) = 1
  gcd(x, x+2y) = 1 or 2
  gcd(x+y, x+3y) = 1 or 2
  gcd(x, x+3y) = 1 or 3.

Furthermore, gcd(x, x+2y) and gcd(x+y, x+3y) can't both be 2.  Now 
explain why this implies that there are integers a, b, c, and d 
(which are pairwise relatively prime) such that one of the following 
must be true:

  (1) x =  a^2, x+y =  b^2, x+2y =  c^2, x+3y =  d^2
  (2) x =  a^2, x+y = 2b^2, x+2y =  c^2, x+3y = 2d^2
  (3) x = 2a^2, x+y =  b^2, x+2y = 2c^2, x+3y =  d^2
  (4) x = 3a^2, x+y =  b^2, x+2y =  c^2, x+3y = 3d^2
  (5) x = 3a^2, x+y = 2b^2, x+2y =  c^2, x+3y = 6d^2
  (6) x = 6a^2, x+y =  b^2, x+2y = 2c^2, x+3y = 3d^2

You'll see this same technique also used in a different problem here:

  Consecutive Integer Proof
    http://mathforum.org/library/drmath/view/65589.html 

Now, three of these cases are easy to eliminate using modular 
arithmetic.  For example, in (3), you have

  3b^2 - 4a^2 = d^2.

Considering this equation mod 3, you'll find that the only solutions 
have a and d both divisible by 3.  But then 3b^2 is divisible by 9, 
so b is also divisible by 3.  This means that both x and y are 
divisible by 3, contrary to our hypothesis.  You can use a similar 
argument to rule out (2) and (4).

Cases (5) and (6) are actually very similar, by reversing the order of 
{a, b, c, d}, so I'll just treat them as one.  That means that we 
still need to worry about cases (1) and (5).

Well, we can't eliminate either of them by modular arithmetic, because
they both have solutions.  The quadruple

  (1, 1, 1, 1)

solves (1), and

  (1, 1, 1, 0)

solves (5).

Well, in each case, you have an intersection of two quadratic surfaces
with a rational point, and this is equivalent to an elliptic curve (or
some would say that it *is* an elliptic curve).

For example, the equation

  a^2 = b^2 - d^2

implies that

  a^2 = (b - d)(b + d)

and since b and d are relatively prime, that means that either

  b + d = r^2, b - d = s^2, a = rs

or

  b + d = 2r^2, b - d = 2s^2, a = 2rs.

The other equation defining (5) is

  c^2 = b^2 + 3d^2.

Substituting our formulas, we get either

  (c/2)^2 = r^4 - r^2*s^2 + s^4,

which I suspect you'll find very familiar, or

  c^2 = 2(r^4 - r^2*s^2 + s^4),

which is easy to see has no solutions (recall that r and s can't both 
be even, and conclude that c^2 is divisible by 2 but not 4).

The familiar equation has (as we determined previously) eight families
of solutions, all of which give either d = 0 or a = 0, and these do
not give positive numbers in your arithmetic sequence.

You can do a similar analysis on case (1), and I think you'll find 
that a^2 = b^2 = c^2 = d^2, which means that the only solutions to 
your problem are from the constant arithmetic sequence where y = 0.

If you have any questions about this or need more help, please write 
back and show me what you have been able to do, and I will try to 
offer further suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 07/20/2006 at 05:57:10
From: sheryl
Subject: can you show me the proof please..

Thank you Dr. Math for a very impressive answer to my question.  It
helped me a lot!