Associated Topics || Dr. Math Home || Search Dr. Math

### 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.

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!
```
Associated Topics:
College Number Theory

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search