No Integer Solution

Date: 10/21/2001 at 03:51:21
From: Tomas Prochazka
Subject: z^2-2y^2=51

I don't know how to prove that the following equation has no solution 
in Z (integers):

   z^2 - 2y^2 = 51

Another equation was

   z^2 - 4y^2 = 51
 (z-2y)(z+2y) = 51

and both () must be integers and 51 = 1*3*17 =>
z-2y = 3
z+2y = 17
...

This is easy, but what to do with the first one? I think that
(z-2^(1/2)y)(z+2^(1/2)y) cannot be used because brackets aren't 
integers, so factoring 51 won't help.

Thanks, 
Tom

Date: 10/21/2001 at 05:51:08
From: Doctor Mitteldorf
Subject: Re: z^2-2y^2=51

Dear Tom,

There is, actually at least one pair of squares with a difference of 
51, and it's not hard to find. If you write a list of squares,
1 4 9 16 25 36 ...  you'll notice that the difference between 
successive squares is a complete series of odd integers:
3 5 7 9 11 ...

How far will you have to go until you have two successive squares have 
a difference of 51?

There's one more pair that you'll find pretty easily by trial-and-
error: 10 and 7. Could you have predicted this from an equation?  How 
about trying (x+3)^2 - x^2 = 51?

Are there any others? What happens if you write (x+5)^2 - x^2 = 51?

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

Date: 10/21/2001 at 05:59:17
From: Tomas Prochazka
Subject: Re: z^2-2y^2=51

Differences between squares are odd numbers; when you compare x^2 
and (x+1)^2 the difference is 2x+1, which is an odd number; if it is 
x^2 - y^2 = 51 then you're right, and it is very simple.

But I have x^2 - 2*y^2 = 51 and y is 2*, so I think it is more 
difficult. I tried to write a program which tries numbers from -1000 
to 1000 and it didn't find any solution.

Tomas Prochazka

Date: 10/21/2001 at 07:18:40
From: Doctor Mitteldorf
Subject: Re: z^2-2y^2=51

Tom -

Just now in the shower, I was thinking about an interesting
generalization of the question you raised. Here's a challenge for you:

We just saw that 51 can be expressed as the difference of two squares 
in two different ways. What's the smallest number that can be 
expressed as the difference of two squares in three different ways?

The kind of algebraic thinking we've been doing should put us on a 
fast track to an answer. Will you write to me with yours?
- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/   

Date: 10/21/2001 at 07:43:26
From: Tomas Prochazka
Subject: Re: z^2-2y^2=51

Dr. Mitteldorf,

I tried to find a number that is the difference of 2 roots whose 
smallest difference (x-n)^2 - x^2 where n is smallest in three ways 
and I think 45 can be the result:

(x+1)^2-x^2 = 2x+1
(x+3)^2-x^2 = 6x+9
(x+5)^2-x^2 = 10x+25
(x+n)^2-x^2 = 2nx+n^2

so that number must be odd (2x+1) and after dividing it by 6 the 
remainder must be 3; the first two numbers that fulfill this 
requirement are 15 and 45. But for 15 n = -1 and n = 1, which gains 
the same result twice (4^2 - 1^2 = 15), so 45 won:

23^2-22^2 = 45
  9^2-6^2 = 45
  7^2-2^2 = 45

but I still don't know what to do with:

x^2 - 2*y^2 = 51

Tom

Date: 10/23/2001 at 16:40:18
From: Doctor Mitteldorf
Subject: Re: z^2-2y^2=51

Dear Tomas,

I'm sorry - I misread your question. Thanks for writing back. Your
analysis of the challenges that I posed was very smart, and I think 
you are right, that 45 is the smallest number that can be expressed as 
the difference of two squares in two different ways.

The case of x^2 - 2y^2 = 51 is more difficult. I think I have a proof
from mod 3 arithmetic.

What is x mod 3?

Suppose x mod 3 = 0. Then y mod 3 must also be 0, so that the
difference x^2 - 2y^2 is also divisible by 3. But this cannot be, 
because x^2 and y^2 would both be divisible by 9; hence x^2 - 2y^2 is 
divisible by 9, but 51 is not.

Suppose x mod 3 = 1. Then x^2 mod 3 = 1, and 2y^2 mod 3 must = 1, so
y^2 mod 3 must = 2. But 2 is not a square mod 3. There is no way to
satisfy y^2 mod 3 = 2.

Suppose x mod 3 = 2. Then x^2 mod 3 = 1, and the same argument 
applies.

There are no other cases.

Please check this proof for me - I think it is right, but I am not 
sure. 

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

Date: 10/24/2001 at 17:55:34
From: Tomas Prochazka
Subject: Re: z^2-2y^2=51

I think you're right. I tried to write a program that checked [X,Y] 
from <-1000;1000>x<-1000;1000> and no two numbers were solution.

Tom