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### 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
```
Associated Topics:
High School Number Theory

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