# Homework Help: No natural numbers that satisfy x^2 - y^2 = 2

1. Mar 22, 2012

### lisamane

I'm trying to prove that there are no natural numbers x and y that satisfy the equation x^2 - y^2 = 2.

I tried to solve it by contradiction and so I assume that x and y are rational numbers and both x and y can be written in the form (a/b) where it's in its simplest form and a and b are both integers and b isn't equal to zero.

Therefore let x = (a/b) and y = (c/d)

Therefore (a^2/b^2) - (c^2/d^2) = 2

Then multiply by (b^2)(d^2)

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

I don't know how to go further from here or if I am even on the right track.

2. Mar 22, 2012

### epenguin

I'd say you are not on the right track because you are trying to do it for rational numbers and the question was about natural numbers. (Which are the ornery numbers 1, 2, 3, and you probably know others. )

3. Mar 22, 2012

### Dick

I would think about the factored form (x-y)(x+y)=2.

4. Mar 22, 2012

### HallsofIvy

"Ornery" numbers?

5. Mar 22, 2012

### lisamane

So using (x-y)(x+y) = 2 I thought of:

if x and y are equal then left hand side is equal to zero (contradiction)

if y is greater than x
I thought that the left hand side would always be negative but this won't be the case as say
x = -3 and y = 1 , then left hand side is equal to 8 which is not negative, so I can't really use this case.

Is this more on the correct track?

6. Mar 22, 2012

### Dick

Sure. But if x and y are natural numbers then x>0 and y>0 and they are integers. If (x+y)>2 then why can't that work? So the only possibility would be x=1 and y=1. Does that work?

Last edited: Mar 22, 2012
7. Mar 22, 2012

### epenguin

Let me say again I think you need to look up the definition of natural numbers, -3 isn't a natural number. http://en.wikipedia.org/wiki/Natural_number

I don't think you need to use the factorisation. To get 2 you need x>y and from there it is very simple.

8. Mar 22, 2012

### epenguin

I have used them in many unpublished calculations.

9. Mar 23, 2012

### lisamane

Bah, I forgot that natural numbers aren't negative. Would you include zero as a natural?

So from (x-y)(x+y) = 2,

Case 1: ( x = y )
Left side = (0)(x.x) = 0 ≠ 2

Case 2: ( y > x )
Left side is negative integer multiplied by positive integer which is equal to a negative integer which ≠ 2

Case 3: ( x > y )
This is what I'm stuck on but I tried doing it like this.

When both x and y are less than or equal to 2:
y = 1, x = 2: Left side = (1)(3) = 3 ≠ 2
When both greater than 2:
then let x = p + 2 and y = q + 2
(p - q)(p + q + 4) = p.p - p.q + p.q - q.q + 4p + 4q
= p.p - q.q + 4(p+q)
I was trying to show that when they are both greater than two than the result will be greater than 2 but I dont know how to.
These are the only three possible cases?

10. Mar 23, 2012

### vela

Staff Emeritus

11. Mar 23, 2012

### epenguin

I say again forget the factorisation.

Let y = x + a where a>1

Later, once you have got it for natural numbers you may well be able to state and prove something for a wider class of numbers - that is what Polya in 'How to Solve It' would recommend.

12. Mar 23, 2012

### lisamane

I'm sorry, I don't really know how to go about anyone's suggestions so continuing with Case 3 I thought since the last is p^2 - q^2 + 4(p+q) then from the equation p^2 - y^2 = 2 so the equation becomes 2 + 4(p + q) and therefore since all positive, all are greater than two unless p and q. Doesn't that work?

13. Mar 23, 2012

### epenguin

Except that when I use American expressions I have certainly heard they seem to be not understood it was in my mind all along to say 'Be a country girl!' meaning I think you are being unnecessarily complicated.

Taking 1 step of my last suggestion you really can't do anything with

(x + a)2 - x2 = 2 ?

14. Mar 23, 2012

### lisamane

Earlier you said let y = x + a which if you put in the equation it should be

x2 - (x + a)2 = 2

How did it come to (x + a)2 - x2 = 2 ?

15. Mar 23, 2012

### epenguin

A so I did, well I have done more than hint so you will have to supply what I should have said that will make it come to something like that, get the idea, nothing complicated.

16. Mar 23, 2012

### checkitagain

As an alternative, I would show that it isn't true for any integers, and then

(The following method is not meant to be shorter than others already mentioned.)

You could prove by contradiction each of four cases to complete it:

$$1st \ case: \ \ \ \ x \ \ even, \ \ y \ \ even$$

$$2nd \ case: \ \ \ \ x \ \ even, \ \ y \ \ odd$$

$$3rd \ case: \ \ \ \ x \ \ odd, \ \ y \ \ even$$

$$4th \ case: \ \ \ \ x \ \ odd, \ \ y \ \ odd$$

-----------------------------------------------------

For example, let m, n belong to the set of integers

For instance, the second case could be:

Then x = 2m and y = 2n + 1.

Substitute these into the original equation and see if
you can arrive at a contradiction.

17. Mar 24, 2012

### JG89

Look at it this way: $x^2 - y^2 = (x-y)(x+y)$, right? So suppose that x and y ARE two natural numbers whose product is 2, then we have $(x-y)(x+y) = 2$. There are only two natural numbers whose product is 2, namely the numbers 1 and 2.

So either $(x-y) = 1$ and $(x+y) = 2$ OR $(x-y) = 2$ and $(x+y) = 1$.

Solve both those systems for x and y to get a contradiction.