# Integers. I need help on this one please?

1. Aug 26, 2008

### sutupidmath

Integers. I need help on this one please???

Hi, i am trying to analyze the following problem, but i am new at abstract algebra so i am not sure whether i am reasoning properly.
Problem:

Why are there no integers x and y with x^2-y^2=34.

Here is how i am reasoning:

x^2-y^2=(x-y)(x+y)=34. so we notice that x-y and x+y are both integral divisors (or factors )of 34.
We know that all factors of 34 are 1,2, 17,34, -1,-2,-17,-34.

This way we have to chose x-y and x+y in such a manner that their product to equal 34. such possibilities are

x-y=1 and x+y=34 , x-y=2 and x+y=17, etc. we take all the cases, i am not typing all of them, but none of these systems of equation has a solution for both x and y integers. so this way we conclude that there are no integers x and y such that x^2-y^2=34.

I want to know whether there is another more elegant way of showing this, rather than what i have performed here?

If you can give some suggestions i would appreciate it. But remember, all we have done so far is 10 pages, we havent even yet gone to common divisors. so don't use too advanced math tools.

2. Aug 26, 2008

### Ben Niehoff

Re: Integers. I need help on this one please???

One thing you can do is find the average of your two divisors:

$$\frac{(x-y) + (x+y)}{2} = x$$

So, are there any pairs of numbers whose product is 34 and whose average is an integer? The pairs (1, 34) and (2, 17) each have one odd number and one even number, so their averages can't be integers.

Alternatively, you can take the difference of your factors:

$$(x + y) - (x - y) = 2y$$

So, you need factors whose difference is an even number.

Either way, what you need are two factors of the same parity. If you look at the prime factorization of 34, you have 2*17; it should be clear that there is no way to split these factors into two groups such that each has the same parity.

This depends entirely on the fact that 34 has exactly one power of 2 in its prime factorization. If it were 2^2 * 17, then it could be split (2, 2*17). Or if it were 3*5*17, it could be split (3*5, 17), etc.

3. Aug 26, 2008

### mathwonk

Re: Integers. I need help on this one please???

do you know modular arithmetic? working mod 4, you are trying to solve x^2 - y^2 = 2. now there are only 2 squares, 0 and 1, and this means there is no solution mod 4. so there is no integer solution.

4. Aug 26, 2008

### snipez90

Re: Integers. I need help on this one please???

I don't really think that your approach is that inelegant. In fact much harder problems can be solved using your method, which is (perhaps informally) called the decomposition method. It's probably the most basic method for solving a diophantine equation. In this case the factorization is clear so all you had to do was examine the cases.

There are 8 systems of two equations, which could be listed quickly. Let's consider one of these systems. Let's choose x-y = 1 and x+y = 34. There are many parity arguments you can make, as Ben Niehoff mentioned. You could add the two systems to get 2x = 35. The LHS is divisible by 2 but the RHS is not $$\Rightarrow$$ no solutions. Or we could argue like this. Since x+y = 34, x and y must both be even or both be odd, otherwise, our sum would be odd. But x and y can't both be even or both be odd, or else the difference x - y is even $$\Rightarrow$$ no solutions.

By similar arguments, it's pretty quick to go through all 8 systems. Of course as mathwonk suggested, modular arithmetic is another viable approach but I guess it all depends on how familiar you are with number theory.

Anyways, here is a much more complicated problem that could be solved using the decomposition method.

Find all integral solutions to the equation

$$(x^2+1)(y^2+1) + 2(x-y)(1-xy) = 4(1+xy)$$