# X^2-y^2=275 (find all solutions for x & y)

1. Oct 29, 2005

### heartless

Hi all,

AFAIK this problem was on a math competition this year, I was just wondering around how to solve it. I didn't know a formula for solving it, and what I did
was: I tried to square all numbers from 18 to n until I find some solutions that would satisfy an equation. I found 18 and 7, however there were 3 right solutions for this and I think that one of them was above 100 (x or y equals 100<), of course it's a little crazy to count all of them in 10 minutes, so I asked my math teacher for some clues, and he gave me one:
x^2-y^2=275 = (x-y)(x+y)=275
Now it look much simplier for me, but again I can't make anything useful out of it, in my point of view we must try to count the solutions in the same way as I did before without knowing teacher's clue. And here is my question:
Can anyone give me another important clue or try to explain that one above?
Also if you can, may I have some tutorials or "keywords" for google on those kind of problems?, I don't really know what to search if branch of math. is unknown to me. At last if you know any faster solutions where answer comes out in matter of minutes, aprox. - 2-5 minutes, can you write something about it?

Thank you all.

2. Oct 29, 2005

### Muzza

I'm assuming we're only talking about integer solutions (this kind of equation is called a Diophantine equation. They are generally very hard to solve). But as you said,

(x - y)(x + y) = 275 = 5*5*11,

and since we have unique factorization into primes, this must mean that

x - y = 5
x + y = 5*11

or

x - y = -1
x + y = -5*5*11

etc (there are a few possibilities). These are linear equation systems and it should be easy to solve them and check which give integer solutions.

3. Oct 30, 2005

### uart

Yes there are only three factor pairs, each pair giving one valid solution.

1 x 275 = 275
5 x 55 = 275
11 x 25 = 275

4. Oct 31, 2005

### heartless

Thanks guys for answers, I just realized today in a bus solution for it however it's still not what it should be, suppose x^2-y^2=945934, now it'll take a little bit too long to solve, after all there must be a faster way to do that.
Thanks again ;-)

5. Oct 31, 2005

### HallsofIvy

Well, a little bit harder.

Recognising that 945934= 2*11*19*31*73 helps.

There are 25 distinct solutions.

Last edited by a moderator: Oct 31, 2005
6. Nov 1, 2005

### uart

Actually I think it's 2^(n-1) possible factor pairs for the case where there are n distinct prime factors, so 16 factor pairs to consider in this case. (hopefully someone will correct me if I'm worng here).

Note that not all factor pairs will neccessarily yeild a solution (though no other integer solutions are possible except for those yielded by testing the factor pairs). To explain this let me denote p and q as a factor pair. (that is pq = 945934 in this case).

Solving (x-y)=p and (x+y)=q gives,

2x = p+q

If all the prime factors are odd then all the factor pairs are also odd hence p+q is always even so the resulting x,y solutions will always be integers, this was the case in the first example you posed.

If however "2" happens to be one of the prime factors then we may now get the situation where one factor of the factor pair is even while the other of the pair is odd. Now p+q is odd and there is no corresponding integer solution for x and y. Indeed in the last problem you posed all possible factor pairs must contain one even and one odd factor and hence there are no integer solutions at all.

Note that the above should not lead you to conclude that if the number on the RHS of your equation is even that there can never be any solution. So long as the prime factor of "2" is repeated more than once then we can have factor pairs where both factors are even and hence integer solutions will follow.

Last edited: Nov 1, 2005
7. Dec 13, 2010

### al-mahed

if x^2-y^2=275, then x^2-y^2 = (x-y)(x+y) =275 = 11*25 = (18-7)(18+7) = 18^2 - 7^2

as for 945934 it is not divisible by 4, so it cannot be written as a difference of two squares