Find all pairs of positive integers x and y that satisfy the equation.

1/x - 1/y = 1/2005

(y-x)/xy=1/2005

2005(y-x) = xy

Is this a good method?

Printable View

- Sep 1st 2005, 07:05 AMardnasSatisfying the equation
Find all pairs of positive integers x and y that satisfy the equation.

1/x - 1/y = 1/2005

(y-x)/xy=1/2005

2005(y-x) = xy

Is this a good method? - Sep 1st 2005, 02:26 PMrgep
So far so good.

- Sep 2nd 2005, 01:41 AMardnas
But I couldn't find a way to continue. Do you have any hints?

- Sep 2nd 2005, 12:48 PMrgep
Get everything over on to one side:

xy - 2005(y-x) = 0, that is

xy + 2005x - 2005y

Try to factorise: (x + ?)(y + ?)

(x-2005)(y+2005) = 2005^2 - Sep 10th 2005, 05:20 AMardnas
I tried to continue and put factors in, but they don't seem to work.

Can you explain a bit further if you can?

Or are there other easier methods?

Thanks! - Sep 10th 2005, 07:35 AMhemzaok
So you have (x-2005) * (y+2005) = 2005^2. The key now is to see that the equation tells you that (x-2005) and (y+2005) are factors of 2005^2. Moreover, when multiplied together, they give 2005^2.

2005^2 in prime factors factorization : 2005^2 = 5^2 * 401^2 (I think 401 is prime)

So you don't have many choices (you choose from those factors some for (x-2005) and keep the rest for (y+2005)) :

1) (x-2005) = 1 and so, (y+2005)=5^2 * 401^2. We get :**x=2006 and y=4018020**

2) (x-2005) = 5 and so, (y+2005)=5 * 401^2. We get :**x=2010 and y=802000**

3) (x-2005) = 5^2 and so, (y+2005)=401^2. We get :**x=2030 and y=158796**

4) (x-2005) = 401 and so, (y+2005)=5^2 * 401. We get :**x=2406 and y=8020**

5) (x-2005) = 5*401 and so, (y+2005)=5*401. We get : x=4010 and y=0(we reject it because y=0 and so is not positive - if the teacher wanted the y=0, he/she would have said give me x and y non-negative. So positive means >0).

6) (x-2005) = 401^2 and so, (y+2005)=5^2. We get : x=162806 and y=-1980 (we reject it because y<0)

And so we must regect all other combos for x because it gives (y+2005) <2005 which gives y<0 (not wanted).

So the main work was the**factorization done by "rgep"**. After that, you look for factors.

For a very similar example see : ryanchow (need help!!!) in the urgent math homework section - Sep 10th 2005, 06:35 PMardnas
Thanks for your explanation. I've tried this method before, but when you put the x and y into the original equation, they don't work.

- Sep 11th 2005, 12:21 PMhemzaSorry
Hi,

I am sorry. There was a little sign mistake in rgep factorization. Here is what he did:

"Get everything over on to one side:

xy - 2005(y-x) = 0, that is

xy + 2005x - 2005y

Try to factorise: (x + ?)(y + ?)

(x-2005)(y+2005) = 2005^2 "

But it should be :

2005(y-x) = xy (from your calculations)

2005y-2005x-xy=0

We add 2005^2 for both sides: 2005^2 +2005y- 2005x-xy=2005^2

We factorize : 2005(2005+y)-x(2005+y)=2005^2

We factorize again : (2005-x)(2005+y)=2005^2 (you see there is a little sign change)

I you use after that the same method I used on my previous reply for finding x and y you'll get the answers wanted.

Sorry, I should have checked the factorization before.

If you have difficulty finding the positive integers x and y satisfying the equation, respond to this "reply" and I'll help you with it. - Sep 12th 2005, 05:37 AMardnas
Thanks so much!