# Solving A Kind Of Diophantine Equation

I have seen many questions like this $\large{\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{n}}$ where n is any positive integer, Find all the ordered pair (x,y).

So How to solve these kinds of questions?

Lets Start.

See,

$\large{\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{n}}$

$\implies \large{\dfrac{x+y}{xy} = \dfrac{1}{n}}$

$\implies \large{ x+y = \dfrac{xy}{n}}$

$\implies \large{ nx+ny = xy}$

$\implies \large{ nx+ny-xy = 0}$

$\implies \large{ x(n-y)+ny = 0}$

$\implies \large{ x(n-y)+ny -n^2= -n^2} \space \space \space \text{Adding -n^2 to both sides}$

Now Factorizing it , will give

$\implies \large{(x-n)(y-n)=n^2}$.

Say $$\large{n=p_{1}^{a_{1}} p_{2}^{a_{2}}...}$$ where $$p_{1},p_{2}$$ are primes.

So total number of divisors of $$n = (a_1 +1)(a_2 +1)..(a_n+1)$$

So Number of pairs = No. of divisors = $$(a_1 +1)(a_2 +1)..(a_n+1)$$

PRACTICE EXAMPLE

• $$\large{\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{2015}}$$ , Find all the ordered pair (x,y).

• $$\large{\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{1890}}$$ , Find all the ordered pair (x,y).

Lets END Here.

Hope it helps

Note by Md Zuhair
1 year ago

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@Md Zuhair I think that the number of ordered pairs are the number of factors of $$n^2$$

So the no. of ordered pairs would be $$(2a_1+1)(2a_2+2)..(2a_n+1)$$

- 7 months, 3 weeks ago

Hey, can you elaborate a bit more? Like for the practice example 1, no of divisors is 3..Then how do you proceed ?

- 1 year ago

$\large{\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{2015}}$

$\implies \large{\dfrac{x+y}{xy} = \dfrac{1}{2015}}$

$\implies \large{ x+y = \dfrac{xy}{2015}}$

$\implies \large{ 2015x+2015y = xy}$

$\implies \large{ 2015x+2015y-xy = 0}$

$\implies \large{ x(2015-y)+2015y = 0}$

$\implies \large{ x(2015-y)+2015y -2015^2= -2015^2} \space \space \space \text{Adding -2015^2 to both sides}$

Now Factorizing it , will give

$\implies \large{(x-2015)(y-2015)=2015^2}$.

Now see that if the in the place of $$2015^2$$ it was $$2 \times 3$$ (SAY).

Then $\implies \large{(x-2015)(y-2015)=2*3}$.

So the solution would have been $$x-2015=2 \space \space \space y-2015=3$$ or $$x-2015=3 \space \space \space y-2015=2$$ or $$x-2015=6 \space \space \space y-2015=1$$ or $$x-2015=1 \space \space \space y-2015=6$$.

So here we can see that number of divisors of 6 are 4(namely 1,2,3,6).

So here we have 4 solns.

But what in case of 2015^2?

We will get $$2015^2=p_{1}^{a_1} p_{2}^{a_2}...$$ So

No. of divisors = No of Solutions = $$(a_1+1)(a_2+1)...$$

OK?

- 1 year ago

@Md Zuhair- You also have to account for the negative divisors ,thus the total number of solutions will be $$2(\text{number of divisors})-1$$,Assuming x and y are integers thus incorporating negative values for x and y

where $$(0,0)$$ is the neglected solution

- 11 months, 3 weeks ago

Ok thanks..

- 1 year ago

So let me help u with the 1st one. Try the 2nd on ur own

- 1 year ago

Like the way I proceeded.

- 1 year ago