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
6 months, 2 weeks 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 \((2a_1+1)(2a_2+2)..(2a_n+1)\)

Shreyansh Mukhopadhyay - 1 month, 2 weeks ago

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

Aman Thegreat - 6 months, 2 weeks ago

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\[\large{\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{2015}}\]

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

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

\[\implies \large{ 2015x+2015y = xy}\]

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

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

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

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?

Md Zuhair - 6 months, 2 weeks ago

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@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

Anirudh Sreekumar - 5 months, 3 weeks ago

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Ok thanks..

Aman Thegreat - 6 months, 2 weeks ago

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

Md Zuhair - 6 months, 2 weeks ago

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Like the way I proceeded.

Md Zuhair - 6 months, 2 weeks ago

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