Solving A Kind Of Diophantine Equation

I have seen many questions like this 1x+1y=1n\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,

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

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

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

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

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

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

  ⟹  x(n−y)+ny−n2=−n2   Adding −n2 to both sides\implies \large{ x(n-y)+ny -n^2= -n^2} \space \space \space \text{Adding }-n^2 \text{ to both sides}

Now Factorizing it , will give

  ⟹  (x−n)(y−n)=n2\implies \large{(x-n)(y-n)=n^2}.

Say n=p1a1p2a2...\large{n=p_{1}^{a_{1}} p_{2}^{a_{2}}...} where p1,p2p_{1},p_{2} are primes.

So total number of divisors of n=(a1+1)(a2+1)..(an+1)n = (a_1 +1)(a_2 +1)..(a_n+1)

So Number of pairs = No. of divisors = (a1+1)(a2+1)..(an+1)(a_1 +1)(a_2 +1)..(a_n+1)

PRACTICE EXAMPLE

  • 1x+1y=12015\large{\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{2015}} , Find all the ordered pair (x,y).

  • 1x+1y=11890\large{\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{1890}} , Find all the ordered pair (x,y).

Lets END Here.

Hope it helps

#NumberTheory

Note by Md Zuhair
3 years, 8 months ago

No vote yet
1 vote

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Comments

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 - 3 years, 8 months ago

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Like the way I proceeded.

Md Zuhair - 3 years, 8 months ago

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

Md Zuhair - 3 years, 8 months ago

1x+1y=12015\large{\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{2015}}

  ⟹  x+yxy=12015\implies \large{\dfrac{x+y}{xy} = \dfrac{1}{2015}}

  ⟹  x+y=xy2015\implies \large{ x+y = \dfrac{xy}{2015}}

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

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

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

  ⟹  x(2015−y)+2015y−20152=−20152   Adding −20152 to both sides\implies \large{ x(2015-y)+2015y -2015^2= -2015^2} \space \space \space \text{Adding } -2015^2 \text{ to both sides}

Now Factorizing it , will give

  ⟹  (x−2015)(y−2015)=20152\implies \large{(x-2015)(y-2015)=2015^2}.

Now see that if the in the place of 201522015^2 it was 2×32 \times 3 (SAY).

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

So the solution would have been x−2015=2   y−2015=3x-2015=2 \space \space \space y-2015=3 or x−2015=3   y−2015=2x-2015=3 \space \space \space y-2015=2 or x−2015=6   y−2015=1x-2015=6 \space \space \space y-2015=1 or x−2015=1   y−2015=6x-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 20152=p1a1p2a2...2015^2=p_{1}^{a_1} p_{2}^{a_2}... So

No. of divisors = No of Solutions = (a1+1)(a2+1)...(a_1+1)(a_2+1)...

OK?

Md Zuhair - 3 years, 8 months ago

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Ok thanks..

Aman thegreat - 3 years, 8 months ago

@Md Zuhair- You also have to account for the negative divisors ,thus the total number of solutions will be 2(number of divisors)−12(\text{number of divisors})-1,Assuming x and y are integers thus incorporating negative values for x and y

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

Anirudh Sreekumar - 3 years, 7 months ago

@Md Zuhair I think that the number of ordered pairs are the number of factors of n2n^2

So the no. of ordered pairs would be (2a1+1)(2a2+2)..(2an+1)(2a_1+1)(2a_2+2)..(2a_n+1)

Shreyansh Mukhopadhyay - 3 years, 3 months ago
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