No. of positive integral solutions of fractional functions

[Physics lover]

Homework Statement

Find no. of positive integral solutions of the equation
$\frac {xy} {x+y}=$ $2^4$ $3^5$ $5^4$

Relevant Equations

No. of positive integral solutions of linear equation=$^{n-1} C_ {r-1}$

I know how to find integral solutions of linear equations like x+y=C or x+y+z=C where C is a constant.
But I don't have any idea how to solve these type of questions.I am only able to predict that both x and y will be greater than 2⁴3⁵5⁴.Please help.

 

[haruspex]

I would start by considering an arbitrary prime, p, and how often it divides each of x and y.

 

[Physics lover]

I would start by considering an arbitrary prime, p, and how often it divides each of x and y.

Can you please explain a little bit more.I am not able to make out what are you saying.

 

[haruspex]

Can you please explain a little bit more.I am not able to make out what are you saying.

Suppose some prime p divides x m times, but no more, and divides y n times, but no more. Consider separately the cases of whether p is or is not one of 2, 3 or 5.
What can you say about m and n?

 

[haruspex]

... but I think I see a much easier way.
Multiply out to get rid of the division, collect terms on one side leaving zero on the other, and think about algebraic factorisation.

 

[Physics lover]

Suppose some prime p divides x m times, but no more, and divides y n times, but no more. Consider separately the cases of whether p is or is not one of 2, 3 or 5.
What can you say about m and n?

I think I got it.I wrote the xpression as-:
(x-$2^4$$3^5$$5^4$)(y-$2^4$$3^5$$5^4$)=$2^8$$3^{10}$$5^8$
And now I have to find the no. of divisors of this no.
Is it correct?

 

[haruspex]

I think I got it.I wrote the xpression as-:
(x-$2^4$$3^5$$5^4$)(y-$2^4$$3^5$$5^4$)=$2^8$$3^{10}$$5^8$
And now I have to find the no. of divisors of this no.
Is it correct?

Yes, but make sure you do not double count,

 

[Physics lover]

Yes, but make sure you do not double count,

Can you check it please.I am getting 891 as my answer.

 

[Physics lover]

Yes, but make sure you do not double count,

i think there will be no double counting here.
For example,consider a solution (x,y)=(a,b).
Now (x,y)=(b,a) will be considered a different solution,right?

 

[SammyS]

i think there will be no double counting here.
For example,consider a solution (x,y)=(a,b).
Now (x,y)=(b,a) will be considered a different solution,right?

Yes, I would consider those different.

But, what if a = b ?

 

[Physics lover]

Yes, I would consider those different.

But, what if a = b ?

yeah I forgot that.There's only one possibility for a =b.So 891-1=890.Is it correct now?

 

[haruspex]

yeah I forgot that.There's only one possibility for a =b.So 891-1=890.Is it correct now?

No, that's the wrong way about. In your 891, you have only counted the factorisation (2⁴3⁵5⁴)² once.
If you consider (a,b) to be a different solution from (b,a) when a and b are different then 891 is the answer. If you consider them the same you would halve, but add 1 first because you have not counted (a,a) twice: (891+1)/2=446.

It is not clear to me which the question wants. I would have guessed 446.

 

[willem2]

Homework Statement:: Find no. of positive integral solutions of the equation
$\frac {xy} {x+y}=$ $2^4$ $3^5$ $5^4$

Doesn't this need gcd(x,y) = 1? If (x,y) is a solution, so is (kx, ky) for all k.

 

[haruspex]

Doesn't this need gcd(x,y) = 1? If (x,y) is a solution, so is (kx, ky) for all k.

Umm... you might want to check that statement.

 

[Physics lover]

No, that's the wrong way about. In your 891, you have only counted the factorisation (2⁴3⁵5⁴)² once.
If you consider (a,b) to be a different solution from (b,a) when a and b are different then 891 is the answer. If you consider them the same you would halve, but add 1 first because you have not counted (a,a) twice: (891+1)/2=446.

It is not clear to me which the question wants. I would have guessed 446.

Thanks.I got it.
Your guess is wrong.The books gives 891 as the answer.