6
$\begingroup$

Find the number of solutions of $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{1995}$ in $\mathbb N^2$.

My Attempt:

This can be simplified to $1995(x+y)=xy$ and then further to $$y={1995x\over x-1995}$$

Since $1995=3\cdot5\cdot7\cdot19$, it has $16$ divisors

Now since $x>0$, This yields total of $16$ solutions (the answer according to book). however, Mathematica found $81$ solution

$\endgroup$
0

3 Answers 3

Reset to default
9
$\begingroup$

First, note $\frac{1}{1995}>\frac{1}{x}$ and so $x>1995$ and, likewise, $y>1995$. Next, we have: $$ 1995(x+y)=xy\quad\implies\quad (x-1995)(y-1995)=1995^2=3^25^27^219^2. $$ Write $x-1995$ as $3^i 5^j 7^k 19^l$. Then, each quadruple $(i,j,k,l)$ uniquely pins down $x$ and $y$. There are 3 options each for $i,j,k,l$ (i.e. $\{0,1,2\}$), so there are $3^4=81$ solutions in total. Your book appears to be in error.

$\endgroup$
4
$\begingroup$

$81$ looks correct to me. It is the number of divisors of $1995^2$

Take any divisor $d$ of $1995^2$.

Then $d$ also divides $1995^2+1995d = 1995(1995+d)$.

So I can set $x=1995+d$, a positive integer and have $y=\dfrac{1995^2}{d}+1995=\dfrac{1995x}{x-1995}$ also a positive integer giving $\dfrac{1}{x}+\dfrac{1}{y}=1995$. In other words

$$\dfrac{1}{1995+d}+\dfrac{1}{\frac{1995^2}{d}+1995}=1995$$

will be one of the $81$ solutions

$\endgroup$
0
$\begingroup$

Write your equation in the form $$y=1995+3980025\, \left( x-1995 \right) ^{-1}$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .