How many solutions are there in $\mathbb{N} \times \mathbb{N}$ to the equation $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{1995}?$
How would you solve this? I have tried but am not sure how I should proceed with this.
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$y=x=3990$ is a solution. – LeeNeverGup Nov 25 '13 at 14:32
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Just found : http://math.stackexchange.com/questions/403036/natural-number-solutions-to-fracxyxy-n-equivalent-to-frac-1x-frac-1y – lab bhattacharjee Nov 26 '13 at 15:35
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After multiplying the equation by 1995xy, you get 1995y + 1995x = xy.
Solving for y, you get y = (1995x)/(x - 1995). The function is undefined for x = 1995, but starting with x = 1996 solutions start to appear. (1996,3982020), (1998,1328670), (2000,798000), (2002,570570), (2004,444220), (2010,267330), and (2014,211470). The solutions continue infinitely to the right, but get further and further apart. At x = 3990, y = 3990. The solutions continue (4200,3800), (4522,3570), and (4788,3420). The solutions also continue infinitely to the left starting with (1994,-3978030), (1992,-1324680), (1990,-794010), (1988,-566580), (1986,-440230), (1980,-263340), and (1976,-207480).
John Butnor
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Let $p,q,k$ be 3 non-zero integers such that $n=kpq$, if $x \ne y$, the general solution to the equation $\frac1x+\frac1y=\frac{1}{n}$ is given by the couples: $$(x,y)=(kp(p+q), kq(p+q))=(y,x)$$ $k,p,q$ can be any combinations of the primes of $n$,including $1$ (although not considered a prime) such that $n=kpq$.
In your case $n=1995=1.3.5.7.19$, WLOG, if we pick $k=1$, the possible cases are: (1(1+1995), 1995(1+1995)); (3(3+665), 665(3+665)); (5(5+399), 399(5+399)); ...etc.
For $x=y$, the result is rather obvious $x=y=2.1995$.