# How many solutions are there to this equation?

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## Summary:

are there any positive integers for x,y,a,b, when xyab are all different numbers? and furthermore, if there are, is there an infinite number of solutions?
(x/y)+(ab)=(a-b)(x+y)

are there any positive integers for x,y,a,b, when xyab are all different numbers? and furthermore, if there are, is there an infinite number of solutions?

anuttarasammyak
Gold Member
I assume x/y=integer and a>b so that the both side of the equation should be positive integer.
Let a=b+c and x/y=n-1 the equation is quadratic for b
$$b^2+cb-(1+(cy-1)n)=0$$
Solving this
$$b=-\frac{c}{2} \pm \frac{\sqrt{c^2+4(1+(cy-1)n)}}{2}$$
so
$$c^2+4(1+(cy-1)n)=p^2$$
where p is a integer. I am not sure it reduces the difficulty of the problem.

Last edited:
mfb
Mentor
Clearly x must be a multiply of y. Let's set x=yz where z is again an arbitrary positive integer. This simplifies the equation: z+(ab)=(a-b)y(z+1)

From here on it's easy to find solutions. Consider a-b=1:
z+b(b+1)=y(z+1)
Consider y=2:
b(b+1)=z+2
There is an infinite set of solutions already. Pick whatever you want for b (except 1), there is a suitable z.

jim mcnamara and anuttarasammyak