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Characterize all positive integers $x$, $y$, and $z$ such that:

$$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{z}$$

For example, $\dfrac{1}{x+1} + \dfrac{1}{x(x+1)} = \dfrac{1}{x}$.

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migrated from mathoverflow.net Feb 26 '15 at 20:30

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The solutions are parameterized by $$x = p^2+pq, y= q^2+pq, z = pq$$ where $p,q$ are arbitrary integer numbers, chosen so that $x,y,z$ are all unequal zero.

The reason for this is that $1/x + 1/y = 1/z$ describes a projective curve in 2-dimensional projective space ${\mathbf P}^2$, given by the homogeneous equation $$(1) \quad y z + x z - x y = 0$$

The equation is of degree $2$ and so every line $l$ intersects it in two points $P_l$ and $Q_l$. Keeping $P=P_l$ fixed and with rational coordinates, the points $Q_l$ are also with rational coordinates when $l$ is a line with rational slope. Furthermore every rational point $Q$ defines a rational line $l$ with $Q = Q_l$. So the lines $l$ through $P$ with rational slope trace out the rational points of $C$, called $C({\mathbb Q})$. As the curve $C$ is projective we have $C({\mathbb Z}) = C({\mathbb Q})$. So the rational and the integer solutions to (1) are identical after multiplying away denominators.

Now take as point $P=(0,0,1)$ and do the calculation explicity on the affine patch $z=1$ where (1) becomes $y+x-xy = 0$. Making the ansatz $x=w,y=p/q \,w$ and solving for $w$ we find the solutions $0, (p+q)/p$. So $x=(p+q)/p$ and $y=(p+q)/q$ are the rational solutions of the affine equation. Bringing this to a common denominator $pq$ we receive the homogeneous solution from the beginning.

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  • $\begingroup$ Does this need to add in non-relatively prime solutions? with $(p,q)=(2,1)$, this captures $(x,y,z)=(6,3,2)$. But how does this capture $(x,y,z)=(12,6,4)$? $\endgroup$ – alex.jordan Feb 26 '15 at 21:55
  • $\begingroup$ You are right, one has to multiply each solution from above with an integer factor to trace out all (x,y,z). Additionally one can assume gcd(p,q) = 1 as p/q is the slope of the line l. So one arrives at the solution given above by Barry Cipra. $\endgroup$ – Jürgen Böhm Feb 26 '15 at 22:13
  • $\begingroup$ If you take $\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq}$ and divide through by $p+q$ you get $\frac{1}{p^2 + pq} + \frac{1}{q^2 + pq} = \frac{1}{pq}$. I was aware that this was suficient, now I see that, with the obvious modifications, it is also necessary. Thanks. $\endgroup$ – steven gregory Feb 26 '15 at 22:34
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${1\over x}+{1\over y}={1\over z}$ if and only if it has the form

$${1\over a(a+b)d}+{1\over b(a+b)d}={1\over abd}\quad\text{with}\quad \gcd(a,b)=1$$

(all variables being understood as positive integers).

The "if" part is trivial to verify. The "only if" part comes by setting $x=ga$ and $y=gb$ with $g=\gcd(x,y)$, which gives

$${1\over x}+{1\over y}={a+b\over gab}$$

and noting that $a+b$ is relatively prime to both $a$ and $b$.

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An idea : you have

$$\frac{1}{x}+\frac{1}{y} = \frac{x+y}{xy}$$

It imply that $x+y | xy$

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  • $\begingroup$ Indeed, it's equivalent. $\endgroup$ – Jack M Feb 26 '15 at 20:44
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$\frac{1}{x} + \frac{1}{y}=\frac{x+y}{xy}$ then we have to found $(x,y)\in \mathbb N^{2}$ such that $x+y|xy$ then $k(x+y)=xy$ if and only if $x=\frac{ky}{y-k}$ then $k<y$ and $y-k|ky$ , then fixed y you can find a solution choosing $y-k$ as a divisor of y.(Note that if y is prime we not have solution)

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For the equation.

$$\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$$

The easiest way is to factor the square.

$$p^2=ks$$

Then decisions can be recorded.

$$x=p$$

$$y=s-p$$

$$z=p-k$$

For the more General case formula there. Number of solution for $xy +yz + zx = N$

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