# A Short Equation in Reciprocals

Find all triples $(x, y, z)$ of positive integers such that

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

Solution Find all triples $(x, y, z)$ of positive integers such that

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

### Solution

The equation is equivalent to $\displaystyle z = \frac{xy}{x + y}.$ Let $d = \mbox{gcd}(x, y).$ Then

$x = dm,\space y = dn,$ with $\mbox{gcd}(m, n) = 1.$

It follows that $\mbox{gcd}(mn, m + n) = 1$ so that

$\displaystyle z = \frac{dmn}{m + n},$

which implies $(m + n) | d,$ i.e., $d = k(m + n),$ $k$ a positive integer.

Thus we obtain the solutions:

$x = km(m + n),$ $y = kn(m + n),$ $z = kmn,$

where the three parameters $k, m, n$ are positive integers.

### Remark

Assume $a, b, c$ are positive integers with no common factor that satisfy $\displaystyle \frac{1}{a} + \frac{1}{b} = \frac{1}{c},$ then $a + b$ is a square!

Indeed, from the foregoing solution, $k = 1,$ $a = m(n + m),$ $b = n(m + n),$ $a + b = (m + n)^{2}.$

Let positive integer $a, b, c$ satisfy $\displaystyle \frac{1}{a} + \frac{1}{b} = \frac{1}{c}.$ Then $a^{2} + b^{2} + c^{2}$ is a square!

Indeed,

\begin{align} a^{2} + b^{2} + c^{2} &= k^{2}[m^{2}(m + n)^{2} + n^{2}(m + n)^{2} + m^{2}n^{2}]\\ &= k^{2}[(m + n)^{4} - 2mn(m + n)^{2} + m^{2}n^{2}]\\ &= k^{2}[(m + n)^{2} - mn]^{2}. \end{align}

When I posted the problem at the Interactive Mathematics and Miscellany facebook page, Leo Giugiuc independently came up with the following solution:

This is trivial if $c=1.$ Assume $c\gt 1.$ We have $c(a+b)=ab.$ Let $d=\mbox{gcd}(a,b).$ Then $a=du,$ $b=dv,$ and $\mbox{gcd}(u,v))=1.$ Thus $c(u+v)=duv.$ But $\mbox{gcd}(u,u+v)=\mbox{gcd}(v,u+v)=1,$ implying that both $u$ and $v$ divide $c:$ $c=tuv,$ from which $t(u+v)=d$ and so $t|d.$ But then $t|d$ and $t|c$ combine to prove that $t|\mbox{gcd}(\mbox{gcd}(a,b),c),$ i.e., $t|\mbox{gcd}(a,b,c)$ which is $1,$ so $t=1.$ In other words, $u+v=d$ and, therefore, $a+b=du+dv=d^2.$

Eugene Lee offered a shortcut, with additional insight. From $\displaystyle a + b = \frac{a b}{c},$ the primes factors of $c$ are either in $a$ or in $b$ but not in both $(gcd(a,b,c)=1),$ so $c = c_1 c_2,$ $a = u c_1,$ $b = v c_2,$ with $\mbox{gcd}(u,c) = \mbox{gcd}(v,c) =1.$ Thus $u c_1 + v c_2 = u v.$ So $u$ divides $v c_2,$ hence $v.$ Similarly $v$ divides $u c_1,$ hence $u.$ Therefore $u = v,$ and $a+b = u^2.$ Eigene also observed that, since $u=v,$ $u c_1 + v c_2 = u v$ reduces to $c_1 + c_2 = u,$ so that the triple of expressions $c = c_1 c_2,$ $a = u c_1 = (c_{1}+c_{2})c_1,$ $b = v c_2 = (c_{1}+c_{2})c_2$ gives a general, two-parameter solution to the equation .

In the discussion, Daniel Hardiski observed that $abc$ is also a square, to which Leo replied with $abc=du\cdot dv\cdot uv=(duv)^2.$ He further found another square, viz., $(a-b)^2+(2c)^2.$ His proof is short. Using: $a = m(m + n),$ $b = n(m + n),$ $c = mn.$

\begin{align} (a-b)^2+(2c)^2&=[(m-n)(m+n)]^{2}+4m^{2}n^{2}\\ &=m^{4}-2m^{2}n^{2}+n^{2}+4m^{2}n^{2}\\ &=m^{4}+2m^{2}n^{2}+n^{2}=(m^{2}+n^{2})^{2}, \end{align}

showing that $a-b$ and $2c$ are the legs of a Pythagorean triangle.

Naturally, all this can be extended to prove $a^{2}+b^{2}+c^{2}$ is a square:

\begin{align} a^{2} + b^{2} + c^{2} &= (a+b)^{2}-2ab+c^2\\ &= (a+b)^{2}-2(a+b)c+c^2\\ &= (a+b-c)^2. \end{align}

### References

1. T. Andreescu, D. Andrica, I. Cucurezeanu, An Introduction to Diophantine Equations, Birkhäuser, 2010, pp. 22-23 • Diophantine Equations
• Finicky Diophantine Equations
• Diophantine Quadratic Equation in Three Variables
• An Equation in Reciprocals
• Minus One But What a Difference
• Two-Parameter Solutions to Three Almost Fermat Equations
• Chinese Remainder Theorem
• Step into the Elliptic Realm
• Fermat's Like Equation
• Sylvester's Problem
• Sylvester's Problem, A Second Look
• Negative Coconuts
• 