## Fraction Reciprocal Sum

#### Problem

Prove that the sum of a proper fraction and its reciprocal can never be integer.

#### Solution

Given a fraction, `x/y`, we can divide top and bottom by HCF(`x,y`), to get a fraction in its lowest common terms, `a/b`. As `x/y = a/b`, it follows that `y/x=b/a`.

Therefore, `x/y` + `y/x` = `a/b` + `b/a` = `k`, where HCF(`a,b`)=1.

Multiplying through by `ab`, `a`^{2} + `b`^{2} = `kab`.

We can see that the right hand side is divisible by `a`, so the left hand side must be divisible by `a`. Clearly `a`^{2} is divisible by `a`, so `b`^{2} must be divisible by `a`. However, as HCF(`a,b`)=1, this is impossible, unless `a=b`=1, and this would only happen if `x=y`.

Hence the sum of a proper fraction and its reciprocal can never be integer.

This is an interesting alternative perspective on the original problem. That is, showing that the sum of a proper fraction and its reciprocal cannot be integer is equivalent to showing that the sum of squares of two positive integers cannot be a multiple of their product.

Investigate the solutions to the equation, `a`^{2} + `b`^{2} + `c`^{2} = `kabc`.