## 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, a2 + b2 = kab.

We can see that the right hand side is divisible by a, so the left hand side must be divisible by a. Clearly a2 is divisible by a, so b2 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, a2 + b2 + c2 = kabc.

Problem ID: 153 (Feb 2004)     Difficulty: 3 Star

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