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2000 Paper 2 Q1
D: 1600.0 B: 1516.0
proof unit fractions number theory prime numbers factorisation uniqueness algebraic manipulation Erdos-Straus conjecture

A number of the form \(1/N\), where \(N\) is an integer greater than 1, is called a unit fraction. Noting that \[ \frac1 2 =\frac13 + \frac16\\\ \mbox{ and } \frac13 = \frac14 + \frac1{12}, \] guess a general result of the form $$ \frac1N =\frac1a +\frac1b \tag{*} $$ and hence prove that any unit fraction can be expressed as the sum of two distinct unit fractions. By writing \((*)\) in the form \[ (a-N)(b-N)=N^2 \] and by considering the factors of \(N^2\), show that if \(N\) is prime, then there is only one way of expressing \(1/N\) as the sum of two distinct unit fractions. Prove similarly that any fraction of the form \(2/N\), where \(N\) is prime number greater than 2, can be expressed uniquely as the sum of two distinct unit fractions.

Solution: Notice that \(\frac{1}{N} = \frac{1}{N+1} + \frac{1}{N(N+1)}\), so any unit fraction can be expressed as the sum of two distinct unit fractions. \begin{align*} && \frac{1}N &= \frac1a + \frac1b \\ \Leftrightarrow && ab&= Nb+Na \\ \Leftrightarrow && 0 &= (a-N)(b-N)-N^2 \\ \Leftrightarrow && N^2 &= (a-N)(b-N) \end{align*} If \(N\) is prime then the only factors of \(N^2\) are \(1,N\) and \(N^2\). if \(a-N = b-N = N\) then \(a=b\) and we don't have distinct fractions. Therefore \(a-N = 1\) and \(b-N = N^2\) and we obtain the decomposition earlier (and it must be the only solution). \begin{align*} && \frac2N &= \frac1a+\frac1b \\ \Leftrightarrow && 2ab &= Nb+Na \\ \Leftrightarrow && 4ab &= 2Na+2Nb \\ \Leftrightarrow && N^2 &= (2a-N)(2b-N) \end{align*} Therefore for \(a,b\) to be distinct we must have \(2a = N+1\) and \(2b = N+N^2\) as the only possible factorisation. Both of the right hand sides are even so we can write \[ \frac{1}{N} = \frac{1}{\frac{N+1}{2}} + \frac{1}{\frac{N(N+1)}{2}} \] and this is unique

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1993 Paper 1 Q3
D: 1516.0 B: 1516.0
proof number theory Diophantine equation inequality integer solutions systematic case analysis unit fractions

  1. Find all the integer solutions with \(1\leqslant p\leqslant q\leqslant r\) of the equation \[ \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=1\,, \] showing that there are no others.
  2. The integer solutions with \(1\leqslant p\leqslant q\leqslant r\) of \[ \frac{1}{p}+\frac{1}{q}+\frac{1}{r}>1\,, \] include \(p=1\), \(q=n,\) \(r=m\) where \(n\) and \(m\) are any integers satisfying \(1\leqslant m\leqslant n.\) Find all the other solutions, showing that you have found them all.

Solution:

  1. Suppose \(p > 3\) then there are clearly no solutions, since \(\frac1p+\frac1q+\frac1r \leq \frac{1}{4} + \frac{1}{4} + \frac{1}{4} < 1\) Therefore there are 3 cases: \(p = 3 \Rightarrow p = q = r = 3\) \(p = 2\): \begin{align*} && \frac12 = \frac1q + \frac1r \\ \Rightarrow && 0 = qr - 2q-2q \\ \Rightarrow && 4 &= (q-2)(r-2) \\ \end{align*} Therefore \((p,q,r) = (2, 3, 6), (2, 4, 4)\) \(p = 1\) we have a contradiction the other way.
  2. We have already shown \(p < 3\), so we just need to check \(p = 2\) (since \(p=1\) is described in the question). \begin{align*} && \frac12 &< \frac1q+\frac1r \\ \Rightarrow && qr &< 2q+2r \\ \Rightarrow && 4 &> (q-2)(r-2) \\ \end{align*} Therefore we can have \((q-2)(r-2) = 0 \Rightarrow p = 2, q = 2, r = n\) Or we have have \((q-2)(r-2) = 1 \Rightarrow q = 3, r = 3\) Or we can have \((q-2)(r-2) = 2 \Rightarrow q = 3, r= 4\)

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