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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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1988 Paper 1 Q1
D: 1500.0 B: 1500.0
curve sketching logarithms stationary points optimisation number theory exponentials and logarithms integer solutions

Sketch the graph of the function \(\mathrm{h}\), where \[ \mathrm{h}(x)=\frac{\ln x}{x},\qquad(x>0). \] Hence, or otherwise, find all pairs of distinct positive integers \(m\) and \(n\) which satisfy the equation \[ n^{m}=m^{n}. \]

Solution:

TikZ diagram
If \(n^m = m^n \Rightarrow \frac{\ln m}{m} = \frac{\ln n}{n}\) so if \(n < m\) we must have that \(n < e \Rightarrow n = 1,2\). \(n = 1\) has no solutions, so consider \(n = 2\) which obviously has the corresponding solution \(m=4\). Since \(h(x)\) is decreasing for \(x > e\) we know this is the only solution. Hence the only solution for distinct \(m,n\) are \((m,n) = (2,4), (4,2)\)

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