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.
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:
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.
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 \((q-2)(r-2) = 2 \Rightarrow q = 3, r= 4\)