Problem source

\begin{questionparts}
\item 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. 
\item 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. 
\end{questionparts}

Solution source

\begin{questionparts}
\item 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.

\item 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 $(q-2)(r-2) = 2 \Rightarrow q = 3, r= 4$
\end{questionparts}