@putputlaw@qoto.org
For \(|x| < 1\):
\begin{equation}
\log\left( \sum_{n=0}^\infty x^n \right) = \sum_{n=1}^\infty \frac{x^n}{n}
\end{equation}
Find all positive integer pairs \( a,b \) such that for given positive integer \( c \)
\begin{equation}
\frac 1a + \frac 1b = \frac 1c
\end{equation}
An elegant solution by "completing the rectangle":
\begin{alignat}{3}
&\ \frac 1a + \frac 1b &=&\ \frac 1c
\\\iff&\ c(a+b) &=&\ ab
\\\iff&\ c^2 &=&\ ab - c(a+b) + c^2
\\\iff&\ c^2 &=&\ (a-c)(b-c)
\end{alignat}
Now just list inspect the divisors of \(c^2\).
**Putnam 2017 A2**
Prove that the sequence of rational functions defined by
\begin{align}
Q_0 &=\ 1 \\
Q_1 &=\ x \\
Q_n &=\ \frac{Q_{n-1}^2-1}{Q_{n-2}}
\end{align}
is actually a sequence of polynomials with integer coefficients.
**Solution (using a determinant property)**
Define
\begin{align}
P_0 &=\ 1 \\
P_1 &=\ x \\
P_n &=\ xP_{n-1} - P_{n-2}
\end{align}
Clearly \(\{P_n\}\) is a sequence of polynomials with integer coefficients.
Now show that \(\{P_n\}\) and \(\{Q_n\}\) define the same sequence:
\begin{align}
&\ \ -P_nP_{n-2} +P_{n-1}^2
\\&=\
\det\begin{pmatrix}
P_n & -P_{n-1} \\ P_{n-1} & -P_{n-2}
\end{pmatrix}
\\&=\
\det\begin{pmatrix}
x & -1 \\
1 & 0
\end{pmatrix}^n
\\&=\ 1
\end{align}
This shows that
\begin{equation}
P_n = \frac{P_{n-1}^2-1}{P_{n-2}}
\end{equation}
