How prove this sequence $a_{n}=a_{n-1}+\frac{1}{a_{n-1}}$,then $a_{p}+a_{q}\notin Z$ let sequence $\{a_{n}\}$ such
$$a_{1}=\dfrac{1}{2},a_{n}=a_{n-1}+\dfrac{1}{a_{n-1}}$$
show that: for any $(p.q)\neq (1,2),p\neq q$,such $a_{p}+a_{q}\notin Z$
My idea: since
$$a_{2}=\dfrac{5}{2},\Longrightarrow a_{1}+a_{2}=\dfrac{1}{2}+\dfrac{5}{2}=3\in Z$$
$$a_{3}=a_{2}+\dfrac{1}{a_{2}}=\dfrac{5}{2}+\dfrac{2}{5}=\dfrac{29}{10}$$
so
$$a_{1}+a_{3},a_{2}+a_{3}\notin Z$$
so I try
$$a_{p}+a_{q}=a_{p-1}+a_{q-1}+\dfrac{1}{a_{p-1}}+\dfrac{1}{a_{q-1}}=(a_{p-1}+a_{q-1})\left(1+\dfrac{1}{a_{p-1}a_{q-1}}\right)$$
But I can't prove $a_{p}+a_{q}\notin Z,(p,q)\neq(1,2),p\neq q$,
Thank you very much
 A: Define $a_n=\frac{p_n}{q_n}$.  Then by the definition we have $a_{n+1} = \frac{p_{n+1}}{q_{n+1}} = \frac{p_n}{q_n}+\frac{q_n}{p_n} = \frac{p_n^2+q_n^2}{p_nq_n}$, so the recurrence for $a_n$ translates into the twin recurrences $p_{n+1} = p_n^2+q_n^2$, $q_{n+1} = p_nq_n$.  Now, since we have $\gcd(p_1, q_1)=\gcd(1, 2) =1$ then by using the fact that $\gcd(a,b\cdot c)|\gcd(a,b)\cdot\gcd(a,c)$ (and so in particular $\gcd(a,b)=1\implies \gcd(a^2, b)=1$) and the "Euclidean identity" $\gcd(a, b) = \gcd(a, b+c\cdot a)$ for any $c$ , we have $$\begin{align}\gcd(p_{n+1},q_{n+1}) &= \gcd(p_n^2+q_n^2, p_nq_n)\\
&|\ \left(\gcd(p_n^2+q_n^2,p_n)\cdot\gcd(p_n^2+q_n^2, q_n)\right)\\
&=\gcd(q_n^2, p_n)\cdot\gcd(p_n^2, q_n)\\
&=1\cdot 1
\end{align}
$$
with the last step by induction; in other words, the fraction $\frac{p_n}{q_n}$ for this sequence is always in lowest terms.  Furthermore, by (obvious) induction we have get $q_{n+k} = q_n\prod_{i=0}^{k-1}p_{n+i}$, and so in particular $p_n|q_{n+k}$ for all $k\gt 0$; this means that $\gcd(p_n, p_{n+k+1})$ $= \gcd(p_n, p_{n+k}^2+q_{n+k}^2)$ $=\gcd(p_n, p_{n+k}^2)$, and since $\gcd(p_n, p_{n+1})$ $=\gcd(p_n, p_n^2+q_n^2)$ $=\gcd(p_n, q_n^2)=1$ then by another inductive step we get $\gcd(p_n, p_{n+k})=1$ for all $k\gt 0$.
Now, consider $a_n+a_{n+k} = \frac{p_n}{q_n}+\frac{p_{n+k}}{q_{n+k}}$; then by using $q_{n+k} = q_n\prod_{i=0}^{k-1}p_{n+i}$ we can rewrite this as $\displaystyle a_n+a_{n+k} = \frac{1}{q_n}\cdot\left(p_n+\frac{p_{n+k}}{\prod_{i=0}^{k-1}p_{n+i}}\right)$.  But $\gcd(p_{n+k}, \prod_{i=0}^{k-1} p_{n+i})=1$ since each of the individual GCDs is 1, and so $\displaystyle\frac{p_{n+k}}{\prod_{i=0}^{k-1}p_{n+i}}\not\in\mathbb{Z}$ as long as the product isn't $1$ (this is where the trivial case $(n,k) = (1,1)$ works).  This implies that $\displaystyle\left(p_n+\frac{p_{n+k}}{\prod_{i=0}^{k-1}p_{n+i}}\right)\not\in\mathbb{Z}$ (since it's the sum of an integer and a non-integer), and then $a_n+a_{n+k}$ $\displaystyle=\frac{1}{q_n}\cdot\left(p_n+\frac{p_{n+k}}{\prod_{i=0}^{k-1}p_{n+i}}\right)\not\in\mathbb{Z}$ since dividing a non-integer by an integer can't give an integer.
