10. We rewrite the recurrence relation as

$\frac{(n+2) a_{n-1}}{4 a_{n} a_{n+1}}+\frac{1}{a_{n} a_{n+1} a_{n+2}}=\frac{(n+3) a_{n}}{4 a_{n+1} a_{n+2}} \Longleftrightarrow(n+2) a_{n+2}=\frac{(n+3) a_{n}^{2}-4}{a_{n-1}}$

for $n \geq 3$. Setting $n=2$ in the initial relation we obtain $4\left(a_{1}+4\right)=5 a_{1} a_{2}^{2}$, implying that $a_{1} \mid 16$ and $5 \mid a_{1}+4$. Therefore $a_{1}=16$ and $a_{2}=1$ or $a_{1}=1$ and $a_{2}=2$.

Case 1. Let $a_{1}=16$ and $a_{2}=1$. Then $5 a_{5}=6 a_{3}^{2}-4, a_{3} a_{6}=18$ and $7 a_{7}=2 a_{5}^{2}-1$ for $n=3,4$ and 5 , respectively. Since $a_{3} \equiv \pm 2(\bmod 5)$ and $a_{3}$ as a divisor of 18 we conclude that $a_{3}=3$ or $a_{3}=18$. The direct check of both values shows no solutions in this case.

Case 2. Let $a_{1}=1$ and $a_{2}=2$. Then $n=3$ and $n=4$ give $5 a_{5}+2=3 a_{3}^{2}$ and $a_{3} a_{6}=18$, respectively. Again $a_{3} \equiv \pm 2(\bmod 5)$ and we see that $a_{3}=3$ and $a_{6}=6$ or $a_{3}=18$ and $a_{6}=1$. In the second case we obtain $a_{5}=194$ which gives a contradiction with $8 a_{8}=\frac{9 a_{6}^{2}-4}{a_{5}}$.

In the first case $a_{5}=5$ and hence the only possible values are $a_{i}=i$ for $i=1,2, \ldots, 6$. Now easy induction shows that $a_{n}=n$ for all $n$.
Problem 10. Find all sequences of positive integers $\left\{a_{n}\right\}_{n=1}^{\infty}$, such that $a_{4}=4$ and the identity

$$
\frac{1}{a_{1} a_{2} a_{3}}+\frac{1}{a_{2} a_{3} a_{4}}+\cdots+\frac{1}{a_{n} a_{n+1} a_{n+2}}=\frac{(n+3) a_{n}}{4 a_{n+1} a_{n+2}}
$$

holds true for every positive integer $n \geq 2$.

Peter Boyvalenkov