Belarus competition problems

Since the term $a_1$ is involved only in the summand $\gcd(a_1, a_2)$ when calculating $a_4$, without loss of generality we can assume that $a_1 = \gcd(a_1, a_2)$. If the numbers $a_1, a_2$ and $a_3$ are divisible by the same number, then all the numbers in the sequence $(a_n)$ are divided by this number and we can take it out as a common factor. Therefore without loss of generality we assume that $\gcd(a_1, a_2, a_3) = 1$, which is equivalent to $\gcd(a_1, a_3) = 1$. Denote $\gcd(a_2, a_3) = d$, then $a_2 = bd$ and $a_3 = cd$, where $b$ and $c$ are coprime positive integers such that $a_1 \mid b$ and $\gcd(a_1, cd) = 1$. Let's find the first consecutive members of the sequence:
$$a_4 = a_3 + \gcd(a_1, a_2) = a_3 + a_1 = cd + a_1,$$
$$a_5 = a_4 + \gcd(a_2, a_3) = a_4 + d = cd + a_1 + d,$$
$$\begin{aligned} a_6 &= a_5 + \gcd(a_3, a_4) = a_5 + \gcd(cd, cd + a_1) = \\ &= a_5 + \gcd(cd, a_1) = a_5 + 1, \end{aligned}$$
$$\begin{aligned} a_7 &= a_6 + \gcd(a_4, a_5) = a_6 + \gcd(cd + a_1, cd + a_1 + d) = \\ &= a_6 + \gcd(cd + a_1, d) = a_6 + \gcd(a_1, d) = a_6 + 1. \end{aligned}$$
Since three consecutive numbers $a_5, a_6$ and $a_7$ differ by one, i.e. neighboring numbers are coprime, then all subsequent numbers will also differ by one, i.e., for any $n \ge 5$, the equality $a_{n+1} - a_n = 1$ will be true.