Belarus2022

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: $$a_4=a_3+\gcd (a_1,a_2)=a_3+a_1=cd+a_1,$$ $$a_5=a_4+\gcd (a_2,a_3)=a_4+d=cd+a_1+d,$$ $$\begin{array}{rl}{a}_6& =a_5+\gcd (a_3,a_4)=a_5+\gcd (cd,cd+a_1)=\\ & =a_5+\gcd (cd,a_1)=a_5+1,\end{array}$$ $$\begin{array}{rl}{a}_7& =a_6+\gcd (a_4,a_5)=a_6+\gcd (cd+a_1,cd+a_1+d)=\\ & =a_6+\gcd (cd+a_1,d)=a_6+\gcd (a_1,d)=a_6+1.\end{array}$$ 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.