North Macedonia competition problems

Let $p$ be the least prime factor of $b$, and $q$ be the least natural number such that $p\mid(a^q-1)$ (such a number exists because $p\mid(a^b-1)$). From Little Fermat's theorem, we have that $p\mid(a^{p-1}-1)$, which implies $q\mid b$ and $q\mid(p-1)$, and from the minimality of $p$, we get that $q=1$, i.e. $p\mid(a-1)$. Let $b=p^\alpha c$, where $c$ is not divisible by $p$. If $p^2\mid(a-1)$, then $a\equiv 1\pmod{p^2}$, so $a^k\equiv 1\pmod{p^2}$, for every natural number $k$ from $(a^{l(p-1)} + a^{l(p-2)} + \dots + 1)\equiv p\pmod{p^2}$ and $(a^{p^\alpha(c-1)} + a^{p^\alpha(c-2)} + \dots + 1)\equiv c\pmod{p^2}$, $a^{p^\alpha c}-1=(a-1)(a^{p-1} + a^{p-2} + \dots + 1)\dots$
$\dots (a^{p^{\alpha-1}}(p-1) + a^{p^{\alpha-1}}(p-2) + \dots + 1)(a^{p^{\alpha}(c-1)} + a^{p^{\alpha}(c-2)} + \dots + 1)$
so the degree of $p$ in $\frac{a^b-1}{a-1}$ is $\alpha$ and therefore we get that $p^{\alpha(a-1)}\mid(a-1)$
$(p^{\alpha a}\mid(a^b-1))$. The last is not possible because for $\alpha \ge 1$, $p \ge 2$ and $a \ge 2$, it holds that $p^{\alpha(a-1)} > a-1$.

This implies that, the degree of $p$ in $a-1$ must be $1$. If $p > 2$, we will prove that the degree of $p$ in $a^{p^k}-1$ is $k+1$, for every natural number $k$. For $k=0$, the statement is true. Let the statement be true for every $k < n$. For $k=n$ from the following equalities
$$a^{p^n} - 1 = (a^{p^{n-1}})^p - 1 = (a^{p^{n-1}} - 1)(a^{p^{n-1}(p-1)} + a^{p^{n-1}(p-2)} + \dots + 1)$$
and
$$a^{lp^{n-1}} - 1 = (a^{p^{n-1}} - 1)(a^{p^{n-1}(l-1)} + a^{p^{n-1}(l-2)} + \dots + 1) \equiv p^n d_n l \pmod{p^{n+1}}$$
we get that
$$\begin{aligned} a^{p^n} - 1 &\equiv p^n d_n \left( p^n d_n (p-1+p-2+\dots+1) + p \right) \equiv \\ &\equiv p^{n+1} d_n \left( p^n d_n \frac{p-1}{2} + 1 \right) \pmod{p^{2n+1}} \end{aligned}$$
(where $d_n = \frac{a^{p^{n-1}}-1}{p^n}$ and $d_n$ and $p$ coprime by the assumption) with which we've proved the statement by induction. From the equality
$$a^{p^{\alpha}c} - 1 = (a^{p^{\alpha}} - 1)(a^{p^{\alpha}(c-1)} + a^{p^{\alpha}(c-2)} + \dots + 1)$$
and from $a^{p^{\alpha}(c-1)} + a^{p^{\alpha}(c-2)} + \dots + 1 \equiv c \pmod{p}$, we get that the degree of $p$ in $a^b - 1$ is $\alpha + 1$, and on the other hand it must be greater or equal to $\alpha a$, which is possible only for $\alpha = 1$ and $a=2$, but this case is not possible because of $p \nmid (a-1)$.

Now, the case when $p=2$ remains. From the equality (the same as above for $p>2$)
$$a^{2^{\alpha}c} - 1 = (a-1)(a+1)\dots(a^{2^{\alpha-1}} + 1)(a^{2^{\alpha}(c-1)} + a^{2^{\alpha}(c-2)} + \dots + 1).$$
The fact that $c$ is odd implies that $a^{2^{\alpha}(c-1)} + a^{2^{\alpha}(c-2)} + \dots + 1$ is an odd number as a sum of an odd number of odd numbers ($a$ has to be odd, because $2|(a-1)$). From $2|(a-1)$ and $2|(a+1)$, we get that $4|(a^{2k}-1)$, for every natural number $k$, but this implies that $4$ is not a divisor of $a^{2k}+1$. This implies that the degree of $2$ in $\frac{a^b-1}{a^2-1}$ is $\alpha-1$, so because $2^{\alpha a}|(a^b-1)$, we get that $2^{\alpha(a-1)+1}|(a^2-1)$, and because $4$ can be a divisor of only one of $a+1$ and $a-1$, we get that $2^{\alpha(a-1)} \le a+1$, which is possible if and only if $\alpha=1$ or $a=3$. Let $r$ be the least prime factor of $c$, $c=r^\beta d$, $r$ and $d$ are coprime and $s$ is the least natural number for which $r|(3^s-1)$. This implies that $s|b$ and $s|(r-1)$ (the same as before for $p$). This is possible only if $s=1$ or $s=2$ (those are the only divisors of $b$, less than $r$). In both cases $r|(3^2-1)$, which is possible only for $r=2$, a contradiction with the choice of $r$. We get that the unique solution is $b=2$ and $a=3$ ($b^a=8=a^b-1$).