Junior Balkan MO

Solution:
First of all, observe that if $p, q$ are both odd, then the left hand side of the given equation is odd and the right hand side is even so there are no solutions in this case. In other words, one of these numbers has to be equal to $2$ so we can discuss the following two cases:

- $p=2$

In this case the given equation becomes
$$2^{a}=1+5 q^{b}$$
Note that $q$ has to be odd. In addition, $2^{a} \equiv 1 \pmod{5}$. It can be easily shown that the last equation holds if and only if $a=4c$, for some positive integer $c$. Now, our equation becomes $2^{4c}-1=5q^{b}$, which can be written into its equivalent form
$$(4^{c}-1)(4^{c}+1)=5q^{b}$$
Since $q$ is odd, it can not divide both $4^{c}-1$ and $4^{c}+1$. Namely, if it divides both of these numbers then it also divides their difference, which is equal to $2$, and this is clearly impossible. Therefore, we have that either $q^{b} \mid 4^{c}-1$ or $q^{b} \mid 4^{c}+1$, which implies that one of the numbers $4^{c}-1$ and $4^{c}+1$ divides $5$. Since for $c \geq 2$ both of these numbers are greater than $5$, we only need to discuss the case $c=1$. But in this case $5q^{b}=15$, which is obviously satisfied only for $b=1$ and $q=3$. In summary, $(p, q, a, b)=(2,3,4,1)$ is the only solution in this case.

- $q=2$

In this case obviously $p$ must be an odd number and the given equation becomes
$$p^{a}=1+5 \cdot 2^{b}$$
First, assume that $b$ is even. Then $2^{b} \equiv 1 \pmod{3}$, which implies that $1+5 \cdot 2^{b}$ is divisible by $3$, hence $3 \mid p^{a}$ so $p$ must be equal to $3$ and our equation becomes
$$3^{a}=1+5 \cdot 2^{b}$$
From here it follows that $3^{a} \equiv 1 \pmod{5}$, which implies that $a=4c$, for some positive integer $c$. Then the equation $3^{a}=1+5 \cdot 2^{b}$ can be written into its equivalent form
$$\frac{3^{2c}-1}{2} \cdot \frac{3^{2c}+1}{2}=5 \cdot 2^{b-2}$$
Observe now that $3^{2c} \equiv 1 \pmod{4}$ from where it follows that $\frac{3^{2c}+1}{2} \equiv 1 \pmod{2}$. From here we can conclude that the number $\frac{3^{2c}+1}{2}$ is relatively prime to $2^{b-2}$, so it has to divide $5$. Clearly, this is possible only for $c=1$ since for $c>1$ we have $\frac{3^{2c}+1}{2}>5$. For $c=1$, we easily find $b=4$, which yields the solution $(p, q, a, b)=(3,2,4,4)$.

Next, we discuss the case when $b$ is odd. In this case,
$$p^{a}=1+5 \cdot 2^{b} \equiv 1+5 \cdot 2 \equiv 2 \pmod{3}$$
The last equation implies that $a$ must be odd. Namely, if $a$ is even then we can not have $p^{a} \equiv 2 \pmod{3}$ regardless of the value of $p$. Combined with the condition $a>1$, we conclude that $a \geq 3$. The equation $p^{a}=1+5 \cdot 2^{b}$ can be written as
$$5 \cdot 2^{b}=p^{a}-1=(p-1)(p^{a-1}+p^{a-2}+\cdots+1)$$
Observe that
$$p^{a-1}+p^{a-2}+\cdots+1 \equiv 1+1+\cdots+1=a \equiv 1 \pmod{2}$$
so this number is relatively prime to $2^{b}$, which means that it has to divide $5$. But this is impossible, since $a \geq 3$ and $p \geq 3$ imply that
$$p^{a-1}+p^{a-2}+\cdots+1 \geq p^{2}+p+1 \geq 3^{2}+3+1=13>5$$
In other words, there are no solutions when $q=2$ and $b$ is an odd number.

In summary, $(a, b, p, q)=(4,4,3,2)$ and $(a, b, p, q)=(4,1,2,3)$ are the only solutions.

Solution:
Analogously as in the first solution we conclude that at least one of the numbers $p$ and $q$ has to be even. Since these numbers are prime, this implies that at least one of $p$ and $q$ must be equal to $2$. Therefore it is sufficient to discuss the following two cases:

- $p=2$

In this case the given equation then becomes
$$2^{a}=1+5q^{b}$$
From here, it follows that $q$ is an odd number. In addition, $2^{a} \equiv 1 \pmod{5}$, which implies that $a=4c$, for some positive integer $c$. Then the above equation can be written in its equivalent form
$$(2^{c}-1)(2^{c}+1)(2^{2c}+1)=5q^{b}$$
Since $2^{c}-1, 2^{c}$ and $2^{c}+1$ are three consecutive integers, one of them must be divisible by $3$. Clearly it is not $2^{c}$ implying that one of the numbers $2^{c}-1$ and $2^{c}+1$ is divisible by $3$. This implies that $3 \mid (2^{c}-1)(2^{c}+1)(2^{2c}+1)$ so $3 \mid 5q^{b}$, hence $q$ must be equal to $3$ and we are left with solving the equation
$$(2^{c}-1)(2^{c}+1)(2^{2c}+1)=5 \cdot 3^{b}$$
Note that $2^{2c}+1 \equiv 2 \pmod{3}$ so from the above equation it follows that $2^{2c}+1$ must be equal to $5$, which implies that $c=1$. For $c=1$ we have $b=1$, so we get $(a, b, p, q)=(4,1,2,3)$ as the only solution in this case.

- $q=2$

In this case the given equation becomes
$$p^{a}=1+5 \cdot 2^{b}$$
so clearly $p$ must be an odd number.

If $a$ is odd then we have
$$5 \cdot 2^{b}=(p-1)(p^{a-1}+p^{a-2}+\cdots+p+1)$$
The second bracket on the right hand side is sum of $a$ odd numbers so it is an odd number. Due to the condition $a>1$ we must have $a \geq 3$. But then
$$p^{a-1}+p^{a-2}+\cdots+p+1 \geq p^{2}+p+1 \geq 3^{2}+3+1>5$$
so we do not have solutions in this case. Therefore it remains to discuss the case when $a$ is even.

Let $a=2c$ for some positive integer $c$. Then we have the following equation
$$(p^{c}-1)(p^{c}+1)=5 \cdot 2^{b}$$
Note that $p^{c}-1$ and $p^{c}+1$ are two consecutive even numbers so one of them is divisible by $2$ but not by $4$. Looking into the right hand side of the above equation, we conclude that this number must be equal to either $2$ or $5 \cdot 2=10$. In other words, either $p^{c}-1 \in \{2,10\}$ or $p^{c}+1 \in \{2,10\}$ yielding the following possible values for $p^{c}$: $1,3,9,11$. Clearly $p^{c}=1$ is impossible, whereas $p^{c}=3$ implies that $(p^{c}-1)(p^{c}+1)$ is not divisible by $5$ so there are no solutions if $p^{c}=3$. Similarly, for $p^{c}=11$ we have that $(p^{c}-1)(p^{c}+1)$ is divisible by $3$ so it can not be equal to $5 \cdot 2^{b}$ for any positive integer $b$. Finally, if $p^{c}=9$, we have solution $(a, b, p, q)=(4,4,3,2)$.

In summary, $(a, b, p, q)=(4,4,3,2)$ and $(a, b, p, q)=(4,1,2,3)$ are the only solutions.