26th Junior Balkan Mathematical Olympiad | JBMO 2022 - Problem 3

Problem 3

Find all quadruples of positive integers $(p, q, a, b)$ , where $p$ and $q$ are prime numbers and $a > 1$ , such that $p^a = 1 + 5q^b.$

Solution

i) $p=2$ and $q\ge 3$

$\pmod 5$ gives us $4|a$ so $a\ge 4$ and $a$ is even. We have $5q^b=(2^{\frac a2}-1)(2^{\frac a2}+1)$ . Clearly, $q$ cannot divide $2^{\frac a2}-1$ and $2^{\frac a2}+1$ simultaneously. Also, both numbers are greater than $1$ as $a\ge 4$ . Hence, one of them is $5$ .

$2^{\frac a2}-1=5\Rightarrow 2^{\frac a2}=6$ , contradiction.

$2^{\frac a2}+1=5\Rightarrow 2^{\frac a2}=4\Rightarrow a=4\Rightarrow q^b=2^{\frac a2}-1=3$ . Hence, $\boxed{(p,q,a,b)=(2,3,4,1)}$ .

ii) $q=2$ and $p\ge 3$

If $a$ is odd, then $a\ge 3$ and we have $(p-1)(p^{a-1}+\cdots +p+1)=5\cdot 2^b$ . Since $a$ is odd, we know that $p^{a-1}+\cdots +1$ is odd as well. Hence, $5\ge p^{a-1}+\cdots +1\ge p^2+1\ge 10$ , contradiction. So $a$ is even.

$\pmod 3$ tells us $p=3$ . Then $\pmod 5$ tells us $4|a$ .

We have $5\cdot 2^b=(3^{\frac a2}-1)(3^{\frac a2}+1)$ . As $4|a$ , we get $2^1|| 3^{\frac a2}+1$ so $3^{\frac a2}+1\in \{2,10\}$ . This gives us $p=3, a=4$ and $5\cdot 2^b=80$ . So $\boxed{(p,q,a,b)=(3,2,4,4)}$ .