Find $a$ and $b$ be natural numbers, such that $(2a + 3)(3a + 4) = 35^b$
One can see $a=1, b=1$ is a solution. The challenge is to prove there are no other solutions. My intuition is that $2a + 3 = 5^b$, and $3a + 4 = 7^b$ - otherwise, $5$ would divide both terms, therefore their difference - so, $5 | a + 1, 7| a+ 1\implies 35 | a + 1,$ then LHS would not be divisible by $35$, contradiction. So, the first parenthesis is $5^b$ and second is $7^b$.