<s>[INST] Solve for natural numbers: $(2a+3)(3a+4) = 35^b$ 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 $7^b$.
 [/INST] hint
$$3(2a+3)-2(3a+4)=1$$
so, By Bezout's Theorem, $ 2a+3 $ and $ 3a+4 $ are relatively primes.
on the other hand
$$(2a+3)(3a+4)=5^b.7^b$$
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