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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 $7^b$.

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2 Answers 2

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So is there a question here? If so, here is a hint, combined w what you noted in your OP:

$7^b > 2×5^b$ for all $b \ge 3$. This leaves only $b=1,2$ to check.

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  • $\begingroup$ Thanks! Didn't know how to prove that there were no solutions for b>2 $\endgroup$
    – Mircea
    Commented Feb 8, 2021 at 22:08
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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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