I´m struggling with this question and it would be really helpful if you can give me a hint on how to solve it.
Let $D=\gcd(a,b)$, $a = \alpha'\times D$ and $b = \beta' \times D$, all nonzero naturals.
Find the possible $D,\alpha,\beta$ that satisfies $D^2\alpha'(1+\beta')=29106$
I've tried decomposing $29106=2^1 3^2 7^2 11^1$, so $D^2$ can only be $3^2, 7^2$ or $3^2 7^2$.
Another thing that I should mention is that $\alpha'$ and $\beta'$ are coprimes.
Thank you and have a nice day!
You have one extra case for $D^2$, which can be $\color{red}{1}, 3^2, 7^2$ or $3^27^2$. Notice that with $D\in \{1,3,7,21\}$, we have $D$ odd and at most one of $a,b$ even.
Now consider $e := D^2\alpha'(1+\beta')$, where we need to have $e= 2\cdot 3^{\color{red}{3}} \cdot 7^2 \cdot 11$
Can $a$ be even? No, since that would make $\alpha '$ even and $\beta '$ odd, so $(1+\beta ')$ would be even and $e$ would be divisible by $4$
Can $b$ be even? No, since then all multiplicative terms for $e$ are odd.
So $a$ and $b$ are odd. We need $(\beta'+1)$ to be even and a factor of $29106$, which gives $24$ initial possibilities for $\beta'$, starting $\{1,5,17,53,13,41,125,\ldots\}$ each of which will then give various detailed options on how $\alpha'$ and $D$ can be assigned against the remaining factors.