Find the smallest integer $n\geq1$ such that $35n$ is a perfect square and $n/7$ is a pefect cube. What I have so far: we express the prime factorizations of $35$ and $7$ as $5\cdot7$ and $7$, respectively. Then $n$ must be of the form $n=5^{\alpha}7^{\beta}$. Thus we see that $$ 35n=5^{\alpha+1}7^{\beta+1} $$ and $$ \frac{n}{7}=5^{\alpha}7^{\beta-1}. $$ From here we proceed to see that if $35n$ is a perfect square, then $\alpha+1$ and $\beta+1$ are even, thus $\alpha$ and $\beta$ themselves must be odd, and so, $\alpha\equiv b\equiv1\bmod2$. Clearly, $n$ must be divisible by $7$ for $n/7$ to be an integral quantity. (This is where I'm stuck..)
I know $\alpha$ and $\beta-1$ are odd and that $n$ must be divisible by 7, however, I'm not sure what to fill in for the question marks below based on these conditions.
$$ \alpha\equiv?\bmod3\quad\text{and}\quad\beta\equiv?\bmod3 $$
Once I know why and how to fill in those question marks, I know it's a matter of solving a system of congruences via the chinese remainder theorem to finish the problem.