Find the smallest integer $n\geq1$ such that $35n$ is a perfect square and $n/7$ is a pefect cube. 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.

Hint $\ \ n = 7a^3,\,\ b^2\! = 35n = 5a(7a)^2\!\!\color{#c00}{\iff}\! a = 5c^2\!\!\iff\! n =7(5c^2)^3\! = 7\cdot 5^3\cdot  c^6 $
where $\color{#c00}\iff$ follows by comparing the parity of exponents in (unique) prime factorizations, to deduce $\,b^2\! = 5a\, d^2\Rightarrow\,5a\,$ is a square, so $\,a\,$ is $\,5\,$ times a square (again, by exponent parity).
