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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.

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    $\begingroup$ If you want $n/7$ to be a cube, then $\alpha \equiv 0\mod 3$ and $\beta-1\equiv 0\mod 3$. $\endgroup$ – rogerl May 12 '14 at 20:03
  • $\begingroup$ @rogerl I've been guessing on that, but I'm not sure why, could you elaborate? $\endgroup$ – Black Milk May 12 '14 at 20:04
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    $\begingroup$ @BlackMilk If $n/7$ is to be a cube, then every prime in its prime factorization must appear to a power that is a multiple of three, right? So both $\alpha$ and $\beta-1$ must be divisible by three. $\endgroup$ – rogerl May 12 '14 at 20:06
  • $\begingroup$ The answer is $n=875$. $\endgroup$ – user26486 May 12 '14 at 20:08
  • $\begingroup$ @rogerl Right, so then $\beta$ must satisfy $\beta\equiv1\bmod 3$ $\endgroup$ – Black Milk May 12 '14 at 20:10
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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).

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$$\alpha\equiv1\bmod2$$ and $$\alpha\equiv 0 \bmod3$$ the smallest $\alpha $ is $3$. Also $$\beta\equiv 1\bmod2$$ and $$\beta\equiv 1\bmod3$$ the smallest $\beta$ is $1$. Therefore $n=5^{\alpha}7^{\beta}=5^3*7$

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Let the perfect square be $k^2$ and the perfect cube $l^3$. Then we have $$\begin{cases}35n=k^2\\n=7l^3\end{cases}\implies 35\cdot 7l^3=k^2\implies 5l^3=\left(\frac{k}{7}\right)^2$$

The number $\frac{k}{7}$ is an integer since $49\mid k^2$.

Hence $5l^3=c^2$ for some $c\in\mathbb N$. By substituting $l$ with $1,2,3,4,5,\ldots$ we can see that the least possible $l$ is $5$. Therefore $n=7l^3=7\cdot 5^3=875$. Not an elegant solution, though.

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First, in $n$ is a solution so is $nm^6$ for any $n$ so you need to do more to justify the form $5^a7^b$.

Let $n=5^a7^bc$ with $(c,5)=(c,7)=1$ so that $$35n=5^{a+1}7^{b+1}c$$ is a square, which means $c$ is a square and $a+1, b+1$ are even.

Also $$\frac n7=5^a7^{b-1}c$$

is a cube, which means $c$ is a cube (hence a sixth power) and that $a, b-1$ are divisible by 3.

You want the smallest possible value. So $c$ is the smallest possible positive integer sixth power, hence is equal to $1$. Then $b-1$ is a non-negative even multiple of $3$ - lowest value is $0$ so $b=1$. And $a$ is the lowest possible positive integer which is and odd multiple of $3$, so $a=3$ whence $$n=5^3\cdot 7 =875$$

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