There are infinitely many $N$ such that $\frac{N}{2}$ is a perfect square, $\frac{N}{3}$ is a perfect cube, and $\frac{N}{5}$ is a perfect fifth power 
Show that there are infinitely many $N$ such that $\frac{N}{2}$ is a perfect square, $\frac{N}{3}$ is a perfect cube, and $\frac{N}{5}$ is a perfect fifth power. 

A hint is given with this question, which is: $7^{30}$ is a perfect square, a perfect cube and a perfect fifth power.
Can anyone solve this? Thanks!
 A: The general form of this kind of number is $2^{15}3^{10}5^{6}x^{30}$, for all values of $x$.  
When $x=1$, this number is the product of powers of 2, 3, and 5 just under some integer, eg $32768$, $59049$ and $15625$ are the largest powers under, say $60,000$.
The general working for such a number is to treat the indeces by chinese remainder.  So, for example, should we seek something where $N/2$ is square, $N/3$ is cube, and $N/5$ is a fifth power, we would note that, eg  it is a product of some powers of 2, 3, and 5, eg $2^a 3^b 5^c$.
We have then that 2 divides $a-1, b, c$, 3 divides each of $a, b-1, c$ and 5 divides each of $a, b, c-1$.  We get then that 15 divides a, 10 divides b, and 6 divides c.  


*

*We find now a multiple of 15 that 2 divides a-1:  result = $a=15$

*We find then that 10 that 3 divides b-1, result $b = 10$

*We find a result where 5 divides c-1: result = $6$.


We then note that every number that is a power of 30, is already a square, a cube, and a fifth power, ie $x^{2 \cdot 3 \cdot 5} = x^{30}$, and thus if $N/2$ is square, so is $(N\cdot x^{30})/2$ is also square, and if $N/3$ is a cube, so is $(N\cdot x^{30})/3$, and if $N/5$ is a fifth power, so is $(N\cdot x^{30})/5$, is true for all $x$.  
In the worked example, we include a final row, for all $x$, where the power of $x$ is a multiple of 2, 3, 5, and 7, and therefore satisfies all of the conditions above.
If we wish to further add $N/7$ as a seventh power, then we would need to look for some set where $2 | (a-1, b, c, d) $,  $3 | (a, b-1, c, d)$, etc.  This means that eg $b, c, d$ are even, $a, c, d$ are multiples of 3, etc.  Here, we write the modulus in list form, where the modulus is taken over the corresponding point in the mod function.
We see that in the first row, $a$ must give a remainder of 0, when dividided by 3, 5, 7, so it must be a multiple of the product of $3 \cdot 5 \cdot 7$, that leaves a remainder of $1$, when divided by $2$.  


*

*a = 1, 0, 0, 0,  mod(2, 3, 5, 7), is a multiple of 105, gives $a=105$

*b = 0, 1, 0, 0,  mod(2, 3, 5, 7), is a multiple of 70, gives $b=70$

*c = 0, 0, 1, 0,  mod(2, 3, 5, 7), is a multiple of 42, gives $c=126$

*d = 0, 0, 0, 1,  mod(2, 3, 5, 7), is a multiple of 30, gives $d=120$

*e = 0, 0, 0, 0,  mod(2, 3, 5, 7), is a multiple of 210, gives $e=210$


So we can write this as $2^{105} 3^{70} 5^{126} 7^{120} x^{210}$, for all values of $x$.  Note that we don't need to write $x^{210y}$, since the range of $x$ already includes $x^y$.
A: I think the hint was meant to be used like this :
If $N$ is such a number, then $7^{30}N$ is another such number. Hence if you can get one solution, you get infinitely many solutions.
A: Hint:
Since you are dealing with some factors of $2,3,5$ it is reasonable to look for the solution in the form $N=2^{\alpha}3^{\beta}5^{\gamma}.$ The condition, that $N/2$ is a perfect square leads to $\alpha-1=0(\mod 2),$ $\beta=0(\mod 2)$ and $\gamma=0(\mod 2).$ Do the same for the other powers too.
A: I've seen this recently. Here is my solution. Let $n$ be of form, $n=2^\alpha 3^\beta 5^\gamma$. Note that, 


*

*$n/2$ being perfect square implies $\alpha\equiv 1\pmod{2}$, $\beta,\gamma\equiv 0\pmod{2}$.

*$n/3$ being perfect cube implies $\alpha\equiv 0\pmod{3}$, $\beta\equiv 1\pmod{3}$, and $\gamma\equiv 0\pmod{3}$.

*Finally, $n/5$ being perfect fifth power implies $\alpha\equiv 0\pmod{5}$, $\beta\equiv 0\pmod{5}$, and $\gamma\equiv 1\pmod{5}$. 


Thus, we are looking for triples $(\alpha,\beta,\gamma)$ obeying the conditions above simultaneously. However, for each element (say $\alpha$), we have three congruences with modulo coprime numbers, therefore, by Chinese remainder theorem, there exists infinitely many such triples, thus, infinitely many such $n$'s. 
