Find the natural numbers so that $n=2a^2 ,n=3b^3 ,n=5c^5$.

Question

Well here it is i spend almost 3 hours on this one!! Find the general form of the natural numbers that are twice a square ,tripple of a cube and 5 times a 5-ith power.Who is the smaller of them?.What i understood is .Find natural numbers so that $n=2 a^2$, $n=3 b^3$, $n=5 c^5$ where $a,b,c$ natural numbers.! My progress so far is that i took the canonical form for a supposed $N$ also said that $2, 3, 5$ must be in the prime factorization and also used a lemma where it says for a number $A$ to be a $n$-th power of number all of the power's from its prime factorization must be divisible by that power.So what I got is $n$ to be $n=2^{z_1} \cdot 3^{z_2} \cdot 5^{z_5} \cdot \dotsb \cdot p_k^{z_i}$.... where $z_1>=15$ and $z_1-1$ is even $z_2>=10$ and $z_2=3k+1$, $z_3>=6$ and $z_3=5l+1$ and the rest $z_i$ must be of the form $30 h_i$!That is all, I'm kinda lost and don't know how to continue and corner it so i get a complete for and not a so general one.Help please.(the $z_i$ are the power's of the prime numbers in the prime factorization)

Answer 1

The smallest such number is clearly of the form $2^a3^b5^c$, and once you work out the values of $a,b,c$ that work then whatever number $n$ you get as a minimal solution, you will have that $n m^{30}$ is also a solution for any $m \geq 1$. This characterizes all solutions by the Chinese remainder theorem.

To figure out the valid smallest power of $2$, note that the power $a$ must be odd because your number is $2$ times a square. Furthermore $a$ must be divisible by $3$ and $5$. So $a$ is an odd number divisible by $15$. So the smallest possible $a$ is $15$. Similarly you can work out the conditions on $b$ and figure out that $b = 10$, and that $c = 6$. Thus the set of all solutions is $2^{15}3^{10}5^6 m^{30}$ for any $m \geq 1$, and the smallest such solution is when $m = 1$.