Show that there are infinitely many positive integers A such that 2A is a square, 3A is a cube and 5A is a fifth power.

Question

Show that there are infinitely many positive integers A such that 2A is a square, 3A is a cube and 5A is a fifth power.

Using some arithmetic, I felt that if $A = 2^{15k}3^{20k}5^{24k}$ then it might be possible. I have a vague feeling that this can be proved using Unique Factorisation.

I would love hints. Please do not use the Chinese Remainder Theorem here(I felt that you could use it.)

Answer 1

Consider $A=2^{15}3^{20}5^{24}k^{30}$

Consider $\left(2A\right)^\frac12=\left(2^{16}3^{20}5^{24}k^{30}\right)^\frac12=2^83^{10}5^{12}k^{15}$

Consider $\left(3A\right)^\frac13=\left(2^{15}3^{21}5^{24}k^{30}\right)^\frac13=2^53^75^8k^{10}$

Consider $\left(5A\right)^\frac15=\left(2^{15}3^{20}5^{25}k^{30}\right)^\frac15=2^33^45^5k^6$

As $k$ can be any integer then there are an infinite number of solutions.

Answer 2

Let $A=2^{15}3^{20}5^{24}7^{30k}$. This works for every $k\geq 1$.

Answer 3

Your approach is reasonable, but for $k=2$ we have $2A=2^{31}\cdot$ something, which is not a square. Your base case works. Then you need to multiply by something that is a $30^{\text{th}}$ power to maintain the property you seek. In essence the statement that you multiply by a $30^{\text{th}}$ power is using the Chinese Remainder theorem-we get that by multiplying $2,3,5$ together.