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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.)

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  • $\begingroup$ But are the square roots,cube roots also integers then only $0$ seems to be a solution $\endgroup$ – Archis Welankar Mar 17 '16 at 14:30
  • $\begingroup$ If $x^5$ is an integer then is the fifth root only integer then there arent infinite solutions $\endgroup$ – Archis Welankar Mar 17 '16 at 14:33
  • $\begingroup$ Hint: your example works for $k=1$. Then multiply by something that is a square, a cube, and also a fifth power. $\endgroup$ – Catalin Zara Mar 17 '16 at 14:34
  • $\begingroup$ @Dhruv, $2^{60}=\left(2^{30}\right)^2$. Isn't it a perfect Square? $\endgroup$ – user249332 Mar 17 '16 at 14:39
  • $\begingroup$ @CatalinZara A mistake by me. The number should have been $2^{31}$ which is not a perfect square. $\endgroup$ – TheRandomGuy Mar 17 '16 at 14:41
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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.

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

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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.

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  • $\begingroup$ Well, he can consider all positive integers $k$ such that $k\equiv 1\mod 30$. So, there are infinitely many such integers $A$. $\endgroup$ – Ángel Valencia Mar 18 '16 at 7:53
  • $\begingroup$ @ÁngelValencia: absolutely. That is multiplying by $30^{30k}$, but we can also multiply by $7^{30k}$ as Marco Flores says. $\endgroup$ – Ross Millikan Mar 18 '16 at 14:47
  • $\begingroup$ What I say is that you have always a number $A=2^{15k}3^{20k}5^{24k}$ with the property given if $k$ satisfies $k\equiv 1\mod 30$. $\endgroup$ – Ángel Valencia Mar 18 '16 at 15:32
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    $\begingroup$ @ÁngelValencia: but there are many more of the form $2^a3^b5^c$ It should be $a \equiv 15, b \equiv 20, c \equiv 25 \pmod {30}$ You don't have to use the same $k$ in your expression. Your expression is sufficient to solve the problem in that it shows there are an infinite number. $\endgroup$ – Ross Millikan Mar 18 '16 at 15:46

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