Question

Prove that there are infinitely many positive integer as such that 2a is a square, 3a is a cube and 5a is a fifth power.

Answer 1

Solution :-

First we observe that 2,3,5 divide A.

So we may take $A=2^{\alpha }{3}^{\beta }{5}^{\gamma }$

Consisdering $2A,3A$ and $5A$, we observe that $\alpha +1,\beta ,\gamma$

are divisible by 2; $\alpha ,\beta +1,\gamma$ divisible by 3; and

$\alpha ,\beta ,\gamma +1$ are divisible by 5.

We can chhose $\alpha =15+30n$, $\beta =20+30n$ and $\gamma =24+30n$.

As n varies over the set of natural numbers,

We get infinite set of numbers.