If $x$ and $y$ are integers and $12x = 5y^2$,
- What could be the possible multiples of $x$ ?
- What can we infer about $y$ ?
I have approached this problem like this :
\begin{align} x = \dfrac{5y^2}{12} \quad \quad \text{here } \ 12 = 2 \cdot 2 \cdot 3\\ x = \dfrac{5y^2}{2 \cdot 2 \cdot 3} \end{align}
I'm able to identify that $y^2$ should definitely have $2 \cdot 2\cdot 3$ so that $x$ can be an integer. But then I'm not able to think of the next step. Like I'm not able to explain to myself.
Can someone help me regarding this and how should I approach these problems.
Since $5$ is relatively prime to $(12)$, $y^2$ will have to pick up the slack on $12 = 2^2 \times 3$.
So, $y$ must be a multiple of $2$, so that $4|y^2.$
However, since $y^2$ also must be a multiple of $3$, $y$ must be a multiple of $3$.
So, you have deduced that $y$ must be a multiple of $6$.
At this point, there are two ways to conclude the problem.
The first is to realize that if $y = 6k$, then $12x = 180k^2$, so at a minimum, $x$ must be a multiple of $15$.
The alternative approach is to recognize that $5y^2$ will necessarily be divisible by both $5$ and $3^2$, with $12$ only being divisible by $3$, not $3^2.$
So, $x$ has to pick up the slack, and be divisible by both $3$ and $5$.