Question

Let $n={p_1}^{r_1}{p_2}^{r_2}...{p_k}^{r_k}$, where, $p_1,p_2,..,p_k$are distinct primes and each $r_i\ge 0$. Prove that n is a perfect square if and only if each $r_1$ is even.

Short answer

It is proved that if all the $r_i$ are even then m can be defined as

$m=\prod _{i=1}^k{p_i}^{\frac{r_i}{2}}$such that $n=m^2$.

Step-by-step solution

Define m

By definition,

If n is a perfect square, then there is some m ,and here, $m^2=n$,where$m|n$.

Consider the $m=\prod _{i=1}^k{p_i}^{s_i}$.

Then, $m^2=n$ is equivalent to $2s_i=r_i$for all $1\le i\le k$. And all the $r_i$ are even.

Step 2: Prove that n is a perfect square if and only if each r1 is even

If all the $r_i$ are even, then m can be defined as

$m=\prod _{i=1}^k{p_i}^{\frac{r_i}{2}}$such that $n=m^2$.

Therefore, it is proved that n is a perfect square if and only if each $r_i$ is even.