Exercise 1.15

Proof. Suppose $a=b^2,b\in \mathbb{Z}$. Then ${\mathrm{ord}}_p(a)=2{\mathrm{ord}}_p(b)$ is even for all primes $p$.

Conversely, suppose that ${\mathrm{ord}}_p(a)$ is even for all primes $p$. We must also suppose $a>0$. Let $a=\underset{p}{\prod <!--\; FUNCTION\; APPLICATION\; -->}{p}^{a(p})$ the decomposition of $a$ in primes. As $a(p)$ is even, $a(p)=2b(p)$ for an integer $b(p)$ function of the prime $p$. Let $b=\underset{p}{\prod <!--\; FUNCTION\; APPLICATION\; -->}{p}^{b(p)}$. Then $a=b^2$.

With a similar proof, we obtain the following generalization for each integer $a\in \mathbb{Z},a>0$ :

$a=b^n$ for an integer $b\in \mathbb{Z}$ iff $n\mid {\mathrm{ord}}_p(a)$ for all primes $p$. □