Proof. We first note that if $m$ has a primitive root, then $\lambda(m)=\phi(m)$. From problem 5 of Section 6.1 , we know that $\phi(m)$ is even, so that $\phi(m) / 2$ is an integer, if $m>2$. Euler's Theorem tells us that
$$a^{\phi(m)}=\left(a^{\phi(m) / 2}\right)^{2} \equiv 1(\bmod m)$$
for all integers $a$ with $(a, m)=1$. From problem 7 of Section 8.3, we know that when $m$ has a primitive root, the only solutions of $x^{2} \equiv 1(\bmod m)$ are $x \equiv \pm 1(\bmod m)$. Hence,
$$a^{\phi(m) / 2} \equiv \pm 1(\bmod m)$$

This implies that
$$\lambda_{0}(m) \leqslant \phi(m) / 2$$

Now let $r$ be a primitive root of modulo $m$ with $\pm 1$ - exponent $e$. Then
$$r^{e} \equiv \pm 1(\bmod m)$$
so that
$$r^{2 e} \equiv 1(\bmod m)$$

Since $\operatorname{ord}_{m} r=\phi(m)$, Theorem 8.1 tells us that $\phi(m) \mid 2 e$, or equivalently, that $(\phi(m) / 2) \mid e$. Hence, the maximum $\pm 1-$ exponent $\lambda_{0}(m)$ is at least $\phi(m) / 2$. However, we know that $\lambda(m) \leqslant \phi(m) / 2$. Consequently, $\lambda_{0}(m)=\phi(m) / 2=\lambda(m) / 2$