 Solution

Let $G={\mathbb{F}}_p^{\times }$, and let $m$ be the least common multiple of the orders of its elements. For every prime power $q^a$ dividing $m$, some element of $G$ has order divisible by $q^a$, and a suitable power of it has order exactly $q^a$. Multiplying these elements over the distinct primes produces, because their orders are coprime, an element of order $m$.

Every element of $G$ is a root of $X^m-1$. A nonzero polynomial of degree $m$ over a field has at most $m$ roots, so $p-1\le m$. On the other hand, every element order divides $p-1$ by Lagrange's theorem, so $m\le p-1$. Hence $m=p-1$, and

The squaring homomorphism has kernel $\{1,-1\}$ because $p$ is odd. Its image $H$ therefore has order $(p-1)/2$ and index two. If $p=3$, then $3=0$ is already a square. Otherwise $2,3\in G$. In the two-element quotient $G/H$, either $2H=H$, or $3H=H$, or both are the nontrivial coset, in which case $6H=H$. Thus one of $2,3,6$ is a square modulo $p$.

Finally,Whichever of $2,3,6$ is a square supplies a root, soThis is the index-two square-class argument for three related residues.