[proofplan] We apply [Euler's Criterion](/theorems/1715) to each of $ab$, $a$, and $b$, then use exponent rules modulo $p$ to match the two sides. The resulting congruence mod $p$ between integers in $\{-1, 0, 1\}$ upgrades to an equality because these three values are pairwise distinct mod $p$ (here we use $p \ge 3$, i.e., $p$ is an odd prime). [/proofplan]

[step:Apply Euler's Criterion to each Legendre symbol] By [Euler's Criterion](/theorems/1715), for any $c \in \mathbb{Z}$ and odd prime $p$, \begin{align*} \left(\tfrac{c}{p}\right) \equiv c^{\frac{p-1}{2}} \pmod p. \end{align*} Applying this to $c = ab$, $c = a$, and $c = b$ respectively: \begin{align*} \left(\tfrac{ab}{p}\right) &\equiv (ab)^{\frac{p-1}{2}} \pmod p, \\ \left(\tfrac{a}{p}\right) &\equiv a^{\frac{p-1}{2}} \pmod p, \\ \left(\tfrac{b}{p}\right) &\equiv b^{\frac{p-1}{2}} \pmod p. \end{align*} [/step]

[step:Combine using multiplicativity of powers] Since $\mathbb{Z}/p\mathbb{Z}$ is a commutative ring, $(ab)^k = a^k b^k$ for every $k \ge 0$, and in particular for $k = (p-1)/2$: \begin{align*} (ab)^{\frac{p-1}{2}} = a^{\frac{p-1}{2}} b^{\frac{p-1}{2}}. \end{align*} Chaining congruences from the previous step, \begin{align*} \left(\tfrac{ab}{p}\right) \equiv (ab)^{\frac{p-1}{2}} = a^{\frac{p-1}{2}} b^{\frac{p-1}{2}} \equiv \left(\tfrac{a}{p}\right)\left(\tfrac{b}{p}\right) \pmod p. \end{align*} [/step]

[step:Upgrade the congruence mod $p$ to equality in $\mathbb{Z}$]Both sides of the congruence \begin{align*} \left(\tfrac{ab}{p}\right) \equiv \left(\tfrac{a}{p}\right)\left(\tfrac{b}{p}\right) \pmod p \end{align*} lie in $\{-1, 0, 1\} \subset \mathbb{Z}$. We claim these three values are pairwise distinct modulo $p$. The differences between any two are $\pm 1$ or $\pm 2$; none of these is divisible by $p$ because $p \ge 3$: indeed $p \mid 1$ is impossible for $p > 1$, and $p \mid 2$ would force $p = 2$, contradicting $p$ odd. Hence the residue classes of $-1, 0, 1$ modulo $p$ are distinct, and two elements of $\{-1, 0, 1\}$ are congruent mod $p$ iff they are equal as integers. Consequently \begin{align*} \left(\tfrac{ab}{p}\right) = \left(\tfrac{a}{p}\right)\left(\tfrac{b}{p}\right), \end{align*} completing the proof.[/step]

[guided]We have the congruence \begin{align*} \left(\tfrac{ab}{p}\right) \equiv \left(\tfrac{a}{p}\right)\left(\tfrac{b}{p}\right) \pmod p \end{align*} from Step 2, but the theorem asserts **equality** in $\mathbb{Z}$, not merely congruence. The bridge is that both sides are elements of $\{-1, 0, 1\}$ by definition of the Legendre symbol. So we need: if $u, v \in \{-1, 0, 1\}$ with $u \equiv v \pmod p$, then $u = v$. Equivalently: the three integers $-1, 0, 1$ represent distinct residue classes in $\mathbb{Z}/p\mathbb{Z}$. The pairwise differences are \begin{align*} 0 - (-1) = 1, \qquad 1 - 0 = 1, \qquad 1 - (-1) = 2. \end{align*} For these differences to be divisible by $p$, we would need $p \mid 1$ (impossible for any prime) or $p \mid 2$. The second forces $p = 2$, which contradicts the standing hypothesis that $p$ is odd. So all three differences are non-zero mod $p$, and the three residue classes are distinct. This is a delicate but essential point: **oddness of $p$** is used here in the upgrade from congruence to equality. For $p = 2$, the Legendre symbol is not usually defined (and when extended, multiplicativity would require separate checking). The case $p$ odd is the full content of the theorem. It is worth noting that if either $p \mid a$ or $p \mid b$, then both sides of the conclusion vanish: $\left(\tfrac{a}{p}\right) = 0$ (say) forces the right-hand side to be $0$, and $p \mid ab$ forces the left-hand side to be $0$. The nontrivial content is when $(a, p) = (b, p) = 1$, where both sides lie in $\{\pm 1\}$ and the argument above reduces to the simpler fact that $1 \not\equiv -1 \pmod p$ (again because $p \ne 2$).[/guided]