Lagrange's theorem (number theory)

In number theory, Lagrange's theorem is a statement named after Joseph-Louis Lagrange about how frequently a polynomial over the integers may evaluate to a multiple of a fixed prime p. More precisely, it states that for all integer polynomials $$\textstyle f \in \mathbb{Z}[x]$$, either:
 * every coefficient of $f$ is divisible by p, or
 * $$ p \mid f(x) $$ has at most $deg f$ solutions in $\{1, 2, ..., p\}$,

where $deg f$ is the degree of $f$.

This can be stated with congruence classes as follows: for all polynomials $$\textstyle f \in (\mathbb{Z}/p\mathbb{Z})[x]$$ with p prime, either:
 * every coefficient of $f$ is null, or
 * $$ f(x)=0 $$ has at most $deg f$ solutions in $$ \mathbb{Z}/p\mathbb{Z} $$.

If p is not prime, then there can potentially be more than $deg f(x)$ solutions. Consider for example p=8 and the polynomial f(x)=x-1, where 1, 3, 5, 7 are all solutions.

Proof
Let $$\textstyle f \in \mathbb{Z}[x]$$ be an integer polynomial, and write $g &isin; (Z/pZ)[x]$ the polynomial obtained by taking its coefficients $mod p$. Then, for all integers x,

$$ f(x) \equiv 0 \pmod{p} \quad\Longleftrightarrow\quad g(x) \equiv 0 \pmod{p} $$.

Furthermore, by the basic rules of modular arithmetic,

$$ f(x) \equiv 0 \pmod{p} \quad\Longleftrightarrow\quad f(x \pmod{p}) \equiv 0 \pmod{p} \quad\Longleftrightarrow\quad g(x \pmod{p}) \equiv 0 \pmod{p} $$.

Both versions of the theorem (over Z and over Z/pZ) are thus equivalent. We prove the second version by induction on the degree, in the case where the coefficients of f are not all null.

If $deg f = 0$ then f has no roots and the statement is true.

If $deg f ≥ 1$ without roots then the statement is also trivially true.

Otherwise, $deg f ≥ 1$ and f has a root $$ k\in\mathbb{Z}/p\mathbb{Z} $$. The fact that Z/pZ is a field allows to apply the division algorithm to f and the polynomial X-k (of degree 1), which yields the existence of a polynomial $$\textstyle g \in (\mathbb{Z}/p\mathbb{Z})[x]$$ (of degree lower than that of f) and of a constant $$\textstyle r \in \mathbb{Z}/p\mathbb{Z}$$ (of degree lower than 1) such that

$$ f(x) = g(x) (x-k) + r.$$

Evaluating at x=k provides r=0. The other roots of f are then roots of g as well, which by the induction property are at most $deg g ≤ deg f - 1$ in number. This proves the result.