Lagrange's theorem (number theory) explained

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

stylef\inZ[x]

, either:

p\midf(x)

has at most solutions in,

where is the degree of .

This can be stated with congruence classes as follows: for all polynomials

stylef\in(Z/pZ)[x]

with p prime, either:

f(x)=0

has at most solutions in

Z/pZ

.

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

Proof

Let

stylef\inZ[x]

be an integer polynomial, and write the polynomial obtained by taking its coefficients . Then, for all integers x,

f(x)\equiv0\pmod{p}\Longleftrightarrowg(x)\equiv0\pmod{p}

.

Furthermore, by the basic rules of modular arithmetic,

f(x)\equiv0\pmod{p}\Longleftrightarrowf(x\pmod{p})\equiv0\pmod{p}\Longleftrightarrowg(x\pmod{p})\equiv0\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 then f has no roots and the statement is true.

If without roots then the statement is also trivially true.

Otherwise, and f has a root

k\inZ/pZ

.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

styleg\in(Z/pZ)[x]

(of degree lower than that of f) and of a constant

styler\inZ/pZ

(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 in number. This proves the result.

References

. William J. LeVeque . Topics in Number Theory, Volumes I and II . Dover Publications . New York . 2002 . 1956 . 978-0-486-42539-9 . 1009.11001 . 42 .