Website owner:  James Miller

Prove. Let p/q be a rational fraction in lowest terms and

            a₀xⁿ + a₁x^n-1 + a₂x^n-2 + ... + a_n-1x + a_n = 0

be a polynomial equation with integral coefficients. If p/q is a root of the equation, then p is a factor of a_n and q is a factor of a₀.

Proof. Let the given equation be

            a₀xⁿ + a₁x^n-1 + a₂x^n-2 + ... + a_n-2x² + a_n-1x + a_n = 0

If p/q is a root, then

            a₀ (p/q)ⁿ + a₁ (p/q)^n-1 + a₂ (p/q)^n-2 + ... + a_n-2 (p/q)² + a_n-1 (p/q) + a_n = 0

Multiplying both members by qⁿ, we obtain

1)        a₀pⁿ + a₁p^n-1q + a₂p^n-2q² + ... + a_n-2p²q^n-2 + a_n-1pq^n-1 + a_nqⁿ = 0

Let us now write 1) as

            a₀pⁿ + a₁p^n-1q + a₂p^n-2q² + ... + a_n-2p²q^n-2 + a_n-1pq^n-1 = -a_nqⁿ

Written in this form, it is clear that p, being a factor of every term of the left member of the equation, must divide a_nqⁿ. Because p/q is expressed in lowest terms, no factor of p will divide q. Consequently, p must divide a_n. Thus the first part of our assertion is proved.

Let us now write 1) as

            a₁p^n-1q + a₂p^n-2q² + ... + a_n-2p²q^n-2 + a_n-1pq^n-1 + a_nqⁿ = -a₀pⁿ

Using the same type reasoning as above, it follows that q must divide a₀.