Let
\[f(x) = a_0 + a_1 x + ... + a_n x^n\]
be a polynomial with integer coefficients. Suppose a prime \[p\]
divides each of \[a_0, a_1, ..., a_{n-1}\]
(every coefficient except coefficient of \[x^n\]
), and that \[p^2\]
does not divide \[a_0\]
. Then \[f(x)\]
has no factors with integer coefficients.Proof
Suppose
\[f = g h\]
for polynomials \[g, h\]
with integer coefficients. Look at this factorization modulo \[p\]
: we get \[f(x) = a_n x^n\]
, so \[g(x) = b_d x^d\]
, \[h(x) = c_e x^e\]
for some constants \[b_d, c_e\]
and for some integers \[d, e\]
with \[d e = n\]
. This implies the constant term of \[g(x)\]
is a multiple of \[p\]
, and similarly for the constant term of \[h(x)\]
, hence \[p^2\]
divides the constant term of \[f(x)\]
, a contradiction.Example
Take
\[f(x)=x^3+4x^2+10x+14\]
2 divides every coefficient except the coefficient of
\[x^3\]
and \[2^2=4\]
does not divide the constant term 14. Therefore \[f(x)\]
does not factorise.