Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose we have a polynomial with degree $n$ and all the coefficient are integers, with the leading coefficient $+1$ or $-1$. Are the roots only either integer or complex number?

share|cite|improve this question

If by "complex" you mean "non-real", the answer is no. For example, $x^2-2$ satisfies your conditions, and it has two noninteger real roots.

What is true is that any rational root must be an integer; this follows from the Rational Root Theorem, since given $$f(x) = a_nx^n+\cdots + a_1x+a_0,\qquad a_i\in\mathbb{Z},\ a_n\neq 0,$$ if a rational number $r=\frac{p}{q}$ written in lowest terms ($p,q\in\mathbb{Z}$, $\gcd(p,q)=1$) is a root of $f(x)$, then $p|a_0$ and $q|a_n$. In particular, if $a_n=\pm 1$, then $q=\pm 1$, so $r\in\mathbb{Z}$.

That is, the roots of the polynomial you describe are either integers, or irrational.

share|cite|improve this answer
or non-real ... – David Mitra Mar 19 '12 at 17:21
@David Above, "irrational" is used not only for reals but generally, i.e. it simply means "not rational", i.e. $\not\in \mathbb Q$. This is quite common usage in university-level algebra. So there is no need to add the remark "or non-real". – Bill Dubuque Mar 19 '12 at 17:29
@David See also my 2006/3/8 sci.math post where I note that if one searches for "irrational algebraic" you'll find such usage by many eminent mathematicians: e.g. Conway, Gelfond, Manin, Ribenboim, Shafarevich, Waldschmidt (esp. in diophantine approximation, e.g. Thue-Siegel-Roth theorem, Gelfond-Schneider theorem, etc). See also other posts in that thread "Is i irrational"? – Bill Dubuque Mar 19 '12 at 17:39
@BillDubuque Yes, that makes sense of course. In fact, all the usual definitions of "irrational" I can find are "a real number is irrational if..."; whereas here "irrational" can be interpreted in a literal sense, that is, "not rational", as you mentioned in your first comment. Should I delete the comment? – David Mitra Mar 19 '12 at 17:51
@David Probably the remarks will prove helpful to other readers who may not be aware of this more general denotation of "irrational". – Bill Dubuque Mar 19 '12 at 18:01

This is simply the Rational Root Test specialized to monic polynomials, i.e. leading coefficient $= 1$. The RRT is usally proved by evaluating at a reduced fraction $\rm\:a/b,\:$ clearing denominators, then noting this implies $\rm\:b\ |\ a^n,\:$ contra $\rm\:gcd(a,b) = 1,\:$ via unique factorization, or Euclid's Lemma.

However, it is worth remarking that this has an alternative proof by induction on degree, by constructively proving that: if $\rm\:a/b\:$ is a root of a monic $\rm\:f_n\in \mathbb Z[x]\:$ of degree $\rm\:n,\:$ then $\rm\:a/b\:$ is a root of a monic $\rm\:f_{n-1}\in \mathbb Z[x]\:$ of degree $\rm\:n-1.\:$ Thus, by induction, $\rm\:a/b\:$ is a root of a linear monic $\rm\:x - n\in\mathbb Z[x],\:$ hence $\rm\:a/b = n\in \mathbb Z.\:$ For example, below is the reductive proof of irrationality of cube-roots to square-roots. The general case is no more difficult, see here.

THEOREM $\ $ If $\rm\ r^3\: =\: m\in \mathbb Z\ $ then $\rm\ r\in \mathbb Q\ \Rightarrow\ r\in\mathbb Z$

Proof $\quad \rm r = a/b \in \mathbb Q,\ \ \gcd(a,b) = 1\ \Rightarrow\ ad-bc \;=\; 1\;$ for some $\:\rm c,d \in \mathbb{Z}\;\;$ by Bezout.

Thus $\rm\ 0\: =\: (a-br)\: (dr^2+cr) \: =\: r^2 + ac\ r - bdm \ $ so $\rm\ r\in\mathbb Z\ $ by the quadratic case. $\ $ QED

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.