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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?

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2 Answers 2

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.

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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 books.google.com 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
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@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

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