If a number a has no prime factors, what numbers can a be? Can someone explain to me why a must equal 0,1, or -1?
EDIT: Can someone give me a hint for an alternative way of solving this without using Rational Root Theorem?
 A: Let’s see whether I can help here.

*

*I hope you know that every integer $\ge2$ is divisible by a prime.


*Therefore, every integer $\le-2$ is divisible by a prime.


*Consequence? The only integers that can possibly be without prime divisors are $-1,0,1$.
If any question about any of these three, get back to us.

EDIT — Addendum
You have asked for another proof. Here’s one that you may find unsatisfactory, ’cause it depends totally on the special form of this polynomial $f(x)=x^3+x+1$.
You want to show that there is no integer $q$ such that $f(q)=0$.
Certainly the case $q=0$ gives us no trouble, since $f(0)=1$.
Now consider integer $q\ge1$. Then $q^3\ge1$ also, so $q^3+q+1\ge1$, and is thus not zero.
Finally, the case $q\le-1$. Then $q^3\le q\le-1$, and $q+1\le0$, so that $q^3+q+1\le-1$, not zero.
There you have it, a proof that does not use “arithmetic” in the sense that mathematicians use the word, i.e. questions involving primes and related matters. It does use inequalities, and if your course is a start-from-zero treatment of number theory, your TA might not find this acceptable, especially if you haven’t dealt with inequalities in the course.
A: The number $0$ is divisible by every prime number (in fact, by every nonzero integer). Also, the Fundamental Theorem of Arithmetic guarantees that every positive integer greater than $1$ (and hence also every negative integer less than $-1$) is divisible by at least one (positive) prime number. So, $1$ and $-1$ are the only integers that are not divisible by any prime number.
