TITLE: Irreducible polynomials with constrained coefficients
QUESTION [18 upvotes]: Over at the Cafe, after reading about TWF 285, I asked more-or-less 

About how many polynomials with coefficients in $\{\pm 1\}$ and of degree $d$ are irreducible?

and that's what I want to ask here.
The first no-go analysis: since if $P$ is reducible, then it is reducible mod $3$, we get that the number of reducible such is $O(\frac{d-1}{d}3^d)$; but that's clearly too large already to help much; reducing mod $2$ we can't distinguish polynomials anymore!

REPLY [20 votes]: Your problem is hard, but here are some things that can actually be proved!
Let $S_d$ be the set of polynomials of degree $d$ with all $d+1$ coefficients in $\{\pm 1\}$.
1) $|S_d| \gg 2^d/d$ as $d \to \infty$ through even integers.
Proof: By Theorem 2 in a paper by Brillhart, Filaseta, and Odlyzko, $f$ will be irreducible if
(a) $f(2)$ is prime,
(b) $f(1) \ne 0$, and
(c) all complex zeros of $f$ have absolute value less than $3/2$.
The condition that $d$ is even guarantees (b).
To guarantee (c), restrict attention to $f$ whose first $100$ coefficients are $+1$, and use Rouch\'e's theorem to show that $f$ has the same number of zeros satisfying $|z|<3/2$ as does $x^d+x^{d-1}+\cdots+1$, i.e., all $d$ of them.  The values of these $f$ at $2$ are the odd integers in a certain interval, so the prime number theorem gives the required number of $f$ satisfying (a).  $\square$
2) The Generalized Riemann Hypothesis implies that there are infinitely many $d$ for which every $f$ in $S_d$ is irreducible.
Proof: GRH implies that there are infinitely many primes $p$ for which $2$ is a primitive root.  If $d+1$ is such a prime, then $x^d+x^{d-1}+\cdots+1$ is irreducible mod $2$, so every $f \in S_d$ will be irreducible over $\mathbf{Z}$.  $\square$
3) There exist infinitely many $d$ for which at least $50\%$ of the polynomials in $S_d$ are irreducible.
Proof: Let $d=2^n-1$ for any $n \ge 1$.  If $f \in S_d$, then $f(x+1) \equiv x^d \pmod{2}$. Thus $f(x+1)$ is Eisenstein at $2$ half of the time.  $\square$
A final remark: It is conjectured that the fraction of irreducible polynomials among those of degree $d$ with 0,1 coefficients tends to $1$ as $d \to \infty$: see this paper by Konyagin.  The same is likely true for your problem, as Greg suggests, but it also seems that the best you can hope for at the moment are partial results such as the ones above.