Double roots of random polynomials with integer coefficients

We consider random polynomials whose coefficients are independent and identically distributed on the integers. We prove that if the coefficient distribution has bounded support and its probability to take any particular value is at most $\tfrac12$, then the probability of the polynomial to have a double root is dominated by the probability that either $0$, $1$, or $-1$ is a double root up to an error of $o(n^{-2})$. We also show that if the support of coefficient distribution excludes $0$ then the double root probability is $O(n^{-2})$. Our result generalizes a similar result of Peled, Sen and Zeitouni for Littlewood polynomials.


Introduction
Let n ∈ N and let (ξ j ) 0≤j be a sequence of independent and identically distributed random variables taking values in Z. Define the random polynomial P = P n by P (z) := n j=0 ξ j z j . (1.1) In a previous paper, Peled, Sen and Zeitouni [13] showed that if the random variables are supported on {−1, 0, +1} with max x∈{−1,0,1} P(ξ 0 = x) < 1 √ 3 , then the probability of P to have a double root in the complex plane is same as having a double root at 0, ±1 up to an error of o(n −2 ). In this paper, we extend the result for more general integer-valued random variables. Our main result is the following.   (1.3) Then we have P P has a double root = P P has a double root at either 0, −1 or 1 + o(n −2 ), (1.4) as n → ∞. Moreover, if P(ξ 0 = 0) = 0, then P P has a double root = O(n −2 ).
We make a few remarks about the above theorem.
(a) When P(ξ 0 = 0) = 0, the upper bound in Theorem 1.1 is sharp. When ξ i 's are i.i.d. ±1 symmetric Bernoulli and (n + 1) is divisible by 4, then it was shown in [13] that the probability of having a double root is Θ(n −2 ).
(b) We can have a better error bound if we allow the possibility of having double roots at other low-degree roots of unity. More precisely, our proof can be modified to show that for any fixed d ≥ 1, P P has a double root = P P has a double root at 0 or some roots of unity of degree at most d + o(n −2d ).
(c) The bounded support Condition (1.2) can be weaken with minor modifications of our arguments. We did not pursue that here for the sake of simplicity. On the other hand, we do not know how to relax Condition (1.3) on the size of the maximum of atom and it seems that the current bound 1 2 is a limitation of our proof. In fact, we believe that both conditions are unnecessary and that the result (1.4) should hold for any non-degenerate integer-valued coefficient distribution.
(d) Even though some parts of our proof closely follow the lines of arguments from the paper of Peled, Sen and Zeitouni [13], extending the result to general integer-valued coefficients, however, poses a few significant challenges. For example, to handle high-degree double roots, we need a key anti-concentration estimate for P (±2) given in the form of Theorem 1.2. When the coefficients are ±1-valued, the map (a 0 , . . . , a n ) → i=1 a i 2 i : {−1, 1} n+1 → Z is one-to-one, which immediately implies the bound that P(P (±2) = m) ≤ max x=±1 P(ξ 0 = x) n+1 .
The paper [13] made use of the above simple observation. But for more general integer-valued coefficients, we lose such one-to-one property, which makes proving Theorem 1.2 nontrivial. One consequence of this difficulty is that the result here requires the maximal atom of the coefficient distribution to be at most 1 2 while in [13] atoms up to 1 √ 3 could be handled.
Moreover, the argument used in [13] to deal with low degree roots does not carry over either. In [13], P was always a monic polynomial and hence its roots were algebraic integers. For algebraic integers, one can use some partial result (see, e.g., Dobrowolski [5]) on Lehmer's conjecture to show that any non-zero algebraic integer is either a root of unity or has a conjugate which is a bit far (depending on its degree) away from unit circle. In low degree case, [13] made use of this dichotomy of algebraic integers. In contrast, in our case we also have to deal with non-monic P , so its roots are algebraic numbers in general. Such dichotomy is not available for algebraic numbers. For example, there are algebraic numbers which are not a root of unity and all of its conjugates lie on the unit circle. This requires new methods for handling such roots which are given in sections 5 and 6.
EJP 22 (2017), paper 10. A key instrument in the proof of the theorem, which may be of independent interest, is the following anti-concentration bound.
We proceed as follows. In Section 1.1 we provide some notation and reduce Theorem 1.1 to several key lemmata. In Section 2 we prove Theorem 1.2. Each subsequent section is then dedicated to the proof of one of the key lemmata stated in Section 1.1.

Proof overview
Preliminaries. Recall that a real number α is called algebraic if it is a root of a polynomial with rational coefficients. Let A denote the set of algebraic numbers. The minimal polynomial of α ∈ A is the unique least degree monic polynomial in Q[X] with a root at α. The algebraic degree of α is the degree of the minimal polynomial of α, which we denote by deg(α). A real number α is said to be an algebraic integer if all the coefficients of its minimal polynomial are integers.
We define Λ(α), the house of α, by where α 1 = α, . . . , α deg(α) are the conjugates of α, i.e., the roots of the minimal polynomial of α. We further define the associated minimal polynomial of α in Z[x] to be the unique polynomial in Z[x] of degree deg(α) with a root at α, whose leading coefficient is positive and whose coefficients are coprime. We first consider the probability of having a double root of algebraic degree and prove the following result. Lemma 1.3 (high degree). Given any B > 0, there exists a constant C 0 > 0 such that P(P has a double root α with deg(α) ≥ C 0 log n) = O(n −B ).
The proof of Lemma 1.3 follows the line of arguments given in [13], which, in turn, was based on idea that appeared in a work of Filaseta and Konyagin [7]. However, several modifications are needed when dealing with general integer-valued coefficients. Most crucially, we need a new anti-concentration bound (Theorem 1.2) that consumes the bulk of our effort. Let us point out here that Theorem 1.2 is the only place where Assumption (1.3) is crucially used.
By virtue of Lemma 1.3, we now have to deal with potential double roots with low algebraic degree, more precisely, with degree at most C 0 log n. In the next lemma we show that the probability that P has a root at an algebraic numbers of low degree such that one of its conjugates lying at a distance of at least Ω((log n) −1 ) from the unit circle is negligible. Lemma 1.4 (low degree roots far away from the unit circle). For every B > 0 and C 0 > 0, there exists C 1 > 0 such that P P has a root α : deg(α) ≤ C 0 log n and Λ(α) > 1 + C1 log n = O(e − Bn log n ).
For the proof, we use a simple sparsification of P to bound the root probability for each fixed low-degree algebraic number lying far away from the unit circle and then employ a rather crude union bound. After Lemma 1.4, we next deal with the low degree double roots with small house (i.e. all of their conjugates lying close to the unit circle). We break this into two cases. First we consider the case when the degree of the root is at least 5 and we show that Lemma 1.5 (low degree roots close to the unit circle). For every C 0 > 0 and C 1 > 0, we have P P has a root α : 4 < deg(α) ≤ C 0 log n and Λ(α) ≤ 1 + C1 log n = o(n −2 ).
From a standard application of inverse Littlewood-Offord type results, it follows that for any fixed algebraic number α of degree at least 5, P(P (α) = 0) = O ε (n −5/2+ε ), for any ε > 0. This is shown in Lemma 5.1. More importantly, to prove Lemma 1.5, we need to count the number of algebraic numbers α such that deg(α) ≤ C 0 log n and Λ(α) ≤ 1+ C1 log n .
Towards this direction, we show in Lemma 5.2 that they are at most o(n ε ) in number for any ε > 0. The counting estimate makes heavy use of a result of Dubickas [6].
Finally, the next lemma takes care of the potential double roots that have degree at most 4 (excluding 0, ±1) and have small house. Lemma 1.6 (roots with degree at most 4). For every C 0 > 0 and C 1 > 0, we have P P has a double root α = 0, ±1 : deg(α) ≤ 4 and Λ(α) It is not hard to see that if α is a root of P for large enough n satisfying the conditions that deg(α) = O(1) and Λ(α) = o(1), then it must be a unimodular root, i.e., all of the conjugates of α must lie on the unit circle. Now if α is a root of unity, we closely follow [13] to bound the probability of having a double root α which involves an application of an anti-concentration bound due to Sárközi and Szemerédi [14]. However, when α is unimodular but not a root of unity, we need a new argument to bound the probability of having a double root at α. In fact, in Lemma 6.1 we show that P(P (α) = 0) = O(n −5/2 ).
The argument relies on a powerful anti-concentration bound by Halász [8].
Clearly, the first assertion of Theorem 1.1 is an immediate consequence of lemmata 1.3, 1.4, 1.5, 1.6. To prove the second assertion of Theorem 1.1, note that since P (ξ 0 = 0) = 0, with probability one, P can not have a root at 0. So, we need to show that P(P has a double root at ± 1) = O(n −2 ).
The above bound follows from an application of an inverse Littlewood-Offord result from [18,Theorem 2.5]. For details, see Lemma A.5 in [4] where the same has been proved under the assumption that ξ 0 has bounded (2 + ε) moment. This completes the proof of Theorem 1.1.

Anti-concentration of P (±2)
In this section we prove Theorem 1.2. As an important first step, we will find a very useful a characterization of integer-valued measures with max-atom bounded by 1 2 in terms of mixture of two-point distributions.

Bernoulli mixture
A probability measure µ is said to be a (unbiased) Bernoulli measure if µ = 1 2 δ a + 1 2 δ b , where a = b ∈ Z and δ x is the Dirac measure at x. A countable mixture of unbiased EJP 22 (2017), paper 10.
Bernoulli measures is simply said to be a Bernoulli mixture. In other words, a probability measure µ is a Bernoulli mixture if it can be written as Note that if the distribution of a random variable ξ is a Bernoulli mixture, then there exists a random vector (I, ∆) on Z × N, such that ξ d = I + B∆, where B is a Ber( 1 2 ) random variable, independent from both I and ∆. With a slight abuse of notation, we will also call such a random variable ξ a Bernoulli mixture.
The following proposition gives a useful characterization for Bernoulli mixtures. Clearly, the necessary part of Proposition 2.1 is trivial. Most of the reminder of Section 2 is dedicated to proving the sufficient part.
Let µ be a non-negative positive finite measure on Z. It induces a unique total order (π µ i ) i∈N on Z such that w µ i := µ(π µ i ) are monotone non-increasing (i.e., w µ i ≥ w µ j if i < j) and π µ i < π µ j if w µ i = w µ j . Then µ can be expressed as We write M for the collection of non-negative finite measures µ on the integers (including the null measure), which satisfy w µ 1 ≤ µ(Z)/2. Also, for any non-null finite measure µ on Z, we denote byμ the normalized probability measureμ(·) := µ(·)/µ(Z).
To prove Proposition 2.1 we use the following couple of lemmata. Lemma 2.2. If µ ∈ M is non-null and µ is supported on at most 3 integers, thenμ is a mixture of at most 3 Bernoulli measures.
Proof. We writeμ = w 1 δ π1 + w 2 δ π2 + w 3 δ π3 where w 1 ≥ w 2 ≥ w 3 and 3 i=1 w i = 1. We then give the explicit decomposition: It is now straightforward to check that each of the weights is non-negative and that equality indeed holds.
where either β is the null measure orβ is a Bernoulli measure, and ν ∈ M and satisfies ν(π µ k ) = 0.
Proof of Proposition 2.1. Write µ 1 for the distribution of ξ. Define a decreasing sequence of finite measures (µ i ) i∈N on Z inductively as follows. Suppose µ i has already been defined and µ i ∈ M. An application of Lemma 2.3 to µ i with k = 4 yields the decomposition µ i = β i + µ i+1 with µ i+1 (π µi 4 ) = 0 where β i is either the null measure orβ i is a Bernoulli measure and µ i+1 ∈ M. This defines the measure µ i+1 . Since (µ i ) i∈N is a decreasing sequence of finite measures, it has a limiting measure (possibly null) which we denote All that remains in order to prove the proposition is to show that µ ∞ is supported on at most 3 integers, and then apply Lemma 2.2.
To that end, assume, if possible that, there exists four distinct integers a 1 , Since µ i ↓ µ ∞ , L i ⊇ L i+1 and a 1 , . . . , a 4 ∈ L i for each i. Thus π µi 4 ∈ L i and hence, by the definition of the measure µ i+1 , we have L i ⊆ L i+1 \ {π µi 4 }. This implies that |L i+1 | < |L i | for each i. Since |L 1 | < ∞, this contradicts the fact that |L i | ≥ 4 for each i. Hence, µ ∞ is supported on at most 3 integers.
Using Proposition 2.1 we may reduce Theorem 1.2 to the following proposition. Proposition 2.4. Let (X i ) 1≤i≤n be i.i.d. random variables whose distribution is a Bernoulli mixture. Then there exists ε > 0 such that for n ∈ N large enough and every sign sequence (σ i ) 1≤i≤n with σ i = ±1, the following holds.

Proof of Proposition 2.4
In this section we prove Proposition 2.4. Throughout the section we fix a sign sequence (σ i ) 1≤i≤n with σ i = ±1. In the course of the proof we shall make several claims whose proofs are given in sections 2.2 and 2.3.
For k ∈ Z we write L(k) for the leading power of 2 in the factorization of k, i.e., L(k) = max{l ∈ Z + : 2 l divides k}. Since (X i ) 1≤i≤n are i.i.d. Bernoulli mixtures, following the representation (2.1), we can express X i as The following claim yields an useful upper bound on p max .
be any two sequences of integers. Then It would now suffice to show that for all m ∈ Z we have To see this, it would be enough to show that and therefore the corresponding sums are distinct.
Applying Claim 2.5 we have, Here and in the rest of the proof we let ε be a small positive constant, chosen to satisfy various constraints which are specified along the proof.  By rewriting the LHS of (2.2) as 2) we get that it would be enough to show the existence of ε > 0 such that for large enough n, Multiplying both sides by 2 n/2 it reduces to showing that for n sufficiently large, In order to show (2.4), we use the following lemma.
(b) Furthermore, there exists δ , ε > 0 depending on α and the law of w 1 such that if α ∈ (1/2 − δ , 1/2 + δ ), then . Proving Lemma 2.6 is the main technical step in the proof of Proposition 2.4, and we devote Section 2.3 to its proof.
The following claim captures two technical properties of the bound obtained in Lemma 2.6. Claim 2.7. For n ∈ N, we define a function f n : (0, 1) → R + as Then the following hold.
Part (a) it is therefore enough to show that EJP 22 (2017), paper 10. is strictly monotone increasing. Taking logarithm it is enough to show that is strictly monotone increasing. Differentiate g to get Finally we are fully equipped to demonstrate the existence of ε > 0 such that (2.4) holds. Let δ , ε be as in part (b) of Lemma 2.6 and let c 0 be as in part (a) of Claim 2.7.
. We are thus left with verifying (2.4).
Applying Part (a) of Lemma 2.6 and part (a) of Claim 2.7, we obtain using the inequality log 2 (1 − x) < − x log 2 for x > 0. From part (b) of Lemma 2.6 we obtain Therefore max(I 1 , I 2 ) < 2 −εn and we obtain (2.4), as required.
We remark that if our interest was limited to obtaining the theorem for the case ε = 0, it would have been possible to use only the first part of Lemma 2.6, which is, as will become evident, easier to obtain.

Proof of Lemma 2.6
In this section we prove Lemma 2.6. In the proof we keep using the notation introduced in the previous section. We assume αn ∈ N. Let Z n := {0, 1, 2, . . . , n − 1}.

Proof of item (a)
In order to bound P W = αn , we use  We then intend to show the following.
For every A ⊂ Z n of size |A| = αn, we have U (A) ≤ α n . (2.8) Part (a) of Lemma 2.6 would follow from (2.8) since , (2.9) where the leftmost inequality uses (2.6), the middle one uses (2.8) and a union bound, and the rightmost one follows from the well-known inequality of the binomial coefficient Towards showing (2.8), let A ⊂ Z n be a set of size |A| = αn. Observe that, by the fact that w i 's are i.i.d., we have We further observe that for every a ∈ A, we have (2.11) We now solve the following maximization problem: (2.12) By applying Jensen's inequality to the log function, we get as required.
and observe that P ∀j < i : .
Next, we show that W is concentrated around its expectation. To this end we use the concentration properties of self-bounding functions of independent variables. We write f (w 1 , . . . , w n ) := W , g i (w 1 , . . . , w i−1 , w i+1 , . . . , w n ) := |{w j + j : 1 ≤ j ≤ n, j = i}|, and observe that for all i ≤ n we have, We then apply [3,Theorem 1 & 7], to obtain that for every β > 0 Setting δ = η 2 and ε = η 2 8 and using (2.15) we observe that for every α < 1 This proves (2.14) and hence completes the proof of part (b) of Lemma 2.6.

High algebraic degree
This section is dedicated to the proof of the following proposition, of which Lemma 1.3 is a straightforward consequence. The proof of the proposition relies on the following consequence of Theorem 1.2.

Lemma 3.2.
Let P be the random polynomial as in (1.1). Then there exist constants C, ε > 0 such that for any positive integer k and for a ∈ {−2, 2} we have Proof. Fix a ∈ {−2, 2}. Let k ≥ 1 be an integer and let r be the integer satisfying M 2 r ≤ k 2 < M 2 r+1 . By conditioning on ξ r , ξ r+1 , . . . , ξ n we have where the last equality follows from the fact that r−1 j=0 ξ j a j ≤ M (2 r − 1) deterministically and k 2 ≥ M 2 r by the definition of r. From Theorem 1.2, it follows that there exists a constant ε ∈ (0, 1) such that .

(3.2)
Combining (3.1) and (3.2) with the fact that r > 2 log 2 k − log 2 M − 1, we conclude that We shall also use the following bound on the probability that P has a root in close proximity to ±2, whose proof we postpone to Section 3.1.
Denote by B(z 0 , r) the closed ball in C with center at z 0 and radius r. Finally, we need a preliminary claim, bounding the number of roots far away from the unit circle. Proof. Assume, without loss of generality, that |a n | = 0. Letf (z) = z n f (z −1 ) = n i=0 a i z n−i be the reciprocal polynomial of f . Denote by N (f ) the number of z ∈ C for which f (z) = 0 and |z| ≥ 3 2 . Then N (f ) is also the number of z ∈ C for whichf (z) = 0 and |z| ≤ 2 3 . Noting that |f (0)| = |a n | ≥ 1 we may apply Jensen's formula (see, e.g., [1, Chapter 5.3.1]) and obtain for any r > 2 . Suppose that α is a double root of P . Note that α cannot be a multiple root of f α , since, otherwise, α is also a root of the polynomial f α whose degree is strictly smaller than d, violating the definition of deg(α). This implies that f 2 α divides P in Z[x] (by Gauss's lemma). In particular, the integer P (a) is divisible by f α (a) 2 , for a = ±2. Next we obtain a suitable lower bound for max{|f α (2)|, |f α (−2)|}. Denote by C(α) the set of algebraic conjugates of α (i.e., the set of roots of f α ). Each of these conjugates of α must also be a root of P . So, by Claim 3.4, all but at most 64M of the β ∈ C(α) satisfy |β| ≥ 3 2 . Therefore, we have Let B > 0 be given as in Proposition 3.1 and let K = K(B) > 0 be as given by Lemma 3.3. Let E be the event that there is at least one root of P within a distance of n −K from either −2 or 2. Note that the event E does not depend on α. On the event E c , for suitable constants c 3 , C 2 , C 3 > 0. Proposition 3.1 follows.

Roots near ±2
In this section we prove Lemma 3.3. We shall require the following. Proof. We prove the lemma for the case a = 2, as the argument for the case a = −2 is nearly identical. Set C 1 = log 3 M . Define a subset of indices J as J = {j ∈ {0, 1, . . . , n} : j ≥ n − C 1 log 2 n, and j is divisible by log 2 (2M + 1) }.
Note that for any two different values of the random vector (ξ j ) j∈J in {0, ±1, . . . , ±M } |J| , the values of the sum j∈J ξ j 2 j differ by at least 1 2 2 n−C1 log 2 n = 1 2 n −C1 2 n . Thus if we choose C = C 1 + 1, then for any fixed z ∈ R, there exists at most one value of the random vector (ξ j ) j∈J in {0, ±1, . . . , ±M } |J| such that | j∈J ξ j 2 j − z| ≤ n −C 2 n . Now by Assumption 1.3, we conclude that  By the Mean Value Theorem and the triangle inequality, for any z ∈ C such that |z| ≤ n −1 , We can now bound the derivative of the polynomial P in B(2, n −1 ) by Plugging in the bound (3.5) and (3.7) in (3.6), we deduce that, for any |z| ≤ n −(C+2) and for sufficiently large n, . The lemma is then obtained by taking K = C + 2.

Roots far from the unit circle
In this section we prove Lemma 1.4. We begin by obtaining the following bound on the probability that P has a particular root α far from the unit circle.  is one-to-one. Thus, as P 1 (α) and P 2 (α) are independent, we have P P (α) = 0 = E P P (α) = 0 | P 2 (α)) = The case when |α| < 1 can be handled similarly. This proves the lemma.
We also make the following simple observation. Proof of Lemma 1.4. Let us first estimate the number of algebraic numbers α such that α is a root of some random polynomial P of degree n and deg(α) ≤ C 0 log n. Write f α for the associated minimal polynomial of α in Z[x] and denote, as usual, the leading coefficient of f α by m α . If α 1 = α, α 2 , · · · , α deg(α) are conjugates of α, we can express f α a j x n−j .
Therefore, by Observation 4.1, we have the following crude bound on the coefficients of for some C > 0 depending on C 0 and M . Since a j has to be an integer, there are at most e C log 2 n possibilities for f α (x) for some constant C > 0.

Roots near the unit circle
In this section we prove Lemma 1.5. Recall that for α ∈ A, m α denotes the leading coefficient of the associated minimal polynomial of α in Z[x]. Fix C 0 , C 1 > 0. By Observation 4.1, we need to consider for Lemma 1.5 the following set of potential roots of P .
To prove Lemma 1.5, we employ the following union bound P ∃α ∈ A : P (α) = 0 ≤ |A| · max α∈A P P (α) = 0 and then proceed to provide upper bounds on max α∈A P P (α) = 0 and on the cardinality of the set A. This is done using Lemma 5.1 and Lemma 5.2 below, whose proofs are presented in Sections 5.1 and 5.2 respectively.
Lemma 5.1. Let α be an algebraic number of degree at least 5. Then for every > 0 there exists C > 0 such that For any polynomial f ∈ Z[x], let Λ(f ) be the maximum moduli of the roots of f .
Note that for every algebraic number α ∈ A, its associated minimal polynomial in Z[x] has degree at most C 0 log n and its leading coefficient is bounded by M . Applying Lemma 5.2 to each a ∈ {1, 2, . . . , M } and each degree 1 ≤ d ≤ C 1 log n with b = 1 6 , we obtain that for every > 0, |A| = o(n ).

Each root of low degree is unlikely
In this section we prove Lemma 5.1. The proof follows closely the proof of [10, Lemma 1] adapted for our case (that is, when the random variables ξ i 's are not Bernoulli random variables). The main ingredient of the proof is the 'inverse Littlewood-Offord type theorem" of Tao and Vu [15, Theorem 1.9], whose specialization for our case is the following.
Proof of Lemma 5.1. Let α be an algebraic number of degree d ≥ 5 and let > 0. Assume towards obtaining a contradiction that P P (α) = 0 > n − 5 2 + . Ber( 1 2 ), independent of (I j , ∆ j ) 0≤j≤n . Conditioning on (I j , ∆ j ) 0≤j≤n yields We now apply Theorem 5.3 with A = 5 2 − ε and δ = 1 2 and z j = d j α j to obtain a symmetric GAP Q of rank B ≤ 2A < 5 such that all but O( √ n) many of the coefficients d j α j belong to Q. Therefore, for large enough n, there exists j 0 ∈ {0, 1, . . . , n} for which d j0+k α j0+k ∈ Q for all k = 0, 1, . . . , 4. Since the rank of Q is at most 4, there exists a nontrivial integer linear combination that annihilates the vector (d j0+k α j0+k ) 0≤k≤4 . Hence the algebraic degree of α is at most 4, in contradiction with our assumption. Hence, the lemma follows.

There are not many low degree polynomials with small house
This section is dedicated to the proof of Lemma 5.2.
In the proof we also make use of the classical Newton identities, known also as Newton-Girard formulae (See [9] for a modern proof).
With a slight abuse of notation we write S k (w) = w k 1 +w k 2 +· · ·+w k d , for w = (w 1 , . . . , w d ) ∈ C d .
Let b > 0 be given. We say that a set W ⊂ C d is (a, d) admissible if the following two conditions are satisfied: From (5.8) and from the definition of F a,d , we deduce that the set S is (a, d) admissible.
To conclude the proof we bound the maximal size of any (a, d) admissible set. In [6,Theorem 2], Dubickas obtained such a bound for the special case when a = 1 using an elementary but clever application of volume formulas of polytopes, and classical estimates on the number of Gauss integers in a circle.
We now use Dubickas' result as a blackbox to bound the cardinality of (a, d) admissible set. Let W be an (a, d) admissible set, and write W for the image of W in C ad under the repetition map w ∈ C d →ŵ := (w, w, . . . , w a times ) ∈ C ad .
where the last inequality follows from the fact u = v ∈ W and W is an (a, d) admissible set. Hence W is (1, ad) admissible. Therefore applying Theorem 5.5 with = ad implies |F a,d | = |S| = O b (exp((ad) 2/3+b )), as required.
In this section we will prove Lemma 1.6. Note that there are only finitely many irreducible polynomials in Z[x] of degree at most 4 with leading coefficients bounded by M in absolute value whose roots are all within distance M + 1 from the origin. In particular, for large enough n, all such polynomials whose roots are in distance 1 + C1 log n from the origin, have, in fact, all roots on the unit circle. Thus to prove the lemma it would suffice to show that for any fixed α ∈ A on the unit circle such that deg(α) ≤ 4 and α = ±1, P(α is a double root of P ) = o(n −2 ), . In fact, any polynomial |b k | has all its root on the unit circle [16]. Thus, first begin by addressing roots which lie on the unit circle but which are not root of unity. Lemma 6.1 (unimodular roots that are not roots of unity). Let α ∈ A be such that |α| = 1 but α is not a root of unity (i.e., α m = 1 for all m ∈ N). Then under Assumption 1.3, The proof of Lemma 6.1 is a straightforward application of the following well-known result due to Halász (see [17,Corollary 7.16], [19, Corllary 6.3 and Remark 3.5]). Lemma 6.2 (Halász). Let G be an infinite Abelian group. Let m ≥ 1 and a 1 , a 2 , . . . , a m ∈ G and let ε 1 , ε 2 , . . . , ε m be i.i.d. with P(ε j = 1) = P(ε j = 0) = 1/2. Fix ∈ N and let R be the number of solutions of the equation a i1 + a i2 + · · · + a i = a j1 + a j2 + · · · + a j . Then Proof of Lemma 6.1. By Proposition 2.1, the random variables (ξ j ) 0≤j≤n can be represented as ξ j = I j + ∆ j ε j where (I j , ∆ j ) 0≤j≤n are i.i.d. random vectors taking values in Z×N and (ε j ) 0≤j≤n 's are i.i.d. Ber( 1 2 ), independent of (I j , ∆ j ) 0≤j≤n . Now by conditioning on I j and ∆ j 's, we have where expectations are taken on the vector (I j , ∆ j ) 0≤j≤n . Fix an integer q in the support of the random variable ∆ j and let η := P(∆ j = q) > 0. Let T denote the random set of indices defined by T = {0 ≤ j ≤ n : ∆ j = q}. Again, conditioning on (ε j ) j ∈T , write E max Applying Halász's result (Lemma 6.2) with coefficients (α j ) j∈T in C and = 2, we count the number of solutions of the equation α i + α j = α k + α l , (6.2) where i, j, k, l are arbitrary indices in T . Taking the absolute value on the both sides of (6.2) and using the fact that |α| = 1, we have |1 + α j−i | = |1 + α l−k |, which implies that either α j−i = α l−k or α j−i = α k−l , or equivalently j − i = ±(l − k). In case that j − i = l − k, we may write (6.2) as α i (1 + α j−i ) = α k (1 + α j−i ), or equivalently as (α i − α k )(1 + α j−i ) = 0. Since α is not a root of unity, then 1 + α j−1 = 0. So, we deduce that α i = α k which implies that i = k. Plugging it in back in j − i = l − k, we also have j = l. Similarly, for the case j − i = k − l, we end up with the equation (α i − α l )(1 + α j−i ) = 0, which, in turn, implies that i = l and j = k. Hence, we conclude that R 2 ≤ 2|T | 2 and the claim follows.
From the claim, we obtain that Note that |T | has a binomial distribution corresponding to n + 1 trials and success probability η. From the standard result on the concentration of binomial random variable, we know that there exists a constant c > 0, depending on η, such that P |T | ≤ η 2 n ≤ e −c(n+1) . This completes the proof of the lemma.
Next, we consider roots of unity. The proof of the above lemma is very similar to that of Lemma 1.4 in [13] where the similar bound holds without the additional k exp(−c n k ) term for any non-constant coefficient distribution supported on {−1, 0, 1}. However, for the sake of completeness we include here a proof of Lemma 6.3. The proof of Lemma 6.3 relies heavily on the following classical anti-concentration bound of Sárközi and Szemerédi [14]. Theorem 6.4 (Sárközi and Szemerédi). Let (ε j ) 1≤j≤N be i.i.d. Ber( 1 2 ) random variables. There exists a constant C > 0 such that for any distinct integers (a j ), 1 ≤ j ≤ N , we Proof of Lemma 6.3. Since ξ j is a mixture of Bernoulli distribution, we can proceed along the same way as in the proof of Lemma 6.1 to obtain P P (α) = 0 ≤ E max x P n j=1 ε j j∆ j α j−1 = x (∆ j ) 1≤j≤n , a i α i = z has at most one integral solution (a 0 , . . . , a deg(α)−1 ) for a given z ∈ C. Hence, for any given values of (∆ j ) 1≤j≤n and any given x ∈ C,  Observe that T r is a binomial random variable with number of trials at least n k and success probability η. This gives us the following bound for the left tail of T r . There exists a constant c > 0 such that P |T r | ≤ η 2 n k ≤ e −c n k . The lemma now follows from (6.6) and the fact that deg(α) ≤ k.
The above bound along with the fact that deg(α) ≤ 4 implies the bound k ≤ C 2 for some absolute constant C 2 . On the other hand, since α = ±1, we have deg(α) ≥ 2. Thus, by Lemma 6.3, we deduce that P(α is a double root of P ) = O(n −3 ). Now assume that α is not a root of unity. As a direct consequence of Lemma 6.1, we also have that P(α is a double root of P ) = O(n −5/2 ). This finishes the proof of the bound (6.1) and hence the proof of Lemma 1.6.

Open problems
We conclude the paper with a couple of open problems. So, it follows from the subadditive property that there exists some λ > 0, depending on the law of ξ 0 , such that p n = e −λn(1+o(1)) . However, the exact value of λ is completely unknown. In the special case when the maximum of atom of ξ 0 is at most 1 2 , our Theorem 1.2 only gives a one-sided bound λ > 1 2 log 2. It would be very interesting to investigate the dependence of value of λ on the law of ξ 0 or, say, on the maximum atom of ξ 0 .
(b) It would be very interesting to investigate the minimum distance between adjacent complex roots of a random polynomial. Note that this problem makes sense even if the coefficient distribution is continuous. To best of our knowledge, precise quantitive bounds on the minimum gap are not available even for the i.i.d. Gaussian polynomials.