FACTORING SPARSE POLYNOMIALS

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F

^ACTORING

S

^PARSE

P

^OLYNOMIALS

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Theorem 1 (Schinzel): Let r be a positive integer, and fix non-zero integers a₀, . . . , a_r. Let

F (x₁, . . . , x_r) = a_rx_r + · · · + a₁x₁ + a₀.

Then there exist finite sets S and T of matrices satisfying:

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F (x₁, . . . , x_r) = a_rx_r + · · · + a₁x₁ + a₀.

(i) Each matrix in S or T is an r × ρ matrix with integer entries and of rank ρ for some ρ ≤ r.

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F (x₁, . . . , x_r) = a_rx_r + · · · + a₁x₁ + a₀.

(ii) The matrices in S and T are computable.

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(iii) For every set of positive integers d₁, . . . , d_r with d₁ < d₂ < · · · < d_r, the non-reciprocal part of F (x^d¹, . . . , x^d^r) is reducible if and only if there is an r × ρ matrix N in S and integers v₁, . . . , v_ρ satisfying



 d₁ d₂

...

d_r



 = N



 v₁ v₂ v..._ρ





but there is no r × ρ⁰ matrix M in T with ρ⁰ < ρ and no integers v₁⁰, . . . , v⁰

ρ⁰ satisfying



 d₁ d₂

...

d_r



 = M





 v₁⁰ v₂⁰

...

v⁰

ρ⁰





 .

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the non-reciprocal part of F (x^d¹, . . . , x^d^r) is reducible

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the non-reciprocal part of F (x^d¹, . . . , x^d^r) is reducible F (x₁, . . . , x_r) = a_rx_r + · · · + a₁x₁ + a₀

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the non-reciprocal part of F (x^d¹, . . . , x^d^r) is reducible F (x₁, . . . , x_r) = a_rx_r + · · · + a₁x₁ + a₀

F (x^d¹, . . . , x^d^r) = a_rx^d^r + · · · + a₁x^d¹ + a₀

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F (x₁, . . . , x_r) = a_rx_r + · · · + a₁x₁ + a₀.

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F (x₁, . . . , x_r) = a_rx_r + · · · + a₁x₁ + a₀.

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F (x₁, . . . , x_r) = a_rx_r + · · · + a₁x₁ + a₀.

(ii) The matrices in S and T are computable.

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(iii) For every set of positive integers d₁, . . . , d_r with F (x^d¹, . . . , x^d^r) not reciprocal and d₁ < d₂ < · · · < d_r, the non-cyclotomic part of F (x^d¹, . . . , x^d^r) is reducible if and only if there is an r × ρ matrix N in S and integers v₁, . . . , v_ρ satisfying



 d₁ d₂

...

d_r



 = N



 v₁ v₂ v..._ρ





but there is no r × ρ⁰ matrix M in T with ρ⁰ < ρ and no integers v₁⁰ , . . . , v⁰

ρ⁰ satisfying



 d₁ d₂

...

d_r



 = M





 v₁⁰ v₂⁰

...

v⁰

ρ⁰





 .

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(iii) For every set of positive integers d₁, . . . , d_r with F (x^d¹, . . . , x^d^r) not reciprocal and d₁ < d₂ < · · · < d_r, the non-cyclotomic part of F (x^d¹, . . . , x^d^r) is reducible if and only if there is an r × ρ matrix N in S and integers v₁, . . . , v_ρ satisfying



 d₁ d₂

...

d_r



 = N



 v₁ v₂ v..._ρ





ρ⁰ satisfying



 d₁ d₂

...

d_r



 = M





 v₁⁰ v₂⁰

...

v⁰

ρ⁰





 .

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(iii) For every set of positive integers d₁, . . . , d_r with F (x^d¹, . . . , x^d^r) not reciprocal and d₁ < d₂ < · · · < d_r, the non-cyclotomic part of F (x^d¹, . . . , x^d^r) is reducible if and only if there is an r × ρ matrix N in S and integers v₁, . . . , v_ρ satisfying



 d₁ d₂

...

d_r



 = N



 v₁ v₂ v..._ρ





ρ⁰ satisfying



 d₁ d₂

...

d_r



 = M





 v₁⁰ v₂⁰

...

v⁰

ρ⁰





 .

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Theorem: There is an algorithm with the following prop- erty: Given a non-reciprocal f (x) ∈ Z[x] with N non- zero terms, degree n and height H, the algorithm determines whether f (x) is irreducible in time

c(N, H)(log n)^c⁰^{(N )}

where c(N, H) depends only on N and H and c⁰(N ) depends only on N .

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Theorem: There is an algorithm with the following prop- erty: Given a non-reciprocal f (x) ∈ Z[x] with N non- zero terms, degree n and height H, the algorithm determines whether f (x) is irreducible in time

c(N, H)(log n)^c⁰^{(N )}

where c(N, H) depends only on N and H and c⁰(N ) depends only on N .

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Proof. Let

f (x) = a_rx^d^r + · · · + a₁x^d¹ + a₀.

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Proof. Let

f (x) = a_rx^d^r + · · · + a₁x^d¹ + a₀. Consider

F (x₁, . . . , x_r) = a_rx_r + · · · + a₁x₁ + a₀ so that

F (x^d¹, . . . , x^d^r) = a_rx^d^r + · · · + a₁x^d¹ + a₀.

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Proof. Let

F (x^d¹, . . . , x^d^r) = a_rx^d^r + · · · + a₁x^d¹ + a₀. Begin the algorithm by constructing the finite sets S and T of matrices in Schinzel’s Theorem 2.

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Proof. Let

F (x^d¹, . . . , x^d^r) = a_rx^d^r + · · · + a₁x^d¹ + a₀. Begin the algorithm by constructing the finite sets S and T of matrices in Schinzel’s Theorem 2. Observe that S and T depend on F and not on the d₁, . . . , d_r.

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Proof. Let

F (x^d¹, . . . , x^d^r) = a_rx^d^r + · · · + a₁x^d¹ + a₀. Begin the algorithm by constructing the finite sets S and T of matrices in Schinzel’s Theorem 2. Observe that S and T depend on F and not on the d₁, . . . , d_r, so this takes running time ≤ c₁(N, H).

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Next, the algorithm checks each matrix N in S to see if



 d₁ d₂

...

d_r



 = N



 v₁ v₂ v..._ρ



 for some integers v₁, . . . , v_ρ.

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

 d₁ d₂

...

d_r



 = N



 v₁ v₂ v..._ρ





for some integers v₁, . . . , v_ρ. In other words, v₁, . . . , v_ρ are unknowns and elementary row operations are done to solve the above system of equations.

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

 d₁ d₂

...

d_r



 = N



 v₁ v₂ v..._ρ





for some integers v₁, . . . , v_ρ. In other words, v₁, . . . , v_ρ are unknowns and elementary row operations are done to solve the above system of equations. The rank of N is ρ, so if a solution exists, then it is unique.

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

 d₁ d₂

...

d_r



 = N



 v₁ v₂ v..._ρ





for some integers v₁, . . . , v_ρ. In other words, v₁, . . . , v_ρ are unknowns and elementary row operations are done to solve the above system of equations. The rank of N is ρ, so if a solution exists, then it is unique. This involves performing elementary operations (+, −, ×, and ÷) with entries in N and the d_j (which are ≤ n).

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

 d₁ d₂

...

d_r



 = N



 v₁ v₂ v..._ρ





for some integers v₁, . . . , v_ρ. In other words, v₁, . . . , v_ρ are unknowns and elementary row operations are done to solve the above system of equations. The rank of N is ρ, so if a solution exists, then it is unique. This involves performing elementary operations (+, −, ×, and ÷) with entries in N and the d_j (which are ≤ n). The running time here is ≤ c₂(N, H) log² n.

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Next, the algorithm checks each matrix M in T to see if



 d₁ d₂

...

d_r



 = M





 v₁⁰ v₂⁰

...

v_ρ⁰





 for some integers v₁⁰ , . . . , v⁰

ρ⁰ by using elementary row operations to solve the system of equations for the v_j⁰.

0

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

 d₁ d₂

...

d_r



 = M





 v₁⁰ v₂⁰

...

v_ρ⁰





ρ⁰ by using elementary row operations to solve the system of equations for the v_j⁰. The rank of M is ρ⁰, so if a solution exists, then it is unique.

0

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

 d₁ d₂

...

d_r



 = M





 v₁⁰ v₂⁰

...

v_ρ⁰





This involves performing elementary operations (+, −,

×, and ÷) with entries in M and the d_j (which are ≤ n).

0

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

 d₁ d₂

...

d_r



 = M





 v₁⁰ v₂⁰

...

v_ρ⁰





This involves performing elementary operations (+, −,

×, and ÷) with entries in M and the d_j (which are ≤ n).

The running time is again ≤ c₂(N, H) log² n.

0

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Schinzel’s Theorem 2 now indicates to us whether f (x) has a reducible non-cyclotomic part.

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Schinzel’s Theorem 2 now indicates to us whether f (x) has a reducible non-cyclotomic part. If so, then we output that f (x) is reducible. If not, we have more work to do.

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Recall, we had the following theorem.

Theorem: There is an algorithm that has the following property: given f (x) = P_N

j=1 a_jx^d^j ∈ Z[x] with deg f = n, the algorithm determines whether f (x) has a cyclotomic factor and with running time

exp (2 + o(1))p

N/ log N (log N + log log n)

× log(H + 1)

as N tends to infinity, where H = max_1≤j≤N{|a_j|}.

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Recall, we had the following theorem.

exp (2 + o(1))p

× log(H + 1)

We’ll come back to this.

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exp (2 + o(1))p

× log(H + 1)

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exp (2 + o(1))p

× log(H + 1) split into two parts

exp (2 + o(1))p

N/ log N (log N ) × log(H + 1) exp (2 + o(1))p

N/ log N (log log n)

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exp (2 + o(1))p

c₃(N, H) exp (2 + o(1))p

N/ log N (log log n)

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exp (2 + o(1))p

c₃(N, H) (log n)^c⁴^{(N )}

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≤ c₃(N, H)(log n)^c⁴^{(N )}

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≤ c₃(N, H)(log n)^c⁴^{(N )}

Algorithm Continued: If the non-cyclotomic part of f (x) is irreducible, then use the algorithm in the above theorem.

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≤ c₃(N, H)(log n)^c⁴^{(N )}

Algorithm Continued: If the non-cyclotomic part of f (x) is irreducible, then use the algorithm in the above theorem.

This completes the proof of the theorem.

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A C

^URIOUS

C

ONNECTION WITH THE

O

^DD

C

^OVERING

P

^ROBLEM

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Coverings of the Integers:

A covering of the integers is a system of congruences x ≡ aj (mod m_j)

having the property that every integer satisfies at least one of the congruences.

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Coverings of the Integers:

A covering of the integers is a system of congruences x ≡ aj (mod m_j)

having the property that every integer satisfies at least one of the congruences.

Example 1:

x ≡ 0 (mod 2) x ≡ 1 (mod 2)

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Example 2:

x ≡ 0 (mod 2) x ≡ 2 (mod 3) x ≡ 1 (mod 4) x ≡ 1 (mod 6) x ≡ 3 (mod 12)

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Example 2:

x ≡ 0 (mod 2) x ≡ 2 (mod 3) x ≡ 1 (mod 4) x ≡ 1 (mod 6) x ≡ 3 (mod 12)

0 1 2 3 4 5 6 7 8 9 10 11

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Open Problem:

Does there exist an “odd covering” of the integers, a covering consisting of distinct odd moduli > 1?

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Open Problem:

Erd˝os: $25 (for proof none exists)

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Open Problem:

Erd˝os: $25 (for proof none exists)

Selfridge: $2000 (for explicit example)

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Sierpinski’s Application:

There exist infinitely many (even a positive proportion of) positive integers k such that k × 2ⁿ + 1 is composite for all non-negative integers n.

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Selfridge’s Example: k = 78557

(smallest odd known)

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Polynomial Question: Does there exist f (x) ∈ Z[x]

such that f (x)xⁿ + 1 is reducible for all non-negative integers n?

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such that f (x)xⁿ + 1 is reducible for all non-negative integers n?

Require: f (1) 6= −1

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such that f (1) 6= −1 and f(x)xⁿ + 1 is reducible for all non-negative integers n?

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such that f (1) 6= −1 and f(x)xⁿ + 1 is reducible for all non-negative integers n?

Answer: Nobody knows.

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Schinzel’s Example:

(5x⁹+6x⁸+3x⁶+8x⁵+9x³+6x²+8x+3)xⁿ+12 is reducible for all non-negative integers n

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Comment: For each n, the above polynomial is divisible by at least one of

Φ_k(x) where k ∈ {2, 3, 4, 6, 12}.

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Φ_k(x) where k ∈ {2, 3, 4, 6, 12}.

n ≡ 0 (mod 2) =⇒ f(x)xⁿ + 12 ≡ 0 (mod x + 1)

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Φ_k(x) where k ∈ {2, 3, 4, 6, 12}.

n ≡ 0 (mod 2) =⇒ f(x)xⁿ + 12 ≡ 0 (mod x + 1)

n ≡ 2 (mod 3) =⇒ f(x)xⁿ + 12 ≡ 0 (mod x² + x + 1)

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Φ_k(x) where k ∈ {2, 3, 4, 6, 12}.

n ≡ 0 (mod 2) =⇒ f(x)xⁿ + 12 ≡ 0 (mod x + 1)

n ≡ 2 (mod 3) =⇒ f(x)xⁿ + 12 ≡ 0 (mod x² + x + 1) n ≡ 1 (mod 4) =⇒ f(x)xⁿ + 12 ≡ 0 (mod x² + 1)

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Φ_k(x) where k ∈ {2, 3, 4, 6, 12}.

n ≡ 0 (mod 2) =⇒ f(x)xⁿ + 12 ≡ 0 (mod x + 1)

n ≡ 1 (mod 6) =⇒ f(x)xⁿ + 12 ≡ 0 (mod x² − x + 1)

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Φ_k(x) where k ∈ {2, 3, 4, 6, 12}.

n ≡ 0 (mod 2) =⇒ f(x)xⁿ + 12 ≡ 0 (mod x + 1)

n ≡ 1 (mod 6) =⇒ f(x)xⁿ + 12 ≡ 0 (mod x² − x + 1) n ≡ 3 (mod 12) =⇒ f(x)xⁿ + 12 ≡ 0 (mod x⁴ − x² + 1)

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Theorem. There exists an f (x) ∈ Z[x] with non-negative coefficients such that f (x)xⁿ+ 4 is reducible for all non- negative integers n.

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Theorem. There exists an f (x) ∈ Z[x] with non-negative coefficients such that f (x)xⁿ+4 is reducible for all non- negative integers n.

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Comment: For each n, the first polynomial is divisible by at least one Φ_k(x) where k divides 12.

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Comment: For each n, the second polynomial is divisible by at least one Φ_k(x) where k divides some integer N having more than 10¹⁷ digits.

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Comment: For each n, the second polynomial is divisible by at least one Φ_k(x) where k divides

24367503340863488003⁴¹5³¹7³⁷11²⁹13²³17¹⁶19¹⁸23²³29²⁹31³¹37³⁷41⁴¹.

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Schinzel’s Theorem: If there is an f (x) ∈ Z[x] such that f (1) 6= −1 and f(x)xⁿ + 1 is reducible for all non-negative integers n, then there is an odd covering of the integers.

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T

^URAN

´ ’

^S

C

^ONJECTURE

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Conjecture: There is an absolute constant C such that if f (x) =

r

X

j=0

a_jx^j ∈ Z[x], then there is a

g(x) =

r

X

j=0

b_jx^j ∈ Z[x]

irreducible (over Q) such that

r

X

j=0

|b_j − a_j| ≤ C.

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Conjecture: There is an absolute constant C such that if f (x) =

r

X

j=0

a_jx^j ∈ Z[x], then there is a

g(x) =

r

X

j=0

b_jx^j ∈ Z[x]

irreducible (over Q) such that

r

X

j=0

|b_j − a_j| ≤ C.

Comment: The conjecture remains open. If we take g(x) = P_s

j=0 b_jx^j ∈ Z[x] where possibly s > r, then the problem has been resolved by Schinzel.

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First Attack on Tur´an’s Problem:

Old Theorem: When m is large, either u(x)x^m + v(x) has an obvious factorization or the non-reciprocal part of u(x)x^m + v(x) is irreducible.

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Idea: Consider

g(x) = xⁿ + f (x).

If one can show g(x) is irreducible for some n, then the conjecture of Tur´an (modified so deg g > deg f is al- lowed) is resolved with C = 1.

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Idea: Consider

g(x) = xⁿ + f (x).

If f (0) = 0 or f (1) = −1, then consider instead g(x) = xⁿ + f (x) ± 1.

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Idea: Consider

g(x) = xⁿ + f (x).

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Idea: Consider

g(x) = xⁿ + f (x).

Problem: Dealing with g(x) = xⁿ + f (x) is essentially equivalent to the odd covering problem. So this is hard.

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Second Attack on Tur´an’s Problem:

Idea: Consider

g(x) = x^m ± xⁿ + f (x).

If f (0) = 0, then consider instead

g(x) = x^m ± xⁿ + f (x) ± 1.

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Theorem (Schinzel): For every f (x) =

r

X

j=0

a_jx^j ∈ Z[x], there exist infinitely many irreducible

g(x) =

s

X

j=0

b_jx^j ∈ Z[x]

such that

max{r,s}

X

j=0

|a_j − b_j| ≤ 2 if f (0) 6= 0 3 always.

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Theorem (Schinzel): For every f (x) =

r

X

j=0

a_jx^j ∈ Z[x], there exist infinitely many irreducible

g(x) =

s

X

j=0

b_jx^j ∈ Z[x]

such that

max{r,s}

X

j=0

|a_j − b_j| ≤ 2 if f (0) 6= 0 3 always.

One of these is such that

s < exp (5r + 7)(kf k² + 3).

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Ideas Behind Proof:

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I Consider F (x) = x^m + xⁿ + f (x) with m ∈ (M, 2M ] and n ∈ (N, 2N ] where M and N are large and M > N .

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I Apply result on u(x)x^m + v(x) with u(x) = 1 and v(x) = xⁿ + f (x) to reduce problem to consideration of reciprocal factors.

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I Apply result on u(x)x^m + v(x) with u(x) = 1 and v(x) = xⁿ + f (x) to reduce problem to consideration of reciprocal factors.

I Find a bound on the number of x^m + xⁿ + f (x) with reciprocal non-cyclotomic factors.

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I To bound the x^m + xⁿ + f (x) with cyclotomic factors, set

A = {(m, n) : M < m ≤ 2M, N < n ≤ 2N }, and let A_p ⊂ A (arising from when F (ζ_pk) = 0).

Use a “sieve” argument to estimate the size of A − [

A_p.

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I To bound the x^m + xⁿ + f (x) with cyclotomic factors, set

A = {(m, n) : M < m ≤ 2M, N < n ≤ 2N }, and let A_p ⊂ A (arising from when F (ζ_pk) = 0).

Use a “sieve” argument to estimate the size of A − [

A_p.

I Deduce that some F (x) = x^m+ xⁿ+ f (x) with m ∈ (M, 2M ] and n ∈ (N, 2N ] is irreducible (where M and N are large and M > N ).

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Current Knowledge:

Theorem: Given f (x) =

r

X

j=0

a_jx^j ∈ Z[x], there are

infinitely many irreducible g(x) =

s

X

j=0

b_jx^j ∈ Z[x]

such that max{r,s}

X

j=0

|a_j − b_j| ≤ 5.

s ≤ 4r exp 4kf k²+ 12.

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r

X

j=0

s

X

j=0

b_jx^j ∈ Z[x]

such that max{r,s}

X

j=0

|a_j − b_j| ≤ 5.

s ≤ 4r exp 4kf k²+ 12.

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r

X

j=0

s

X

j=0

b_jx^j ∈ Z[x]

such that max{r,s}

X

j=0

|a_j − b_j| ≤ 3.

s ≤ some polynomial in r.