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† ] C_^ ‰ ۅF7t „‡†9ˆ¨‰Ø~ F=lÇI rxj F5ìGHZlt r_i`j † ] C ë ‰ – r tc ºe¬ J•– r tKc ³ }JF jJi t rxjJ| l ‚-i I r t rx F r_j t F | F2A<• }'– r tc ³ Ë Â ‘<‘<‘  ³  }JF jJi t r_jJ| } r Iut r_j f7t r_j tF | F2AI2CDGc i –btZc lt rxy8„‡†9ˆ¨‰¶r IPAKF5}JHJf rx~Z FǕ-tcJF j ³ ¬ ^ } „‡†9ˆ-‰Er I‹lqtKA3l j I  lt rxi`j€i`y…ˆÝ†9ˆ >À^ ‰2†9ˆ > ê ‰5†#ˆ > ] ‰ Í[^`C

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‰‡‚i I r t rҁ F rxj tKF | F2AI ³ ?
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‰Ø‚-i I r t rx F r_j tF | F7AI ³ ?
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N EWTON P OLYGONS

N EWTON P OLYGONS WITH

RESPECT

TO A PRIME

p

f (x)

=

x6 + 24 x5 + 12 x3 − 18 x + 36

f (x)

=

x6 + 2331x5 + 12 x3 − 18 x + 36

f (x)

=

x6 + 2331x5 + 2231x3 − 18 x + 36

f (x)

=

x6 + 2331x5 + 2231x3 − 2132x + 36

f (x)

=

x6 + 2331x5 + 2231x3 − 2132x + 2232

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=2

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=2

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=2

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=2

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=2

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=2

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=2

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=2

Newton polygon of f (x) with respect to p

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=2

Newton polygon of f (x) with respect to 2

Comment: The slopes of the edges of a Newton polygon increase going from left to right.

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=3

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=3

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=3

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=3

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=3

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=3

0

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=3

0

1

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=3

0

1

1

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=3

0

1

1

2

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=3

0

1

1

2

2

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=3

Newton polygon of f (x) with respect to 3

D UMAS ’ T HEOREM

D UMAS ’ T HEOREM . . .

g(x) = x3 +2x−4, h(x) = x5 −6x4 + 2x2 −12

g(x) = x3 +2x−4, h(x) = x5 −6x4 + 2x2 −12

Newton polygon of g(x) with respect to 2

g(x) = x3 +2x−4, h(x) = x5 −6x4 + 2x2 −12

Newton polygon of h(x) with respect to 2

Newton polygon of g(x)h(x) with respect to 2

Newton polygon of g(x)h(x) with respect to 2

Newton polygon of g(x)h(x) with respect to 2

Newton polygon of g(x)h(x) with respect to 2

Newton polygon of g(x)h(x) with respect to 2

g(x) = x3 +2x−4, h(x) = x5 −6x4 + 2x2 −12

g(x) = x3 +2x−4, h(x) = x5 −6x4 + 2x2 −12 f (x) = x8 −6x7 +2x6 −14x5 +24x4 −8x3 −8x2 −24x+48

f (x) = x8 −6x7 +2x6 −14x5 +24x4 −8x3 −8x2 −24x+48

Newton polygon of f (x) with respect to 2

How is Dumas’ theorem used to establish irreducibility?

f (x) = x10 − 7x8 + 14x7 − 28x5 − 49x2 − 21

f (x) = x10 − 7x8 + 14x7 − 28x5 − 49x2 − 21 Why is f (x) irreducible?

f (x) = x10 − 7x8 + 14x7 − 28x5 − 49x2 − 21 Why is f (x) irreducible? Eisenstein’s Criterion applies

f (x) = x10 − 7x8 + 14x7 − 28x5 − 49x2 − 21

f (x) = x10 − 7x8 + 14x7 − 28x5 − 49x2 − 21

Newton polygon of f (x) with respect to 7

f (x) = x10 − 7x8 + 14x7 − 28x5 − 49x2 − 21

Newton polygon of f (x) with respect to 7 What factors could f (x) have?

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=2

Newton polygon of f (x) with respect to 2

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=2 What factors could f (x) have?

Newton polygon of f (x) with respect to 2

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=3

Newton polygon of f (x) with respect to 3

f (x) = 2030x6 + 2331x5 + 2231x3 − 2132x + 2232 p=3 What factors could f (x) have?

Newton polygon of f (x) with respect to 3

f (x)

=

x6 + 24 x5 + 12 x3 − 18 x + 36 f (x) is irreducible What factors could f (x) have?

Newton polygon of f (x) with respect to 3

M ATH 788F Practice Test Problems

(1) Prove the following: Let f (x) be a monic polynomial in Z[x] for which f (0) 6= 0. Suppose further that f (x) = f1(x)f2(x)f3(x) where each fj (x) is irreducible. If f (x) = (x − α1)(x − α2) · · · (x − αn)

where each αj ∈ C, then |αj | ≥ 1 for at least three values of j ∈ {1, 2, . . . , n}.

(1) Prove the following: Let f (x) be a monic polynomial in Z[x] for which f (0) 6= 0. Suppose further that f (x) = f1(x)f2(x)f3(x) where each fj (x) is irreducible. If f (x) = (x − α1)(x − α2) · · · (x − αn)

where each αj ∈ C, then |αj | ≥ 1 for at least three values of j ∈ {1, 2, . . . , n}.

(1) Prove the following: Let f (x) be a monic polynomial in Z[x] for which f (0) 6= 0. Suppose further that f (x) = f1(x)f2(x)f3(x) where each fj (x) is irreducible. If f (x) = (x − α1)(x − α2) · · · (x − αn)

where each αj ∈ C, then |αj | ≥ 1 for at least three values of j ∈ {1, 2, . . . , n}. f1(x)f2(x)f3(x) = (x − α1)(x − α2) · · · (x − αn)

(1) Prove the following: Let f (x) be a monic polynomial in Z[x] for which f (0) 6= 0. Suppose further that f (x) = f1(x)f2(x)f3(x) where each fj (x) is irreducible. If f (x) = (x − α1)(x − α2) · · · (x − αn)

where each αj ∈ C, then |αj | ≥ 1 for at least three values of j ∈ {1, 2, . . . , n}. f1(x)f2(x)f3(x) = (x − α1)(x − α2) · · · (x − αn)

(1) Prove the following: Let f (x) be a monic polynomial in Z[x] for which f (0) 6= 0. Suppose further that f (x) = f1(x)f2(x)f3(x) where each fj (x) is irreducible. If f (x) = (x − α1)(x − α2) · · · (x − αn)

where each αj ∈ C, then |αj | ≥ 1 for at least three values of j ∈ {1, 2, . . . , n}. Us Too!

f1(x)f2(x)f3(x) = (x − α1)(x − α2) · · · (x − αn)

(1) Prove the following: Let f (x) be a monic polynomial in Z[x] for which f (0) 6= 0. Suppose further that f (x) = f1(x)f2(x)f3(x) where each fj (x) is irreducible. If f (x) = (x − α1)(x − α2) · · · (x − αn)

where each αj ∈ C, then |αj | ≥ 1 for at least three values of j ∈ {1, 2, . . . , n}. f1(x)f2(x)f3(x) = (x − α1)(x − α2) · · · (x − αn)

(1) Prove the following: Let f (x) be a monic polynomial in Z[x] for which f (0) 6= 0. Suppose further that f (x) = f1(x)f2(x)f3(x) where each fj (x) is irreducible. If f (x) = (x − α1)(x − α2) · · · (x − αn)

where each αj ∈ C, then |αj | ≥ 1 for at least three values of j ∈ {1, 2, . . . , n}. Us Too!

f1(x)f2(x)f3(x) = (x − α1)(x − α2) · · · (x − αn)

(1) Prove the following: Let f (x) be a monic polynomial in Z[x] for which f (0) 6= 0. Suppose further that f (x) = f1(x)f2(x)f3(x) where each fj (x) is irreducible. If f (x) = (x − α1)(x − α2) · · · (x − αn)

where each αj ∈ C, then |αj | ≥ 1 for at least three values of j ∈ {1, 2, . . . , n}. Show each fj (x) has a root with absolute value ≥ 1.

fj (x) monic in Z[x], fj (0) 6= 0

fj (x) monic in Z[x], fj (0) 6= 0 fj (x) = (x − α01)(x − α02) · · · (x − α0r )

fj (x) monic in Z[x], fj (0) 6= 0 fj (x) = (x − α01)(x − α02) · · · (x − α0r ) |fj (0)| = |α01| |α02| · · · |α0r |

fj (x) monic in Z[x], fj (0) 6= 0 fj (x) = (x − α01)(x − α02) · · · (x − α0r ) 1 ≤ |fj (0)| = |α01| |α02| · · · |α0r |

fj (x) monic in Z[x], fj (0) 6= 0 fj (x) = (x − α01)(x − α02) · · · (x − α0r ) 1 ≤ |fj (0)| = |α01| |α02| · · · |α0r |

=⇒ there exists j such that |α0j | ≥ 1

(4) Let f (x) = x16 − 8x15 − 4x14 − 2x13 − x12 − x11 − x10 − · · · − x − 1. (a) Consider F (x) = (2x − 1)f (x). Explain why F (x) has exactly one root α satisfying |α| ≥ 1.

(b) Explain why this implies that f (x) is irreducible.

(4) Let f (x) = x16 − 8x15 − 4x14 − 2x13 − x12 − x11 − x10 − · · · − x − 1. (a) Consider F (x) = (2x − 1)f (x). Explain why F (x) has exactly one root α satisfying |α| ≥ 1.

(4) Let f (x) = x16 − 8x15 − 4x14 − 2x13 − x12 − x11 − x10 − · · · − x − 1. F (x) = (2x − 1)f (x)

(4) Let f (x) = x16 − 8x15 − 4x14 − 2x13 − x12 − x11 − x10 − · · · − x − 1. F (x) = (2x − 1)f (x)

(4) Let f (x) = x16 − 8x15 − 4x14 − 2x13 − x12 − x11 − x10 − · · · − x − 1. F (x) = (2x − 1)f (x) = 2x17

(4) Let f (x) = x16 − 8x15 − 4x14 − 2x13 − x12 − x11 − x10 − · · · − x − 1. F (x) = (2x − 1)f (x) = 2x17

(4) Let f (x) = x16 − 8x15 − 4x14 − 2x13 − x12 − x11 − x10 − · · · − x − 1. F (x) = (2x − 1)f (x) = 2x17

(4) Let f (x) = x16 − 8x15 − 4x14 − 2x13 − x12 − x11 − x10 − · · · − x − 1. F (x) = (2x − 1)f (x) = 2x17

(4) Let f (x) = x16 − 8x15 − 4x14 − 2x13 − x12 − x11 − x10 − · · · − x − 1. F (x) = (2x − 1)f (x)

= 2x17 − 17x16

(4) Let f (x) = x16 − 8x15 − 4x14 − 2x13 − x12 − x11 − x10 − · · · − x − 1. F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12

(4) Let f (x) = x16 − 8x15 − 4x14 − 2x13 − x12 − x11 − x10 − · · · − x − 1. F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12 − x11 − · · · − x

(4) Let f (x) = x16 − 8x15 − 4x14 − 2x13 − x12 − x11 − x10 − · · · − x − 1. F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12 − x11 − · · · − x + 1

F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12 − x11 − · · · − x + 1

F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12 − x11 − · · · − x + 1

Lemma 2. Let f (x) and g(x) be polynomials in C[x], and let C = {z ∈ C : |z| = 1}. If the strict inequality |f (z) + g(z)| < |f (z)| + |g(z)| holds for each z ∈ C, then f (x) and g(x) have the same total number of zeroes (counting multiplicity) inside the circle C (i.e., in the interior of the region bounded by C).

F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12 − x11 − · · · − x + 1

G(x) = 17x16

F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12 − x11 − · · · − x + 1

G(x) = 17x16 |F (z) + G(z)|

F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12 − x11 − · · · − x + 1

G(x) = 17x16 |F (z) + G(z)| = |2z 17 − z 12 − · · · − z + 1|

F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12 − x11 − · · · − x + 1

G(x) = 17x16 |F (z) + G(z)| = |2z 17 − z 12 − · · · − z + 1|

≤ 2|z|17 + |z|12 + · · · + |z| + 1

F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12 − x11 − · · · − x + 1

G(x) = 17x16 |F (z) + G(z)| = |2z 17 − z 12 − · · · − z + 1|

≤ 2|z|17 + |z|12 + · · · + |z| + 1 ≤ 2 + 1 + ··· + 1 + 1

F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12 − x11 − · · · − x + 1

G(x) = 17x16 |F (z) + G(z)| = |2z 17 − z 12 − · · · − z + 1|

≤ 2|z|17 + |z|12 + · · · + |z| + 1 ≤ 2 + 1 + · · · + 1 + 1 = 15

F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12 − x11 − · · · − x + 1

G(x) = 17x16 |F (z) + G(z)| = |2z 17 − z 12 − · · · − z + 1|

≤ 2|z|17 + |z|12 + · · · + |z| + 1 ≤ 2 + 1 + · · · + 1 + 1 = 15

< 17|z|17

F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12 − x11 − · · · − x + 1

G(x) = 17x16 |F (z) + G(z)| = |2z 17 − z 12 − · · · − z + 1|

≤ 2|z|17 + |z|12 + · · · + |z| + 1 ≤ 2 + 1 + · · · + 1 + 1 = 15

< 17|z|17 = |G(z)|

F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12 − x11 − · · · − x + 1

G(x) = 17x16 |F (z) + G(z)| = |2z 17 − z 12 − · · · − z + 1|

≤ 2|z|17 + |z|12 + · · · + |z| + 1 ≤ 2 + 1 + · · · + 1 + 1 = 15

< 17|z|17

= |G(z)| ≤ |F (z)| + |G(z)|

F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12 − x11 − · · · − x + 1

G(x) = 17x16 |F (z) + G(z)| = |2z 17 − z 12 − · · · − z + 1|

≤ 2|z|17 + |z|12 + · · · + |z| + 1 ≤ 2 + 1 + · · · + 1 + 1 = 15

< 17|z|17

= |G(z)| ≤ |F (z)| + |G(z)|

F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12 − x11 − · · · − x + 1

G(x) = 17x16 Since G(x) has exactly 16 zeroes in C = {z ∈ C : |z| = 1},

so does F (x).

F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12 − x11 − · · · − x + 1

G(x) = 17x16 Since G(x) has exactly 16 zeroes in C = {z ∈ C : |z| = 1},

so does F (x). Hence, F (x) has exactly one root with absolute value ≥ 1.

F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12 − x11 − · · · − x + 1

G(x) = 17x16 Since G(x) has exactly 16 zeroes in C = {z ∈ C : |z| = 1},

so does F (x). Hence, F (x) has exactly one root with absolute value ≥ 1.

F (x) = (2x − 1)f (x)

= 2x17 − 17x16 − x12 − x11 − · · · − x + 1

G(x) = 17x16 Since G(x) has exactly 16 zeroes in C = {z ∈ C : |z| = 1},

so does F (x). Hence, f (x) has exactly one root with absolute value ≥ 1.

So why is f (x) irreducible?

So why is f (x) irreducible? Lemma 1. Let f (x) be a monic polynomial in Z[x] for which f (0) 6= 0. Suppose further that f (x) has exactly 1 root α (with multiplicity 1) such that |α| ≥ 1. Then f (x) is irreducible.

So why is f (x) irreducible? Lemma 1. Let f (x) be a monic polynomial in Z[x] for which f (0) 6= 0. Suppose further that f (x) has exactly 1 root α (with multiplicity 1) such that |α| ≥ 1. Then f (x) is irreducible. f (x) = x16 − 8x15 − 4x14 − 2x13 − x12 − x11 − x10 − · · · − x − 1

So why is f (x) irreducible? Lemma 1. Let f (x) be a monic polynomial in Z[x] for which f (0) 6= 0. Suppose further that f (x) has exactly 1 root α (with multiplicity 1) such that |α| ≥ 1. Then f (x) is irreducible. f (x) = x16 − 8x15 − 4x14 − 2x13 − x12 − x11 − x10 − · · · − x − 1

So why is f (x) irreducible? Lemma 1. Let f (x) be a monic polynomial in Z[x] for which f (0) 6= 0. Suppose further that f (x) has exactly 1 root α (with multiplicity 1) such that |α| ≥ 1. Then f (x) is irreducible. f (x) = x16 − 8x15 − 4x14 − 2x13 − x12 − x11 − x10 − · · · − x − 1

So why is f (x) irreducible? Lemma 1. Let f (x) be a monic polynomial in Z[x] for which f (0) 6= 0. Suppose further that f (x) has exactly 1 root α (with multiplicity 1) such that |α| ≥ 1. Then f (x) is irreducible. f (x) = x16 − 8x15 − 4x14 − 2x13 − x12 − x11 − x10 − · · · − x − 1

So why is f (x) irreducible? Lemma 1. Let f (x) be a monic polynomial in Z[x] for which f (0) 6= 0. Suppose further that f (x) has exactly 1 root α (with multiplicity 1) such that |α| ≥ 1. Then f (x) is irreducible. f (x) = x16 − 8x15 − 4x14 − 2x13 − x12 − x11 − x10 − · · · − x − 1

(5) For p a prime, prove that the Bernoulli polynomial B(2p−1)(p−1)(x) is irreducible.

(5) For p a prime, prove that the Bernoulli polynomial B(2p−1)(p−1)(x) is irreducible. Idea: Take m = (2p − 1)(p − 1), and show that µ ¶ m X m e m(x) = xj pBj pB j j=0

is a rational number times a polynomial in Eisenstein form with respect to p.

m = (2p − 1)(p − 1),

f (x) =

m X

j=0

µ ¶ m j pBj x j

m = (2p − 1)(p − 1), The Leading Coefficient is

f (x) =

m X

j=0

.

µ ¶ m j pBj x j

m = (2p − 1)(p − 1),

f (x) =

The Leading Coefficient is pBm .

m X

j=0

µ ¶ m j pBj x j

m = (2p − 1)(p − 1),

f (x) =

m X

j=0

µ ¶ m j pBj x j

The Leading Coefficient is pBm . Use that (p − 1)|m implies Bm is a rational number which when reduced has denominator divisible by p but not by p2.

m = (2p − 1)(p − 1),

f (x) =

m X

j=0

µ ¶ m j pBj x j

The Leading Coefficient is pBm . Use that (p − 1)|m implies Bm is a rational number which when reduced has denominator divisible by p but not by p2. The Constant Term is

.

m = (2p − 1)(p − 1),

f (x) =

m X

j=0

µ ¶ m j pBj x j

The Leading Coefficient is pBm . Use that (p − 1)|m implies Bm is a rational number which when reduced has denominator divisible by p but not by p2. The Constant Term is p .

m = (2p − 1)(p − 1),

f (x) =

m X

j=0

µ ¶ m j pBj x j

The Leading Coefficient is pBm . Use that (p − 1)|m implies Bm is a rational number which when reduced has denominator divisible by p but not by p2. The Constant Term is p . So p divides the constant term and p2 does not.

m = (2p − 1)(p − 1),

f (x) =

Other Terms where (p − 1) - j:

m X

j=0

µ ¶ m j pBj x j

m = (2p − 1)(p − 1),

f (x) =

Other Terms where (p − 1) - j:

m X

j=0

µ ¶ m j pBj x j

Use that Bj is a rational number which when reduced has both numerator and denominator not divisible by p.

m = (2p − 1)(p − 1),

f (x) =

Other Terms where (p − 1) - j:

m X

j=0

µ ¶ m j pBj x j

Use that Bj is a rational number which when reduced has both numerator and denominator not divisible by p.

Other Terms where (p − 1)|j:

m = (2p − 1)(p − 1),

f (x) =

Other Terms where (p − 1) - j:

m X

j=0

µ ¶ m j pBj x j

Use that Bj is a rational number which when reduced has both numerator and denominator not divisible by p.

Other Terms where (p − 1)|j: Panic and hope for partial credit .

m = (2p − 1)(p − 1),

f (x) =

m X

j=0

µ ¶ m j pBj x j

Other Terms where (p − 1) - j:

Use that Bj is a rational number which when reduced has both numerator and denominator not divisible by p.

Other Terms where (p − 1)|j: Panic and hope for partial credit or use the lemma on the power of p dividing a factorial.

m = (2p − 1)(p − 1),

f (x) =

Other Terms where (p − 1)|j:

m X

j=0

µ ¶ m j pBj x j

m = (2p − 1)(p − 1),

f (x) =

Other Terms where (p − 1)|j: m = 2p2 − 3p + 1

m X

j=0

µ ¶ m j pBj x j

m = (2p − 1)(p − 1),

f (x) =

m X

j=0

µ ¶ m j pBj x j

Other Terms where (p − 1)|j: m = 2p2 − 3p + 1 = p2 + (p − 3)p + 1

m = (2p − 1)(p − 1),

f (x) =

m X

j=0

µ ¶ m j pBj x j

Other Terms where (p − 1)|j: m = 2p2 − 3p + 1 = p2 + (p − 3)p + 1 s(m) = p−1

m = (2p − 1)(p − 1),

f (x) =

m X

j=0

µ ¶ m j pBj x j

Other Terms where (p − 1)|j: m = 2p2 − 3p + 1 = p2 + (p − 3)p + 1 s(m) = p−1 =⇒ νp(m!) =

m = (2p − 1)(p − 1),

f (x) =

m X

j=0

µ ¶ m j pBj x j

Other Terms where (p − 1)|j: m = 2p2 − 3p + 1 = p2 + (p − 3)p + 1 s(m) = p−1 =⇒ νp(m!) =

m−s(m) p−1

m = (2p − 1)(p − 1),

f (x) =

m X

j=0

µ ¶ m j pBj x j

Other Terms where (p − 1)|j: m = 2p2 − 3p + 1 = p2 + (p − 3)p + 1 s(m) = p−1 =⇒ νp(m!) =

m−s(m) p−1

= 2p−2

m = (2p − 1)(p − 1),

f (x) =

m X

j=0

Other Terms where (p − 1)|j: νp(m!) = 2p − 2

µ ¶ m j pBj x j

m = (2p − 1)(p − 1),

f (x) =

j=0

Other Terms where (p − 1)|j: νp(m!) = 2p − 2

νp(j!) =

j −s(j) p−1

m X

µ ¶ m j pBj x j

m = (2p − 1)(p − 1),

f (x) =

j=0

Other Terms where (p − 1)|j: νp(m!) = 2p − 2

νp(j!) =

j −s(j) p−1

<

j

p−1

m X

µ ¶ m j pBj x j

m = (2p − 1)(p − 1),

f (x) =

m X

j=0

µ ¶ m j pBj x j

Other Terms where (p − 1)|j: νp(m!) = 2p − 2

νp(j!) =

j −s(j) p−1

<

j

p−1

=⇒ νp(j!) ≤

m = (2p − 1)(p − 1),

f (x) =

m X

j=0

µ ¶ m j pBj x j

Other Terms where (p − 1)|j: νp(m!) = 2p − 2

νp(j!) =

j −s(j) p−1

<

j

p−1

=⇒ νp(j!) ≤

j p−1

−1

m = (2p − 1)(p − 1),

f (x) =

m X

j=0

µ ¶ m j pBj x j

Other Terms where (p − 1)|j: νp(m!) = 2p − 2,

νp(j!) ≤

j p−1

−1

m = (2p − 1)(p − 1),

f (x) =

m X

j=0

µ ¶ m j pBj x j

Other Terms where (p − 1)|j: νp(m!) = 2p − 2, ¡

νp(j!) ≤

j p−1

¢ m−j −1 νp (m − j)! ≤ p−1

−1

νp(m!) = 2p − 2, ¡

νp(j!) ≤

j p−1

¢ m−j νp (m − j)! ≤ −1 p−1

−1

νp(m!) = 2p − 2, ¡

νp(j!) ≤

j p−1

¢ m−j νp (m − j)! ≤ −1 p−1 µµ ¶¶ m νp = j

−1

νp(m!) = 2p − 2, ¡

νp(j!) ≤

j p−1

−1

¢ m−j νp (m − j)! ≤ −1 p−1 µµ ¶¶ ¡ ¢ m νp = νp(m!) − νp(j!) − νp (m − j)! j

νp(m!) = 2p − 2, ¡

νp(j!) ≤

j p−1

−1

¢ m−j νp (m − j)! ≤ −1 p−1 µµ ¶¶ ¡ ¢ m νp = νp(m!) − νp(j!) − νp (m − j)! j m−j j − +2 ≥ 2p − 2 − p−1 p−1

νp(m!) = 2p − 2, ¡

νp(j!) ≤

j p−1

−1

¢ m−j νp (m − j)! ≤ −1 p−1 µµ ¶¶ ¡ ¢ m νp = νp(m!) − νp(j!) − νp (m − j)! j m−j j − +2 ≥ 2p − 2 − p−1 p−1

νp(m!) = 2p − 2, ¡

νp(j!) ≤

j p−1

−1

¢ m−j νp (m − j)! ≤ −1 p−1 µµ ¶¶ ¡ ¢ m νp = νp(m!) − νp(j!) − νp (m − j)! j m−j j − +2 ≥ 2p − 2 − p−1 p−1

νp(m!) = 2p − 2, ¡

νp(j!) ≤

j p−1

−1

¢ m−j νp (m − j)! ≤ −1 p−1 µµ ¶¶ ¡ ¢ m νp = νp(m!) − νp(j!) − νp (m − j)! j m ≥ 2p − p−1

νp(m!) = 2p − 2, ¡

νp(j!) ≤

j p−1

−1

¢ m−j νp (m − j)! ≤ −1 p−1 µµ ¶¶ ¡ ¢ m νp = νp(m!) − νp(j!) − νp (m − j)! j (2p−1)(p−1) ≥ 2p − p−1

νp(m!) = 2p − 2, ¡

νp(j!) ≤

j p−1

−1

¢ m−j νp (m − j)! ≤ −1 p−1 µµ ¶¶ ¡ ¢ m νp = νp(m!) − νp(j!) − νp (m − j)! j (2p−1)(p−1) ≥1 ≥ 2p − p−1

M ATH 788F: P RACTICE T EST (1) Prove the following: Let f (x) be a monic polynomial in Z[x] for which f (0) 6= 0. Suppose further that f (x) = f1 (x)f2 (x)f3 (x) where each fj (x) is irreducible. If f (x) = (x − α1 )(x − α2 ) · · · (x − αn ) where each αj ∈ C, then |αj | ≥ 1 for at least three values of j ∈ {1, 2, . . . , n}. (2) Let f (x) = x3 + 22. Determine with proof all primes p for which f (x) is Eisenstein with respect to p. For each such p, find a value of a for which f (x + a) is in Eisenstein form with respect to p. (3) Let

f (x) = x7 + 21x6 − 30x4 − 90x3 + 1350x + 2700.

Using Newton polygons, explain why f (x) is irreducible. (Be careful, and indicate as clearly as possible what information you are obtaining from each Newton polygon you use in your argument.) (4) Let

f (x) = x16 − 8x15 − 4x14 − 2x13 − x12 − x11 − x10 − · · · − x − 1.

For each part below, give all details of your solution. Do NOT refer to any theorems from class (in particular, do not refer to Perron’s Theorem or its proof). You may however use the following lemmas from class: Lemma 1. Let f (x) be a monic polynomial in Z[x] for which f (0) 6= 0. Suppose further that f (x) has exactly 1 root α (with multiplicity 1) such that |α| ≥ 1. Then f (x) is irreducible. Lemma 2. Let f (x) and g(x) be polynomials in C[x], and let C = {z ∈ C : |z| = 1}. If the strict inequality |f (z) + g(z)| < |f (z)| + |g(z)| holds for each z ∈ C, then f (x) and g(x) have the same total number of zeroes (counting multiplicity) inside the circle C (i.e., in the interior of the region bounded by C). (a) Consider F (x) = (2x − 1)f (x). Explain why F (x) has exactly one root α satisfying |α| ≥ 1. (Hint: Expand the product (2x − 1)f (x).) (b) Explain why this implies that f (x) is irreducible. (5) For p a prime, prove that the Bernoulli polynomial B(2p−1)(p−1) (x) is irreducible.

M ATERIAL TO S TUDY FOR M ATH 788F • Difference between irreducibility over Z and over Q • Gauss’ Theorem connecting irreducibility over Q with irreducibility over Z (know proof) • Computing the greatest common divisor of two polynomials • The Sch¨onemann-Eisenstein Criterion (know a proof) • Determining if a polynomial is Eisenstein • Newton polygons (definitely on test) • Do not concern yourselves with the last section of Chapter 2 (Schur’s theorem) • Perron’s Theorem (know proof given the second lemma, that is given Rouch´e’s theorem) • A. Cohn’s and G. P´olya’s Theorem concerning f (10) being prime (know proof of the theorem and the lemma to the theorem) • Bernoulli polynomials (know proof of Theorem 4.2.2) • Be able to do the exercises at the end of the chapters

M ATH 788F T EST

(1) Let f (x) = 45x8 − 30x7 − 15x6 − 2x4 + 30x2 + 150. Using Newton polygons, explain why f (x) is irreducible. Below are two Newton polygons to get you started. You may want to use others.

Newton polygon of f (x) with respect to 2

Newton polygon of f (x) with respect to 3

(2) Let f (x) = x3 + k where k is an arbitrary integer. Suppose that f(x) is Eisenstein with respect to a prime p. Prove that either p = 3 or p is a divisor of k. em (x) = (3) Let m = p2 + p − 2. Recall that pB

m X

pBj

j=0

µ ¶ ¡ ¢ m . Explain why each of the coefficients pBj m j for j

1 ≤ j ≤ m − 1 is a rational number which, when reduced, has its numerator divisible by p. (I am not asking em (x) is a rational number times an Eisenstein polynomial. I am, however, asking you to give you to prove pB em (x) is a rational number times an Eisenstein polynomial.) part of a proof that pB

(4) Let dn dn−1 . . . d0 be the decimal representation of a product of three primes. Let f(x) =

n X

dj xj . Prove

j=0

that f (x) is the product of at most three irreducible polynomials. In other words, show that if f (x) = f1 (x)f2 (x)f3 (x)f4(x) where each fj (x) ∈ Z[x], then fj (x) ≡ ±1 for at least one j ∈ {1, 2, 3, 4}. Prove any lemmas from class you use except you may use Lemma 5, without proof, given in the handout. (5) Let f (x) =

n X

aj xj where an = 1, an−1 = 0, and

j=0

Note that the coefficients should be integers and the constant term non-zero.

an−2 > 1 + |an−3 | + |an−4 | + · · · + |a1 | + |a0 |. Thus, an−2 is positive and greater than the sum of the absolute values of the other coefficients. (a) Show that f (x) has exactly two roots (counting multiplicity) with absolute values ≥ 1. (b) Show that f (x) has no real roots with absolute value ≥ 1. (Hint: If z ∈ R, then the terms an z n and an−2 z n−2 have the same sign. Also, the inequality in this problem is mighty strong.)

(c) Explain why f(x) is irreducible.

L AST L ECTURE OF

M ATH 788F

Final Exam:

Final Exam: Thursday, December 12, 2:00 p.m.

Final Exam: Thursday, December 12, 2:00 p.m. The Final will be in this room.

Final Exam: Thursday, December 12, 2:00 p.m. The Final will be in this room. The Final is optional.

Final Exam: Thursday, December 12, 2:00 p.m. The Final will be in this room. The Final is optional. It can only help your grade.

Final Exam: Thursday, December 12, 2:00 p.m. The Final will be in this room. The Final is optional. It can only help your grade. Material to Know:

Final Exam: Thursday, December 12, 2:00 p.m. The Final will be in this room. The Final is optional. It can only help your grade. Material to Know: Same as what you needed to know for the test.

Final Exam: Thursday, December 12, 2:00 p.m. The Final will be in this room. The Final is optional. It can only help your grade. Material to Know: Same as what you needed to know for the test. Will I be Around?

Final Exam: Thursday, December 12, 2:00 p.m. The Final will be in this room. The Final is optional. It can only help your grade. Material to Know: Same as what you needed to know for the test. Will I be Around? Can be

Final Exam: Thursday, December 12, 2:00 p.m. The Final will be in this room. The Final is optional. It can only help your grade. Material to Know: Same as what you needed to know for the test. Will I be Around? Can be (maybe will be).

Final Exam: Thursday, December 12, 2:00 p.m. The Final will be in this room. The Final is optional. It can only help your grade. Material to Know: Same as what you needed to know for the test. Will I be Around? Can be (maybe will be). Please send me email if you would like to get together (or if you have questions).

W HAT WERE WE DISCUSSING BEFORE OUR TEST ?

W HAT WERE WE DISCUSSING BEFORE OUR TEST ? Laguerre Polynomials

W HAT WERE WE DISCUSSING BEFORE OUR TEST ? Laguerre Polynomials R EALLY ? W HAT ARE THEY ?

W HAT WERE WE DISCUSSING BEFORE OUR TEST ? Laguerre Polynomials R EALLY ? W HAT ARE THEY ? m X (m+α)(m−1+α) · · · (j +1+α)(−x)j j=0

(m − j)!j!

m X (m+α)(m−1+α) · · · (j +1+α)(−x)j j=0

(m − j)!j!

m X (m+α)(m−1+α) · · · (j +1+α)(−x)j j=0

(m − j)!j! (α)

We denote this Lm (x).

m X (m+α)(m−1+α) · · · (j +1+α)(−x)j j=0

(m − j)!j! (α)

We denote this Lm (x). T HEOREM 7.7.2. ∀α ∈ Q − Z−, ∃ finitely many (α) m ∈ Z+ such that Lm (x) is reducible.

α=

u v

6∈ Z−,

v > 0,

gcd(u, v) = 1

α=

u

6∈ Z−,

v > 0,

gcd(u, v) = 1

v   m bj = (m + α)(m − 1 + α) · · · (j + 1 + α) j

α=

u

6∈ Z−,

v > 0,

gcd(u, v) = 1

v   m bj = (m + α)(m − 1 + α) · · · (j + 1 + α) j   m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) = j v m−j

α=

u

6∈ Z−,

v > 0,

gcd(u, v) = 1

v   m bj = (m + α)(m − 1 + α) · · · (j + 1 + α) j   m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) = j v m−j Let g(x) =

m X j=0

bj xj and f (x) =

m X j=0

aj bj xj .

α=

u

6∈ Z−,

v > 0,

gcd(u, v) = 1

v   m bj = (m + α)(m − 1 + α) · · · (j + 1 + α) j   m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) = j v m−j Let g(x) =

m X j=0

g(x) is monic.

bj xj and f (x) =

m X j=0

aj bj xj . Then

α=

u

6∈ Z−,

v > 0,

gcd(u, v) = 1

v   m bj = (m + α)(m − 1 + α) · · · (j + 1 + α) j   m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) = j v m−j Let g(x) =

m X j=0

bj xj and f (x) =

m X

aj bj xj . Then

j=0

g(x) is monic. Assume f (x) has a factor of degree k in [1, m/2].

α=

u

6∈ Z−,

v > 0,

gcd(u, v) = 1

v   m bj = (m + α)(m − 1 + α) · · · (j + 1 + α) j   m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) = j v m−j Let g(x) =

m X j=0

bj xj and f (x) =

m X

aj bj xj . Then

j=0

g(x) is monic. Assume f (x) has a factor of degree k in [1, m/2]. Use Lemma 2.4.2 to obtain a contradiction.

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) 

j

v m−j

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) 

v m−j

j

Let g(x) =

m X j=0

bj xj and f (x) =

m X

aj bj xj . Then

j=0

g(x) is monic. Assume f (x) has a factor of degree k in [1, m/2]. Use Lemma 2.4.2 to obtain a contradiction.

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) 

v m−j

j

Let g(x) =

m X j=0

bj xj and f (x) =

m X

aj bj xj . Then

j=0

g(x) is monic. Assume f (x) has a factor of degree k in [1, m/2]. Use Lemma 2.4.2 to obtain a contradiction. We want a prime p that satisfies certain conditions with g(x).

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) 

v m−j

j

Let g(x) =

m X j=0

bj xj and f (x) =

m X

aj bj xj . Then

j=0

g(x) is monic. Assume f (x) has a factor of degree k in [1, m/2]. Use Lemma 2.4.2 to obtain a contradiction. We want a prime p that satisfies certain conditions with g(x). One of them is that p does not divide the leading coefficient of g(x).

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) 

v m−j

j

Let g(x) =

m X j=0

bj xj and f (x) =

m X

aj bj xj . Then

j=0

g(x) is monic. Assume f (x) has a factor of degree k in [1, m/2]. Use Lemma 2.4.2 to obtain a contradiction. We want a prime p that satisfies certain conditions with g(x). One of them is that p does not divide the leading coefficient of g(x). This is clear.

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) 

v m−j

j

g(x) =

m X j=0

bj xj ,

k ∈ [1, m/2],

p prime

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) 

v m−j

j

g(x) =

m X j=0

C ASES :

bj xj ,

k ∈ [1, m/2],

p prime

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) 

v m−j

j

g(x) =

m X

bj xj ,

k ∈ [1, m/2],

j=0

C ASES : • k > m/ log2 m • k0 ≤ k ≤ m/ log2 m • 2 ≤ k < k0 •k =1

p prime

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) 

v m−j

j

g(x) =

m X

bj xj ,

k ∈ [1, m/2],

j=0

C ASES : • k > m/ log2 m • k0 ≤ k ≤ m/ log2 m • 2 ≤ k < k0 •k =1

p prime

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) 

v m−j

j

g(x) =

m X

bj xj ,

k ∈ [1, m/2],

j=0

BASIC I DEA IN E ACH C ASE :

p prime

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) 

v m−j

j

g(x) =

m X

bj xj ,

k ∈ [1, m/2],

p prime

j=0

BASIC I DEA IN E ACH C ASE : • Want p|(v(m−j)+u) for some j ∈ {0, 1,..., k−1}.

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) 

v m−j

j

g(x) =

m X

bj xj ,

k ∈ [1, m/2],

p prime

j=0

BASIC I DEA IN E ACH C ASE : • Want p|(v(m−j)+u) for some j ∈ {0, 1,..., k−1}. • Want p > v.

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) 

v m−j

j

g(x) =

m X

bj xj ,

k ∈ [1, m/2],

p prime

j=0

BASIC I DEA IN E ACH C ASE : • Want p|(v(m−j)+u) for some j ∈ {0, 1,..., k−1}. • Want p > v. • Then νp(bj ) ≥ 1 for all j ∈ {0, 1,..., m−k}.

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u) 

v m−j

j

g(x) =

m X

bj xj ,

k ∈ [1, m/2],

p prime

j=0

BASIC I DEA IN E ACH C ASE : • Want p|(v(m−j)+u) for some j ∈ {0, 1,..., k−1}. • Want p > v. • Then νp(bj ) ≥ 1 for all j ∈ {0, 1,..., m−k}. • Show slope of right-most edge of N. P. of g(x) is < k1 .

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X j=0

v m−j bj xj ,

k ∈ [1, m/2],

p prime

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X j=0

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

 ν(b0)−ν(bj ) . Slope of right-most edge is max 1≤j≤m j 

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

 ν(b0)−ν(bj ) . Slope of right-most edge is max 1≤j≤m j 

ν(b0)−ν(bj )

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

 ν(b0)−ν(bj ) . Slope of right-most edge is max 1≤j≤m j 

ν(b0)−ν(bj ) ≤ ν((vj +u)(v(j −1)+u)···(v +u))

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

 ν(b0)−ν(bj ) . Slope of right-most edge is max 1≤j≤m j 

ν(b0)−ν(bj ) ≤ ν((vj +u)(v(j −1)+u)···(v +u))

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

 ν(b0)−ν(bj ) Slope of right-most edge is max . 1≤j≤m j 

ν(b0)−ν(bj ) ≤ ν((vj +u)(v(j −1)+u)···(v +u))

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

 ν(b0)−ν(bj ) . Slope of right-most edge is max 1≤j≤m j 

ν(b0)−ν(bj ) ≤ ν((vj +u)(v(j −1)+u)···(v +u))

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

 ν(b0)−ν(bj ) . Slope of right-most edge is max 1≤j≤m j 

ν(b0)−ν(bj ) ≤ ν((vj +u)(v(j −1)+u)···(v +u)) ≤ ν((vj +|u|)!)

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

 ν(b0)−ν(bj ) . Slope of right-most edge is max 1≤j≤m j 

ν(b0)−ν(bj ) ≤ ν((vj +u)(v(j −1)+u)···(v +u)) vj +|u| ≤ ν((vj +|u|)!) < p−1

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

 ν(b0)−ν(bj ) . Slope of right-most edge is max 1≤j≤m j 

ν(b0)−ν(bj ) ≤ ν((vj +u)(v(j −1)+u)···(v +u)) vj +|u| ≤ ν((vj +|u|)!) < p−1

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

 ν(b0)−ν(bj ) . Slope of right-most edge is max 1≤j≤m j 

ν(b0)−ν(bj ) ≤ ν((vj +u)(v(j −1)+u)···(v +u)) vj +|u| ≤ ν((vj +|u|)!) < (v + |u|)k

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

 ν(b0)−ν(bj ) . Slope of right-most edge is max 1≤j≤m j 

ν(b0)−ν(bj ) ≤ ν((vj +u)(v(j −1)+u)···(v +u)) (v + |u|)j ≤ ν((vj +|u|)!) < (v + |u|)k

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

 ν(b0)−ν(bj ) . Slope of right-most edge is max 1≤j≤m j 

ν(b0)−ν(bj ) ≤ ν((vj +u)(v(j −1)+u)···(v +u)) j ≤ ν((vj +|u|)!) < k

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

 ν(b0)−ν(bj ) . Slope of right-most edge is max 1≤j≤m j 

ν(b0)−ν(bj ) ≤ ν((vj +u)(v(j −1)+u)···(v +u)) j ≤ ν((vj +|u|)!) < k

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X j=0

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

j=0

C ASES : • k > m/ log2 m • k0 ≤ k ≤ m/ log2 m • 2 ≤ k < k0 •k =1

p > (v + |u|)k

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

j=0

C ASES : • 2 ≤ k < k0 •k =1

p > (v + |u|)k

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

C ASES : • 2 ≤ k < k0 •k =1 BASIC I DEA IN E ACH C ASE : • Want p|(v(m−j)+u) for some j ∈ {0, 1,..., k−1}.

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

C ASES : • 2 ≤ k < k0, p|(vm+u)(v(m−1)+u) •k =1 BASIC I DEA IN E ACH C ASE : • Want p|(v(m−j)+u) for some j ∈ {0, 1,..., k−1}.

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

C ASES : • 2 ≤ k < k0, p|(vm+u)(v(m−1)+u) • k = 1, p|(vm + u) BASIC I DEA IN E ACH C ASE : • Want p|(v(m−j)+u) for some j ∈ {0, 1,..., k−1}.

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

C ASES : • 2 ≤ k < k0, p|(vm+u)(v(m−1)+u) • k = 1, p|(vm + u) L EMMA 7.7.7. If a, b, c, d ∈ Z with ad − bc 6= 0, then the largest prime factor of (am + b)(cm + d) tends to infinity with m.

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

C ASES : • 2 ≤ k < k0, p|(vm+u)(v(m−1)+u) • k = 1, p|(vm + u) L EMMA 7.7.7. If a, b, c, d ∈ Z with ad − bc 6= 0, then the largest prime factor of (am + b)(cm + d) tends to infinity with m.

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

C ASES : • k = 1, p|(vm + u)

L EMMA 7.7.7. If a, b, c, d ∈ Z with ad − bc 6= 0, then the largest prime factor of (am + b)(cm + d) tends to infinity with m.

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

C ASES : • k = 1, p|(vm + u)

L EMMA 7.7.7. If a, b, c, d ∈ Z with ad − bc 6= 0, then the largest prime factor of (am + b)(cm + d) tends to infinity with m.

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

C ASES : • k = 1, p|m(vm + u)

L EMMA 7.7.7. If a, b, c, d ∈ Z with ad − bc 6= 0, then the largest prime factor of (am + b)(cm + d) tends to infinity with m.

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

j=0

C ASES : • k = 1, p|m(vm + u)

p > (v + |u|)k

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

C ASES : • k = 1, p|m(vm + u) BASIC I DEA IN E ACH C ASE : • Want p|(v(m−j)+u) for some j ∈ {0, 1,..., k−1}.

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

C ASES : • k = 1, p|m(vm + u) BASIC I DEA IN E ACH C ASE : • Want p|(v(m−j)+u) for some j ∈ {0, 1,..., k−1}.

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

C ASES : • k = 1, p|m(vm + u) BASIC I DEA IN E ACH C ASE : • Want p|(v(m−j)+u) for some j ∈ {0, 1,..., k−1}.

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

j=0

C ASES : • k = 1, p|m(vm + u) BASIC I DEA IN C ASE k = 1:

p > (v + |u|)k

 bj =

m (vm+u)(v(m−1)+u) ··· (v(j +1)+u)

g(x) =



j m X

v m−j bj xj ,

k ∈ [1, m/2],

p > (v + |u|)k

j=0

C ASES : • k = 1, p|m(vm + u) BASIC I DEA IN C ASE k = 1: m
• Use that if p|m and p is large, then p| j for small j

and the numerator above for large j.