Polynomial PSet Solutions

AMSP 2010 Algebra 3.5 Days 8-9 Problem Solutions

Polynomial PSet Solutions

Note: Problems A1, A2, B2, B8, C2, D2, E3, and E6 were done in class.

(A) Values and Roots

1. Rearrange to get

(x+ 1)P (x)− x = 0 for x = 0, 1, . . . , n.

Since this equation has roots x = 0, 1, . . . , n, and the left-hand side has degree n+1,

(x+ 1)P (x)− x = kx(x− 1) · · · (x− n)

for some integer k. Then

P (x) =kx(x− 1) · · · (x− n) + x

x+ 1.

The numerator must be divisible by x+ 1 so must have x = −1 as a zero. Plugging

in, we get (−1)n+1(n+ 1)!k − 1 = 0, or k = (−1)n+1

(n+1)!. Plugging in x = n+ 1 gives

P (n+ 1) =

{1, n oddnn+2

, n even

2. This time it’s easier to guess the solution. The polynomial

Q(x) =n∑i=0

(x

i

)(r − 1)i

has degree n and by the Binomial Theorem, satisfies the given conditions. SinceP (x) = Q(x) for n+ 1 values of x, actually P,Q are the same polynomial, and

P (n+ 1) = Q(n+ 1) =

(n+1∑i=0

(n+ 1

i

)(r − 1)i

)− (r − 1)n+1 = rn+1 − (r − 1)n+1.

3. Letting ray OQ1 be the positive real axis, Qi represent the nth roots of unity ωi

in the complex plane. Hence PQi equals |2 − ωi|. The roots of xn − 1 = 0 arejust the nth roots of unity, so xn − 1 =

∏n−1i=0 (x − ωi). Plugging in x = 2 gives∏n

k=1 |PQi| = 2n − 1.

4. The given condition says

f(x)2 = x(x− 1) · · · (x− n)Q(x) + x2 + 1 (1)

for some polynomial f(x) of degree at most n. Plugging x = 0, 1, . . . , n into (1)gives

f(x) = ±√x2 + 1, when x = 0, 1, . . . , n. (2)

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The following is key: Given n + 1 points (x0, y0), . . . , (xn, yn) with distinct x-coordinates, there exists exactly one polynomial f of degree at most n so thatf(xi) = yi for i = 0, 1, . . . , n.

Applying this to (2) we get 2n+1 possibilities for f(x) since we have 2 choices ofsign for each of x = 0, 1, . . . , n. If f(x) is a solution to (2) then so is −f(x); we get2n possibilities for f(x)2. Solve (1) to get 2n possibilities for Q(x):

Q(x) =f(x)2 − x2 − 1

x(x− 1) · · · (x− n)

Each such polynomial is a valid solution because f(x)2 − x2 − 1 is zero at x =0, 1, . . . , n and hence is divisible by x(x− 1) · · · (x− n).

5. Clearing denominators,

5∑i=1

[aix

∏i 6=j,1≤j≤5

(x+ j)

]− (x+ 1)(x+ 2)(x+ 3)(x+ 4)(x+ 5) = 0

for x = 1, 4, 9, 16, 25. Let f(x) denote the LHS. Since f(x) = 0 has the rootsx = 1, 4, 9, 16, 25, we conclude that (x − 1)(x − 4) · · · (x − 25) divides f(x). Sincef(x) has degree at most 5,

f(x) = k(x− 1)(x− 4)(x− 9)(x− 16)(x− 25)

for some constant k. However, equating

f(0) = [−(x+ 1)(x+ 2)(x+ 3)(x+ 4)(x+ 5)]x=0 = −5!

and f(0) = −k · 5!2 gives k = 15!

. Thus

f(x) =1

5!(x− 1)(x− 4)(x− 9)(x− 16)(x− 25).

Then

5∑i=1

ai62 + i

=f(62)

x(x+ 1)(x+ 2)(x+ 3)(x+ 4)(x+ 5)|x=62− 1

62

=187465

6744582.

6. Suppose |p(x) − rx| < 1 for x = 0, 1, . . . , n. Let yi = p(i) for 0 ≤ i ≤ n. By theLagrange Interpolation Formula,

p(x) =n∑i=0

[yi∏j 6=i

x− ji− j

].

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When ri − 1 ≤ yi ≤ ri + 1, p(n + 1) is maximized when we take yi = ri − 1 whenthe product is negative (i.e. when n− i is odd) and yi when the product is positive(i.e. when n− i is even):

p(n+ 1) <n∑i=0

(ri + (−1)n−i)∏j 6=i

(n+ 1)− ji− j

=n∑i=0

(ri + (−1)n−i)(−1)n−i(n+ 1)!

i!(n+ 1− i)!

=n∑i=0

[(n+ 1

i

)ri(−1)n−i +

(n+ 1

i

)]= −((r − 1)n+1 − rn+1) + (2n+1 − 1) by the Binomial Theorem

= rn+1 − 1 + 2n+1 − (r − 1)n+1 ≤ rn+1 − 1

An alternate solution is to use Lagrange Interpolation with n+ 2 points and showthat the coefficient of xn+1 cannot be 0.

7. Given two points (x1, y1) and (x2, y2) the equation

y − y1y2 − y1

=x− x1x2 − x1

=⇒ (y − y1)(x2 − x1) = (x− x1)(y2 − y1)

represents the line passing through these two points (it is a linear equation satisfiedby the coordinares of the two points). It follows that three points (x1, y1), (x2, y2)and (x3, y3) lie on the same line if and only if the condition

(y3 − y1)(x2 − x1) = (x3 − x1)(y2 − y1) (3)

holds. Now suppose that (x1, y1), (x2, y2) and (x3, y3) represent the (changing)coordinates of the three ducks as they waddle along their paths. Each coordinateis a linear function of time t, so (3) is an equation in t of degree at most 2 (i.e., isa quadratic). If such an equation has more than 2 solutions then it must reduce toan identity and thus hold true for all values of t. That is, if the ducks are in a rowat more than two times, then they are always in a row.

8. Note that g(x) is one of the 16 integer divisors of 2008 for each of the 81 integerroots. There must be at least 6 roots of f(x) for which g(x) has the same value.Since g(x) is nonconstant, its degree must be greater than 5.

9. Let p(x) be the monic real polynomial of degree n. If n = 1, then p(r) = r + a forsome real number a, and p(x) is the average of x and x + 2a, each of which has 1real root. Now we assume that n > 1. Let

g(x) = (x− 2)(x− 4) · · · (x− 2(n− 1)).

The degree of g(x) is n− 1. Consider the polynomials

q(x) = xn − kg(x), r(x) = 2p(x)− q(x) = 2p(x)− xn + kg(x).

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We will show that for large enough k these two polynomials have n real roots. Sincethey are monic and their average is clearly p(x), this will solve the problem.

Consider the values of the polynomial g(x) at n points x = 1, 3, 5, . . . , 2n−1. Thesevalues alternate in sign and are at least 1 (since at most two of the factors havemagnitude 1 and the others have magnitude at least 2). On the other hand, there isa constant c > 0 such that for 0 ≤ x ≤ n, we have |xn| < c and |2p(x)−xn| < c. Takek > c. Then we see that q(x) and r(x) evaluated at n points x = 1, 3, 5, . . . , 2n− 1alternate in sign. Thus q(x) and r(x) each has at least n − 1 real roots. Howeversince they are polynomials of degree n, they must have n real roots, as desired.

10. Without loss of generality, suppose n = deg(f) ≥ deg(g). Let r1, . . . , rk be thedistinct roots of f and let s1, . . . , sl be the distinct roots of f − 1.

We claim that k + l ≥ n+ 1. Indeed, suppose

f(x) = (x− r1)p1 · · · (x− rk)pk .

Then(x− r1)p1−1 · · · (x− rk)pk−1 | f ′.

Similarly, iff(x)− 1 = (x− s1)q1 · · · (x− sl)ql ,

then(x− s1)q1−1 · · · (x− sl)ql−1 | f ′.

Since the roots of f and f − 1 are distinct,

(x− r1)p1−1 · · · (x− rk)pk−1(x− s1)q1−1 · · · (x− sl)ql−1 | f ′.

Since f ′ has degree n− 1,

(p1 − 1) + . . .+ (pk − 1) + (q1 − 1) + . . . (ql − 1) ≤ n− 1.

Since p1 + . . . + pk = q1 + . . . + ql = n, this gives (n − k) + (n − l) ≤ n − 1, ork + l ≥ n+ 1.

Now f−g has degree at most n and has at least n+1 distinct roots r1, r2, . . . , rk, s1, . . . , sl,so it must be identically 0, and f = g.

11. Suppose f, g have no common nonconstant factor. Consider f, g as elements ofC(x)[y]. Since C(x) is a field, C(x)[y] is a principal ideal domain. From Gauss’sLemma applied to the ring C[x] and its fraction field C(x), any two polynomialswith no common factor in C[x][y] have no common factor in C(x)[y]. Hence thegreatest common divisor of f, g in C(x)[y] is 1, and we can write

1 = u(x, y)f(x, y) + v(x, y)g(x, y)

for some u, v ∈ C(x)[y]. Multiplying to clear denominators in u, v, we get

p(x) = s(y)f(x, y) + t(y)g(x, y).

Any zero (x, y) of both f and g must satisfy p(x) = 0. Since p is a nonzeropolynomial, it has finitely many zeros, i.e. there are finitely many possibilities forx. Each value gives a polynomial equation in y, which again has finitely many zeros.

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(B) Arithmetic Properties

1. Take n so that P (n) is nonzero. Suppose it is prime. Now P (n) | P (n+ kP (n))−P (n)⇒ P (n) | P (n+ kP (n)) for all k ∈ Z. If P (x) is not composite for all integerx, then P (n + kP (n)) is ±P (n) or 0 for all k ∈ Z. P attains one of these valuesinfinitely many times so is constant.

2. If not, then no two are equal. Without loss of generality, assume that c is betweena and b. Then

|P (a)− P (b)| = |c− b| < |b− a|.However, b− a | P (b)− P (a), a contradiction.

3. We first show that every fixed point x of Q is in fact a fixed point of P ◦P . Considerthe sequence given by x0 = x and xi+1 = P (xi) for i ≥ 0. Assume xk = x0. Then

di := xi+1 − xi|P (xi+1)− P (xi) = xi+2 − xi+1 = di+1

for all i, which together with dk = d0 implies |d0| = |d1| = · · · = |dk|. Supposed1 = d0 = d 6= 0. Then d2 = d as otherwise x3 = x1 and x0 will never occur inthe sequence again. Similarly di = d for all i and xi = x0 + id 6= x0 for all i, acontradition. It follows that d1 = −d0, as claimed. Thus we can assume Q = P ◦P .

If every integer t with P (P (t)) = t also satisfies P (t) = t, the number of solutions isat most deg(P ) = n. Suppose P (t1) = t2, P (t2) = t1, P (t3) = t4, P (t4) = t3, wheret1 is not equal to any of t2, t3, t4. Since t1 − t3 divides t2 − t4 and vice versa, weconclude t1− t3 = ±(t2− t4). Assume t1− t3 = t2− t4, i.e. t1− t2 = t3− t4 = u 6= 0.Since the relation t1 − t4 = ±(t2 − t3) similarly holds, we obtain t1 − t3 + u =±(t1− t3−u), which is impossible. Therefore, we must have t1− t3 = t4− t2, whichgives us P (t1) + t1 = P (t3) + t3 = c for some c. It follows that all integral solutionst of the equation P (P (t)) satisfy P (t) + t = c, and their number does not exceed n.

4. Induct on the degree. If f has degree n with leading coefficient c, then the firstnonzero term in the representation must be c

unpn(x) and f(x) − c

unpn(x) can be

uniquely written as an−1pn−1(x)+. . .+a1p1(x)+a0p0(x) by the induction hypothesis.

5. The assertions (a)⇒ (b) and (c)⇒ (a) are clear ((xi

)are integers for all integers x

and nonnegative integers i).

Suppose (b) holds. First assume that f(x) takes on integer values at 0, 1, . . . , n.We inductively build the sequence a0, a1, . . . so that the polynomial

Pm(x) = am

(x

m

)+ am−1

(x

m− 1

)+ · · ·+ a0

(x

0

)matches the value of f(x) at x = 0, . . . ,m. Define a0 = f(0); once a0, . . . , am havebeen defined, let

am+1 = f(m+ 1)− Pm(m+ 1).

Noting that(

xm+1

)equals 1 at x = m+1 and 0 for 0 ≤ x ≤ m, this gives Pm+1(x) =

f(x) for x = 0, 1, . . . ,m+ 1. Now Pn(x) is a degree n polynomial that agrees withf(x) at x = 0, 1, . . . , n, so they must be the same polynomial.

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Now if f takes on integer values for any n + 1 consecutive values, then by theargument above on the translated function it takes on integer values for all x; inparticular, for x = 0, 1, . . . , n. Use the above argument to get the desired represen-tation in (c).

6. Replacing f(x) by f(x)− f(n) as necessary, it suffices to show

lcm[1, 2, . . . , deg(f)] · f(m)

m∈ Z.

Letting d = deg(f) and writing f as in Problem 5, the above expression equals

lcm(1, 2, . . . , d)d∑i=0

aim

(m

i

)= lcm(1, 2, . . . , d)

d∑i=0

aii

(m− 1

i− 1

).

Each term is an integer because i | lcm(1, 2, . . . , d) for all 1 ≤ i ≤ d.

7. Step 1 Suppose P has degree d. Let Q be the polynomial of degree at most d withQ(x) = qx for 0 ≤ x ≤ d. Since the qx are all integers, Q has rational coefficients,and there exists k so that kQ has integer coefficients. Then m− n|kQ(m)− kQ(n)for all m,n ∈ N0.

Step 2 We show that Q is the desired polynomial.

Let x > n be given. Now

kqx ≡ kqm (mod x−m) for all integers m ∈ [0, d]

Since kQ(x) satisfies these relations as well, and kqm = kQ(m),

kqx ≡ kQ(x) (mod x−m) for all integers m ∈ [0, d]

and hencekqx ≡ kQ(x) (mod lcm(x, x− 1, . . . , x− d)). (4)

Now

lcm(x, x− 1, . . . , x− i− 1) = lcm[lcm(x, x− 1, . . . , x− i), x− i− 1]

=lcm(x, x− 1, . . . , x− i)(x− i− 1)

gcd[lcm(x, x− 1, . . . , x− i), (x− i− 1)]

≥ lcm(x, x− 1, . . . , x− i)(x− i− 1)

gcd[x(x− 1) · · · (x− i), (x− i− 1)]

≥ lcm(x, x− 1, . . . , x− i)(x− i− 1)

(i+ 1)!

so by induction lcm(x, x−1, . . . , x−d) ≥ x(x−1)···(x−d)d!(d−1)!···1! . Since P (x), Q(x) have degree

d, for large enough x (say x > L) we have∣∣∣Q(x)± x(x−1)···(x−d)

kd!(d−1)!···1!

∣∣∣ > P (x). By (4) kqx

must differ by a multiple of lcm(x, x−1, . . . , x−d) from kQ(x); hence qx must differ

by a multiple of x(x−1)···(x−d)kd!(d−1)!···1! from Q(x), and for x > L we must have kqx = kQ(x)

and qx = Q(x).

Now for any y we have kQ(y) ≡ kQ(x) ≡ kqx ≡ kqy (mod x − y) for any x > L.Since x− y can be arbitrarily large, we must have Q(y) = qy, as needed.

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8. For i ∈ N, let

Pi(x) =i∏

k=1

(x− (2k − 1)).

(Define P0(x) = 1.)

Step 1 By Problem 4, any polynomial with integer coefficients can be written inthe form

∑0≤i≤n ciPi(x).

Step 2 Let ai =∑∞

k=0

⌊i2k

⌋. We claim that 2ai | Pi(x) for all i ∈ N and all odd x.

For a prime p and n ∈ Z, denote by vp(n) the exponent of the highest power of pdividing n (by convention vp(0) =∞). For given odd x let f(α) be the number ofvalues of k (0 ≤ k ≤ i− 1) where 2α | x− 1− 2k. Then

v2(Pi(x)) =i−1∑k=0

v2(x− 1− 2k) =∞∑α=1

f(α)

since each k with 2α || x− 1− 2k is counted α times in either sum.

Since any set of 2α−1 consecutive even integers has one divisible by 2α, any setof i consecutive even integers has at least

⌊i

2α−1

⌋integers divisible by 2α. Hence

f(α) ≥⌊

i2α−1

⌋, and v2(Pi(x)) ≥

∑∞α=0

⌊i2α

⌋as desired.

Note a0 = 0, a1 = 1, a2 = 3, a3 = 4, a4 = 7, a5 = 8, and ai ≥ 10 for i ≥ 6.

Step 3 Next, we claim that if P (x) =∑

0≤i≤n ciPi(x) has a complete remaindersequence then c1 is odd. (c0 obviously needs to be odd.) We have 4 | P (4k+i)−P (i)for any integer i; hence r(4k+ 1) ≡ r(1) (mod 4) and r(4k+ 3) ≡ r(3) (mod 4) foreach k. In order for the remainder sequence to be complete, we need r(1) 6≡ r(3)(mod 4). But noting that ai ≥ 2 and Pi(x) ≡ 0 (mod 4) for odd x and i ≥ 2, wehave P (3)− P (1) ≡ c1(P1(3)− P1(1)) ≡ 2c1 (mod 4). Hence c1 is odd.

Step 4 Since for any odd x, Pi(x) is divisible by 2ai , if we mod out ci by 210−ai ,and delete the terms with Pi for i ≥ 6 (where ai ≥ 10), we get a polynomial withthe same remainder sequence as Pi. If P (x) gives a complete remainder sequence,then c0 is odd, so there are 29 choices for it; c1 is odd, so there are at most 28

choices for c1 (mod 29) (a1 = 1); for 2 ≤ i ≤ 5 there are at most 210−ai choices forci (mod 210−ai). Hence the number of complete remainder sequences is at most

29 · 28 ·5∏i=2

210−αi = 29 · 28 · 27 · 26 · 23 · 22 = 235.

9. Step 1 Let k ≥ 1 be an integer and p be a prime. Call

(f(0), . . . , f(i)) (mod pk)

the (i+ 1)-subsignature of f modulo pk.

By Problem 4, each polynomial f ∈ (Z/pkZ)[x] can be uniquely represented as

f(x) =∑i≥0

aixi, ai ∈ Z/pkZ (5)

The following are clear:

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(a) The term aixi does not affect the value of f(0), . . . , f(i− 1).

(b) p - i! for 0 ≤ i < p.

(c) p||i! for p ≤ i < p2.

(d) p|xi when i ≥ p.

(e) p2|xi when i ≥ 2p.

We show that for m ≤ p, the pkm choices for a0, . . . , am−1 give the pkm distinct m-term sequences modulo pk as m-subsignatures. This is clear for m = 1. Suppose thisis true for a certain m < p. Given a sequence (y0, . . . , ym), we can find a0, . . . , am−1in (5) such that f(i) = yi for 0 ≤ i < m. Next note the term amx

m ranges overall of Z/pkZ for x = m since p - m! implies m! is invertible modulo pk. Hencewe can choose am to make f(m) = ym (this does not affect f(0), . . . , f(m − 1) byitem (a) above), and all pk(m+1) distinct (m + 1)-term sequences modulo pk are(m+ 1)-subsignatures.

Hence there are pkp possible p-subsignatures modulo pk. In particular, there are ppossible signatures modulo p and p2 p-subsignatures modulo p2.

Step 3 Now, we show there are p3p 2p-subsignatures (y0, y1, . . . , y2p−1) modulo p2,corresponding to the sequences

(a0 mod p2, . . . , ap−1 mod p2, ap mod p, . . . , a2p−1 mod p). (6)

Again, we induct on the following statement: there are p2ppk (p+ k)-subsignatures(y0, y1, . . . , yp+k−1) modulo p2, corresponding to the coefficients

(a0 mod p2, . . . , ap−1 mod p2, ap mod p, . . . , ap+k−1 mod p). (7)

This is true for k = 0 by Step 2; once it has been established for some k < p, notethat ap+kx

p+k ranges over p residues modulo p2 when x = p + k, as p||(p + k)!.Hence given the coefficients (7) there are exactly p possible values for f(p + k),giving p2ppk+1 possible (p+ k + 1)-subsignatures. The claim follows.

Step 4 The 2p-subsignature modulo p2 completely determines the signature modulop2.

From Step 3, each such subsignature corresponds to a sequence as in (6). Sincep2|xi for i ≥ 2p and p|xi for i ≥ p,

f(x) =∑i≥2p

ai xi︸︷︷︸0 mod p2

+∑

p≤i<2p

ai︸︷︷︸determined mod p

xi︸︷︷︸0 mod p

+∑i<p

ai︸︷︷︸determined mod p2

xi

is determined modulo p2 for each x.

Step 5 If p||n then there are pp possible signatures modulo p and if p2||n thenthere are p3p possible signatures modulo p2. Hence, since a sequence modulo pvp(n)

for every prime p|n uniquely determines the sequence modulo n by the ChineseRemainder Theorem, there are at most∏

p||n

pp

∏p2||n

p3p

=

∏p|n

pp

∏p2|n

p2p

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possible signatures modulo n. All of these are attainable: it suffices to show thatgiven a valid signature sp modulo pvp(n) for every prime p|n, we can find a polynomialthat has signature sp modulo pvp(n) for every prime p|n. There exists a polynomialRp with signature sp; then

f =∑

prime p|n

∏q|n,q 6=p

q2

∏q|n,q 6=p

q2

−1 (mod p2)

Rp

is the desired polynomial.

10. Step 1 A polynomial P (x) has all terms divisible by a prime p if and only ifx(x − 1) · · · (x − (p − 1)) | P (x) in Z/pZ. Call a polynomial p-valid if x(x −1) · · · (x− (p− 1)) - P (x) modulo p.

Step 2 A polynomial is fine if and only if it is p-valid for every prime p. Indeed,the forward assertion follows from Step 1. Conversely, suppose that the polynomialis p-valid for every prime p; then there does not exist a prime p such that x(x −1) · · · (x − (p − 1)) | P (x) in Z/pZ. Given k ∈ N, let p1, . . . , pj be all the primesdividing k. Then, for 1 ≤ i ≤ j, there exists xi ∈ Z/pZ such that P (xi) 6≡ 0(mod pi) for all 1 ≤ i ≤ j. Then any x′ ∈ N with x′ ≡ xi (mod pi) causes P (x′) tonot be divisible by any of p1, . . . , pj and hence relatively prime to k. By the ChineseRemainder Theorem we can find infinitely many such x′, so P (x) is fine.

Step 3 Let p1 < . . . < pm be all primes less than n. In S the proportion of pi-validpolynomials modulo pi is 1− 1

ppii

. Indeed, write

P (x) ≡ x(x− 1) · · · (x− (pi − 1))Q(x) +R(x) (mod pi),

where R(x) is the remainder upon dividing P (x) by R(x). Q(x) can be any polyno-mial of degree n− pi, and for any choice of Q(x), exactly ppii − 1 out of ppii choicesfor R(x) are possible, namely any polynomial of degree less than pi except the zeropolynomial.

Step 4 The equations P (x) ≡ Ri(x) (mod pi) uniquely determine P (x) modulop1p2 · · · pm. (This is a direct application of the Chinese Remainder Theorem.) Sincethe proportion pi-valid polynomials modulo pi is 1 − 1

ppii

, the proportion of poly-

nomials in S that are pi-valid for each i is∏m

i=1

(1− 1

ppii

). Now p1p2 · · · pm|n! so

there are a fixed number of polynomials corresponding to each polynomial modulop1 · · · pm that is pi-valid for each i. Hence the proportion of fine polynomials is at

most∏m

i=1

(1− 1

ppii

).

(C) Divisibility, GCD, and Irreducibility

1. There exist polynomials u, v ∈ Q[x] so that uf +vg = 1. Clearing denominators weget the desired representation. Now if uf+vg = k for f, g ∈ Z[x] then replacing u by(u mod g) and v by (v mod g), we get u′f+v′g = k (why?) with u′, v′ having degreesless than deg(f), deg(g), respectively. Note that u′, v′ are uniquely determined up to

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a constant multiple, since we need u′f ≡ −k (mod g), and u′ is determined modulog up to a constant factor. Letting c be gcd(a0, . . . , am, b0, . . . , bn), we necessarilyhave u1 = ±u/c, and v1 = ±v/c, giving that k = ±ck0.

2. We have uf(n)+vg(n) = k for some u, v ∈ Z and nonzero . Hence gcd(f(n), g(n)) |k.

3. Let fn denote f iterated n times. We will encode the statement of the problem inpolynomials. For a polynomial p(x) = cnx

n + · · ·+ c1x+ c0, we will denote by p(f)the function cnf

n(x) + · · ·+ c1f(x) + c0x. We are given (x3 + 2x4 +x− 4)f = 0 and(x2009−1)f = 0. Since gcd(x3 +2x2 +x−4, x2009−1) = x−1 and hence there existpolynomials u(x), v(x) ∈ Q[x] for which u(x)(x3+2x2+x−4)+v(x)(x2009−1) = x−1.Then

0 = (u(x)(x3 + 2x2 + x− 4)f)(t) + (v(x)(x2009 − 1)f)(t) = ((x− 1)f)(t) = f(t)− t

as desired.

4. First we prove uniqueness. In any nonzero multiple of (x+2)(x+5) = x2 +7x+10,the coefficient of the term with smallest exponent must be divisible by 10.

Any two polynomials both satisfying the given condition must have difference di-visible by (x + 2)(x + 5). Given any two distinct polynomials Q and R havingdifference divisible by x2 + 7x+ 10, one coefficient must differ by at least 10, so thecoefficients of Q and R cannot all be in [0, 10).

Now we show existence. Start with the polynomial n and repeat the followingprocess. If the smallest term with coefficient 10 or greater is anx

n, add xn(x +2)(x + 5)(x − 1) = xn(x3 + 6x2 + 3x2 − 10) to this polynomial (call this firing atposition n). Changing the polynomial by a multiple of (x + 2)(x + 5) does notchange its value at −2,−5. Furthermore, note the sum of coefficients stays thesame. If at some time the polynomial has all coefficients less than 10 we are done.

Suppose this process goes on forever. Consider the sequence of triplets (an, an−1, an−2)where an is the leading coefficient of the polynomial. By our process, n − 3 musthave fired in the previous step; positions less than n−23 will never fire again; termsxi, i < n− 3 have coefficients less than 10. Since there are a finite number of possi-bilities for (an, an−1, an−2) (as their sum is bounded), at some later time the sametriplet must appear twice, say as the coefficients of xn, xn−1, xn−2 at step t1 andas the coefficients of xm, xm−1, xm−2 at step t2 > t1. The coefficients of xn−3, . . . , 1have not changed in the intervening steps. Then the difference between the poly-nomials at these two times is (xm−n − 1)(anx

2 + an−1x+ an−2). Since x2 + 7x+ 10must divide this polynomial but it has no factor in common with xm−n− 1, it mustdivide anx

2 + an−1x+ an−2. This means that anxn + an−1x

n−1 + an−2xn−2 evaluates

to 0 at −2,−5. Hence if we remove these terms at time t1, then we get a polynomialsatisfying the desired conditions.

5. By division with remainder we can write f(x) = q(x)g(x) + r(x) where deg(r) <deg(f). Then

f(n)

g(n)= q(x) +

r(n)

g(n).

10


When |n| is large enough, |r(n)| < |g(n)|. Hence r(n) = 0 for infinitely many n,and in fact r(x) = 0, g | f .

6. Let f(x) = xn + x2 − 1 and g(x) = xm + x− 1. By the previous problem, we needf(x)|g(x). Both f, g have a unique root in (0, 1), since both are increasing and gofrom −1 to 1. This must be the same root since g|f ; call it α.

We use α to show m < 2n. Certainly α > φ where φ is the positive root ofh(x) := x2 − x+ 1 = 0. This is because f is increasing in (0, 1) and f(φ) < h(φ) =0 = f(α). On the other hand, if m ≥ 2n then 1− α = αm ≤ (αn)2 = (1− α2)2 andthe outer terms rearrange to give α(α− 1)(α2 − α + 1) ≥ 0, which requires α ≤ φ,a contradiction.

We show that the only solution with m < 2n is (m,n) = (5, 3). Since xn + x2 −1|(xm−n(xn + x2 − 1) − (xm + x − 1)), xn + x2 − 1|xm−n+2 − xm−n − x + 1. Sincexm−n+2−xm−n−x+1 is not identically 0, and since m ≤ 2n−1, m−n+2 is either mor m+1. In the first case, we must have xm−n+2−xm−n−x+1 = xn+x2−1, which isimpossible. In the second case, we must have (x−1)(xn+x2−1) = xn+1−xn−1−x+1so −xn−1 = −xn + x3 − x2, which can only happen if n = 3, and therefore m = 5.As (x3 + x2 − 1)(x2 − x+ 1) = (x5 + x− 1), this solution works.

7. In light of Problem 1, we just need to find u, v with uf + vg a constant, deg(u) <deg(g), and with the coefficients of f, g together having gcd 1. This constant willbe the desired k0. For polynomials f, g define (f mod g) to be the remainder whenf is divided by g. For an integer d and f ∈ Z we write d||f if d is the maximalpositive integer that divides every coefficient of f . We need the following facts.

(a) (from Gauss’s Lemma) If p, q ∈ Z[x] and d||p, e||q then de||pq.(b) For integer n ≥ 1, integer k, 0 < k < 2n,

(2n

k

)is even. Moreover,

gcd

((2n

1

), . . . ,

(2n

2n − 1

))= 2.

(c) If u, v ∈ Z[x], v is monic, there exist q, r ∈ Z[x] such that u = qv + r,deg(r) < deg(u).

Let P (x) = (xn−1)n(−2)n . Note the roots of xn + 1 are eπk/n for k odd, 0 < k < 2n. For

any root x of xn + 1 = 0,

P (x) =(xn − 1)n

(−2)n=

(−2)n

(−2)n= 1.

So P (x) ≡ 1 (mod xn + 1). Since (−1)n − 1 = 0, x + 1 divides xn − 1. Hence

(x+ 1)n | (xn−1)n(−2)n or

P (x) ≡ 0 (mod (x+ 1)n).

Therefore, for some polynomials f, g,

P (x) = (x+ 1)nf = −(xn + 1)g + 1.

11


We have f(x) =(

xn−1(−2)(x+1)

)n, and

(−2)n =

(xn − 1

x+ 1

)n(x+ 1)n + (−2)ng(x)(xn + 1).

We can write (xn − 1

x+ 1

)n= s(x)(xn + 1) + t(x)

where deg(t) < deg(xn + 1) = n, and find

(−2)n = t(x)(x+ 1)n + [(−2)ng(x) + s(x)(x+ 1)n](xn + 1).

Now we find the d ∈ N such that d||t(x).

Claim 1 2(2r−1)q is the highest power of 2 dividing[(

xn−1x+1

)nmod x2

r+ 1].

We have

xn − 1

x+ 1= (x− 1)(x2 − 1) · · · (x2r − 1)(x2

r(q−1) + · · ·+ x+ 1).

For 0 ≤ i < r

(x2i − 1)2

r−i=

2r−i∑k=0

(2r−i

k

)x2

ik(−1)k ≡2r−i−1∑k=1

(2r−i

k

)x2

ik(−1)k (mod x2r

+ 1).

By item (c), 2|(2r−i

k

)for 1 ≤ k ≤ 2r−i − 1. Let fi(x) = 1

2

∑2r−i−1k=1

(2r−i

k

)x2

ik(−1)k.Note that fi ∈ Z[x] and

fi(1) =1

2((x2

i − 1)2r−i − 2)|x=1 = −1.

We have

(x2i − 1)2

rq = [(x2i − 1)2

r−i]2iq ≡ 22iqfi(x)2

iq (mod x2r

+ 1)

Multiplying these equations for 0 ≤ i < r we get

[(x− 1)(x2 − 1) · · · (x2r−1 − 1)]n = 2(20+21+···+2r−1)q

r−1∏i=0

fi(x)2iq

= 2(2r−1)qr−1∏i=0

fi(x)2iq (mod x2

r

+ 1)

Let w(x) =[(∏r−1

i=0 fi(x)2iq)

(x2r(q−1) + · · ·+ x+ 1)n

]. By item 3, we can write,

w(x) = w1(x)(x2r

+ 1) + w2(x);w1, w2 ∈ Z[x]. Then w(1) is odd so the sum ofcoefficients of w(x) is odd and not all coefficients of w2(x) are even. Hence(xn − 1

x+ 1

)nmod x2

r

+ 1 = [(x− 1)(x2 − 1) · · · (x2r−1 − 1)(x2r(q−1) + · · ·+ x+ 1)]n mod x2

r

+ 1

= 2(2r−1)q[w(x) mod x2r

+ 1]

12


and 2(2r−1)q is the highest power of 2 dividing xn−1x+1

mod x2r

+ 1.

Claim 2 22rq is the highest power of 2 dividing(xn−1x+1

)nmod x2

r(q−1) − · · · − x2r + 1.

We have

x2r(q−1) + · · ·+ x2

r

+ 1 ≡∑

0<i<q−1,i odd

2x2ri (mod x2

r(q−1) − · · · − x2r + 1)

so(xn−1x+1

)nis congruent to

[(x− 1)(x2 − 1) · · · (x2r − 1)(x2r(q−1) + · · ·+ x2

r

+ 1)]n

≡ [(x− 1)(x2 − 1) · · · (x2r − 1)]n2n

( ∑0<i<q−1,i odd

x2ri

)n

(mod x2r(q−1) − · · · − x2r + 1)

and the claim follows.

Let R1 =(xn−1x+1

)nmod x2

r+ 1 and R2 =

(xn−1x+1

)nmod x2

r(q−1) − · · · − x2r + 1. Let

Q =∑

0<i<q,i odd x2ri, P1 =

∑1≤i<q x

2ri, P2 = x2r

+ 1. Then

P1 = QP2 + 1

R1 −R2 = (R1 −R2)(P1 − qP2)

R1 − P1(R1 −R2) = R2 − P2Q(R1 −R2)

Let F = R1−P1(R1−R2). Then F ≡ R1 (mod x2r+1) and F ≡ R2 (mod x2

r(q−1)−· · · − x2

r+ 1). Hence by the Chinese Remainder Theorem, since t(x) ≡ R1

(mod x2r

+ 1), t(x) ≡ R2 (mod x2r(q−1) − · · · − x2

r+ 1), and deg(F ) < n we

have F (x) = t(x). Let G(x) = (−2)ng(x)− s(x)(x+ 1)n. Then

(−2)n = F (x)(x+ 1)n +G(x)(xn + 1).

Note 2(2r−1)q is the highest power of 2 dividing R1, and 22rq|R2, F (x) = R1(x) −P1(x)(R1(x)−R2(x)) = (1− P1(x))R1(x) + P1(x)R2(x). 2 - 1− P1(x) so 2(2r−1)q isthe highest power of 2 dividing S1(x) = (1 − P1(x))R1(x). There exists j so thatthe coefficient uj of xj in S1(x) is not divisible by 2(2r−1)q+1. Let the coefficient ofxj in P1(x)R2(x) be bj. Note 22rq|R2 implies 22rq||bj. Hence 2(2r−1)q+1 - aj + bj, and

2(2r−1)q+1 - F (x). Let F(x) = F (x)

2(2r−1)q and G(x) = G(x)

2(2r−1)q . Then F(x) ∈ Z[x]. Let

F(x) =∑l

i=0 cixi,G(x) =

∑mi=0 dix

i. Then

2q = 22rq−(2r−1)q = F(x)(x+ 1)n + G(x)(xn + 1).

Since (xn+1)|2q−F(x)(x+1)n in Q[x], by divisibility holds in Z[x], i.e. G(x) ∈ Z[x].2 - gcd(c0, . . . , cl) and gcd(c0, . . . , cl, d0, . . . , dm)|2q. Hence the gcd is 1. By Problem1, k0 = 2q.

(D) Algebraic Numbers

1. Let β be the root of greatest absolute value. Suppose by way of contradictionf(α3 + 1) = 0. Since f(x) is irreducible and f(α) = 0, it is the minimal polynomialof α. Since the polynomial g(x) = f(x3 + 1) has α as a root f(x) | f(x3 + 1).Hence f(β) = f(β3 + 1) = 0. Let r = |β|. Now r3 − r − 1 > 0 for r ≥ 3

2, so

|β3 + 1| ≥ |β|3 − 1 > |β|, contradicting the maximality of β.

13


2. Suppose the LHS is not 0. The conjugates of c1a1 + . . . + cnan are among thenumbers c1b1 + . . .+ cnbn where bi is a conjugate of ai. Each of these have absolutevalue at most |c1a′1|+ · · ·+ |cna′n|. The product of all conjugates (including c1a1 +. . .+ cnan) is an integer because it is the constant term of the minimal polynomialof c1a1 + . . . + cnan; hence it is at least 1. The conjugates multiply to an integerwith absolute value at least 1, they number at most d1 · · · dn and each is at most

|c1a′1|+ · · ·+ |cna′n|; hence each is at least(

1|c1a′1|+···+|cna′n|

)d1d2···dn−1.

3. This time we can do better than in Problem 2. Suppose ω1, . . . , ωk are the roots ofunity. We claim that

p−1∏i=1

(x− (ωi1 + . . .+ ωik)) (8)

has integer coefficients since each coefficient. Indeed, we can write each coefficientin the form

a0 + a1ω + · · ·+ ap−2ωp−2, (9)

where ω is a primitive pth root of unity. If we replace ω by ωi for some 1 ≤ i ≤ p−1,then the product in (8) stays the same. Hence (9) stays the same as well, and since1, ω, . . . , ωp−2 satisfy no linear dependency relation, we must have a0 = · · · = ap−2,and (9) is an integer. (A more advanced way of looking at this is the following:the coefficients are in the fixed field of all automorphisms of Q(ω) fixing Q; GaloisTheory allows us to conclude that these coefficients are actually in Q.)

Thus the conjugates of ω1 + · · · + ωk are among the ωi1 + . . . + ωik. Since each hasabsolute value at most k, and their product is a nonzero integer (and hence hasabsolute value at least 1), each must be at least 1

kp−2 .

(E) Cyclotomic and Chebyshev Polynomials

1. Let the side lengths be a0, . . . , ap−1. Then letting ω be the pth root of unity,

ap−1ωp−1 + · · ·+ a0 = 0

Hence ω satisfies the polynomial equation ap−1xp−1 + · · · + a0. Since the minimal

polynomial of ω is Φp = xp−1 + · · ·+ x+ 1, xp−1 + · · ·+ x+ 1 | ap−1xp−1 + · · ·+ a0,i.e. ap−1 = · · · = a0.

2. The factorization of xn − 1 into irreducible (monic) factors P1 · · ·Pk gives a (notnecessarily complete) factorization of 2n − 1. If Pi has degree d then Pi(2) ≤ 3d

sincePi(2) =

∏ω root of Pi

(2− ω)

and all roots have absolute value 1. Letting ω be a primitive nth root of unity,the irreducible factor containing ωi is Φn/gcd(n,i), which has degree ϕ(m) ≤ ϕ(n) forsome factor m of n. Hence it suffices to find n so that 3ϕ(n) ≤ 2

n1993 , or

ϕ(n) ≤ n ln(2)

1993 ln(3).

14


However, for any C > 0 we can find n so that ϕ(n) < Cn. The product∏

p primepp−1

is infinite (rewrite as geometric series, expand to get the harmonic series). Hence∏p prime

p−1p

= 0, and taking n = p1 · · · pk for sufficiently many distinct primes pi

we find that we can make ϕ(n)/n =∏k

i=1pi−1pi

as close to 0 as desired.

3. (A) Let ω = cos(pπ) + i sin(pπ). Then ω is a root of unity and hence an algebraicinteger; similarly ω is as well. Hence 2 cos(pπ) = ω+ω is an algebraic integer; sinceit is rational it is a rational integer. Hence we must have 2 cos(pπ) = 0,±1,±2,giving p = 0, 1

3, 12, 23, 1.

(B)Step 1 If a is nonreal, aa ∈ Q and the irreducible polynomial of a over Q hasdegree n, then the irreducible polynomial of a + a has degree n/2. Indeed, [Q(a) :Q(a + a)] = 2 since a satisfies a quadratic equation in Q[a + a] (namely x2 − (a +a)x+ aa = 0), and a 6∈ R implies a 6∈ Q(a+ a). Since

n = [Q(a) : Q] = [Q(a) : Q(a+ a)][Q(a+ a) : Q],

the desired result follows. In particular, this holds for roots of unity.

Step 2 If cos(pπ) = 12(ω+ω) has degree n over Q, then ω has degree 2n. However, if

2p = qr

in reduced form, then ω is a rth root of unity, and its irreducible polynomialis the cyclotomic polynomial Φr. Hence 2n = ϕ(r). If n = 2, then r = 4, and thesolutions to ϕ(r) = 4 are 5, 10, 12. Hence in addition to the p above, rational p withdenominators 5 or 6 also work.

Try to use this method to describe the factorization of Chebyshev polynomials!

4. It is easy to check that√n+ 1 +

√n is a conjugate of

√n+ 1 −

√n. If cos pπ =

12(√n+ 1−

√n) then 1

2(√n+ 1−

√n) is a zero of some Chebyshev polynomial. But

then 12(√n+ 1+

√n) > 1 is also a zero, a contradiction since all zeros of Chebyshev

polynomials are in [−1, 1].

5. Solution 1 Let cos(kθ) = p ∈ Q, cos[(k + 1)θ] = q ∈ Q. Then

cos θ = cos[(k + 1)θ] cos θ ± sin[(k + 1)θ] sin kθ

= pq ±√

(1− p2)(1− q2)= a±

√b

for some a, b ∈ Q such that b is not a perfect square. Now Tk(a±√b) = Tk(cos(θ)) =

cos(kθ) = p so either a+√b or a−

√b is a zero of the polynomial Tk(x)− p ∈ Q[x].

The minimal polynomial in Q[x] having a +√b (or a −

√b) as a root is R(x) =

[x− (a+√b)][x− (a−

√b)] so R(x)|Tk(x)−p. Now if a∓

√b 6∈ [−1, 1] then we have

Tk(a∓√b) 6∈ [−1, 1] and hence not equal to p, a contradiction. So there must exist

φ so that cos(φ) = a∓√b. By similar reasoning a∓

√b is a root of Tk+1(x)− q = 0.

15


Now cos(kφ) = Tk(a∓√b) = p and similarly cos[(k + 1)φ] = q. In summary,

cos θ = a±√b (10)

cosφ = a∓√b (11)

cos kθ = cos kφ = p (12)

cos[(k + 1)θ] = cos[(k + 1)φ] = q (13)

Now (12) implies kφ = 2πj ± kθ for some j ∈ Z, or

φ =2πj

k± θ (14)

and (13) implies

φ =2πl

(k + 1)± θ (15)

for some l ∈ Z. Since b 6= 0, a+√b 6= a−

√b and cos θ 6= cosφ. If (14) or (16) hold

true for the same sign, then 2πjk

= 2πlk+1

and j(k + 1) = lk; since gcd(k, k + 1) = 1but k - j this is a contradiction. So equating (14) and (16) gives

θ = 2π · m

2k(k + 1)

for some m ∈ Z. Since cos(θ) is the root of a quadratic, we know by Problem 3 thatcos θ =

√3/2, 1/2, (

√5± 1)/2. It is easy to check that θ = π/6 is the only solution.

Solution 2 We claim that if cos kθ and cos lθ are rational for relatively primepositive integers k, l, then either cos θ is rational or θ is a rational multiple of π.

Let cos kθ = p and cos lθ = q as before. Noting that eiφ + 1eiφ

= 2 cosφ, we havethat eiθ is a root of

xk +1

xk= 2p =⇒ x2k − 2pxk + 1 = 0 (16)

xl +1

xl= 2q =⇒ x2l − 2qxl + 1 = 0 (17)

Suppose θ is not a rational multiple of π. Then the numbers e±iθ+2πijk for 0 ≤ j < k

are all distinct. They are the roots of (16). Similarly, e±iθ+2πijl are the roots of (17).

Since θ is not a rational multiple of π and k, l are relatively prime, we have that

the only solution to e±iθ+2πij1k = e±iθ+

2πij2k for 0 ≤ j1 < k and 0 ≤ j2 < l for some

choices of signs is when j1 = j2 = 0. Thus the only roots the two polynomials sharein common are ω = eiθ and ω̄ = −eiθ, and hence the polynomial

g(x) := (x− eiθ)(x− e−iθ)

is the greatest common divisor of polynomials x2k − 2pxk + 1 and x2l − 2qxl + 1,which have rational coefficients. Hence g(x) has rational coefficients, and eiθ+e−iθ =2 cos θ is rational.

Now by Problem 3, the only rational multiples of π that give a rational values forcosine are multiples of π

6. By our claim, θ is a rational multiple of π. Hence kθ and

(k + 1)θ are both multiples of π6, and so must θ. Since cos θ is irrational, θ = π

6.

16


6. Let p(x) = xn + an−1xn−1 + · · ·+ a0. Since x ∈ [−1, 1], we set x = cos(θ).

Step 1 Write p(cos θ) in the form

p(cos θ) = bn cos(nθ) + bn−1 cos((n− 1)θ) + · · ·+ b0.

We know that cosnθ = Tn(cos θ), so this is possible by Problem B4. Moreover,since the leading coefficient of Tn is 2n−1 (by induction), we must have bn = 1

2n−1 .

Step 2 The maximum of h(cos θ) = cn cos(nθ) + · · ·+ c1 cos θ + c0 is at least |cn|.Assume WLOG cn > 0. Suppose by way of contradiction that max |h(cos θ)| < cn.Then

cnTn

(cos

(kπ

n

))− h

(cos

(kπ

n

))= (−1)kcn − h

(cos

(kπ

n

))is positive for even k and negative for odd k in the range 0 ≤ k ≤ n. We concludethat cnTn(x) − h (x) has at least n roots in [0, 1]. However this polynomial hasdegree less than n, contradiction.

Applying Step 2 to p gives the desired result.

7. By induction, the constant term of Tk is 0 for k odd and ±1 for k even, and theleading coefficient is 2k. Hence by Vieta’s formula, the product of the roots of T2nis

cosπ

4n· cos

3π

4n· · · cos

(4n− 1)π

4n=±1

22n−1 .

Since cos π4n· cos 3π

4n· · · cos (2n−1)π

4n= ± cos (2n+1)π

4n· cos (2n+3)π

4n· · · cos (4n−1)π

4n, we get

cosπ

4n· cos

3π

4n· · · cos

(2n− 1)π

4n=

1

2n−12

.

(F) Polynomials in Number Theory

1. By Fermat’s Little Theorem, xp−1 ≡ 1 (mod p). Thus in Z/pZ,

xp−1 − 1 ≡p−1∏i=1

(x− i) (mod p). (18)

Write (x − 1)p−1 =∑p−1

i=0 aixi. Then matching coefficients on both sides of (18)

givesai ≡ 0 (mod p) for all 1 ≤ i < p− 1. (19)

Since p ≥ 5, letting x = p gives

(p− 1)! = (x− 1)p−1 = pp−1 +

(p−2∑i=2

aipi

)+ a1p+ (p− 1)!

since (−1)(−2) · · · (−p + 1) = (−1)p−1(p − 1)! = (p − 1)!. Subtracting (p − 1)! onboth sides,

0 = pp−1 +

(p−1∑j=2

aipi

)+ a1p.

17


Using (19), p3 | aipi for 2 ≤ i ≤ p − 1. Hence, since p ≥ 5, p3 | pp−1 +∑p−1

i=2 aipi.

Since p3 divides the LHS, p3 | a1p and p2 | a1. Now p3 | (kp)p−1 +(∑p−2

i=2 ai(kp)i)

as well and we get

(kp− 1)p−1 = (x− 1)p−1|x=pk

= (kp)p−1 +

(p−1∑j=2

ai(kp)i

)+ a1p+ (p− 1)!

≡ (p− 1)! (mod p3). (20)

Now, (pa

pb

)=

(pa)pb

(pb)!

=

∏ai=a−b+1[(pi)(pi− 1)p−1]∏bi=1[(pi)(pi− 1)p−1]

=ab

b!

[b∏i=1

[p(i+ a− b)− 1]p−1

(pi− 1)p−1

]

=ab

b!

[b∏i=1

[p(i+ a− b)− 1]p−1

(pi− 1)p−1

](21)

By (20), [p(i + a − b) − 1]p−1 ≡ (pi − 1)p−1 (mod p3). Hence (21) becomes(ab

)modulo p3, as needed.

2. Continuing the above, we have by Viete’s formula that

a1 = (p− 1)!

(1 +

1

2+ · · ·+ 1

p− 1

).

We’ve shown that p2|a1, as needed.

3. Rearranging, it suffices to show (p2 − 1)p−1 ≡ (p− 1)! (mod p4)

Continuing the above, if we plug in p2 into the polynomial (x− 1)p−1 we get

(p2 − 1)p = p2(p−1) +

p−2∑i=2

aip2i + a1p

2 + (p− 1)! ≡ (p− 1)! (mod p4).

4. Consider the polynomial

P (x) = (x+ a)(x+ b)(x+ c)− (x− d)(x− e)(x− f) = Sx2 +Qx+R

where Q = ab + bc + ca − de − ef − fd and R = abc + def . Since S|Q,R, itfollows that S|P (x) for every x ∈ Z. Hence S|P (d) = (d + a)(d + b)(d + c). SinceS > d + a, d + b, d + c and thus cannot divide any of them, it follows that S mustbe composite.

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5. Note that for any odd p, p|n! + 1 has a finite number of positive integer solutionsbecause for n ≥ p, p|n!, p - n! + 1. Suppose the solutions to

n! + 1 ≡ 0 (mod p)

are exactly n1 < n2 < · · · < nm with m > 1. Then for each i, 1 ≤ i < m we have

ni+1! ≡ −1 (mod p)

ni! ≡ −1 (mod p)

Dividing these 2 relations gives

(ni + 1)(ni + 2) · · · (ni+1) ≡ 1 (mod p).

Letting k = ni+1 − ni, we see that x = ni is a solution to

(x+ 1)(x+ 2) · · · (x+ k) ≡ 1 (mod p).

For each k, 1 ≤ k < p the polynomial

(x+ 1)(x+ 2) · · · (x+ k)− 1 ∈ Z/pZ[x]

has at most k roots modulo p. Therefore, there are at most k indices i whereni+1 − ni = k, for each k with 1 ≤ k < p.

Now let j be the smallest positive integer such that m ≥ j(j+1)2

. Then (j+1)(j+2)2

≥m ≥ j(j+1)

2and

p > nm − n1 =m−1∑i=1

(ni+1 − ni)

≥j∑i=1

i2

=j(j + 1)(2j + 1)

6

where the second inequality follows from the fact that for a fixed k, 1 ≤ k < p,ni+1 − ni has at most k solutions. Since m ≥ j(j+1)

2=∑j

i=1 j, when the differencesni+1 − ni are written in ascending order, the first is at least 1, the next two are atleast 2, and so on, each time the next i differences are at least i and hence sum toat least i2, up to i = j. Now since

limj→∞

j(j + 1)(2j + 1)/6

((j + 1)(j + 2)/2)3/2

is finite and greater than 0, there exists a constant 0 < c1 < 1 such that thisexpression is greater than c1 for each j ∈ N; letting c = 1

c1we see that

p >j(j + 1)(2j + 1)

6≥ c1

((j + 1)(j + 2)

2

)3/2

≥ c1m3/2

or m < cp2/3. (Requiring c1 < 1 leads to cp2/3 > 1 so this is true even if m = 0, 1.)

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6. We continue from Problems 1-3. The numerator of fp(x)− fp(y) in lowest terms isdivisible by p3 iff

p−1∑k=1

1

(px+ k)2−

p−1∑k=1

1

(py + 1)2≡ 0 (mod p3),

where the inverse is taken modulo p3. First note that since

p | aj = (−1)p−1−j(p− 1)!∑

1≤i1<···<ij≤p−1

1

i1i2 · · · ij

for 1 ≤ j ≤ p− 1 by Vieta’s formula,∑1≤i1<···<ij≤p−1

1

i1i2 · · · ij≡ 0 (mod p).

From Problem 2,p−1∑i=1

1

i≡ 0 (mod p2).

Now

∑1≤i<j<k≤p−1

1

ijk=

( ∑1≤i≤p−1

1

i

)3

−

( ∑1≤i≤p−1

1

i3

)− 3

( ∑1≤i<j≤p−1

1

ij

)( ∑1≤i≤p−1

1

i

).

Hence by Vieta’s,

a3 = (p− 1)!

( ∑1≤i<j<k≤p−1

1

ijk

)≡ 0 (mod p2).

Let P (x) = (x− 1)p. Write

f(t) :=

p−1∏i=1

[t− (px+ i)] = tp−1 + bp−2tp−2 + · · ·+ b1t+ b0.

Then

tp−1 + bp−2tp−2 + · · ·+ b0 = (t− px)p−1 + ap−2(t− px)p−2 + · · ·+ a1(t− px) + a0.

Matching the coefficient of t2 on both sides, using the Binomial Theorem,

bn ≡p−1∑n=2

an

(n

2

)(px)n−2

≡ a2 + a3p+ a4p2 ≡ a2 (mod p3)

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(We used p > 5⇒ p|a4.) By Vieta’s formula,

b2 = (px+ p− 1)p−1

( ∑1≤i<j≤p−1

1

(px+ i)(px+ j)

).

By Wolstenholme,

(px− 1)p−1 ≡ (py − 1)p−1 (mod p3).

Using Wolstenholme and a2 ≡ b2 (mod p3), we get∑1≤i<j≤p−1

(1

(px+ i)(px+ j)

)≡

∑1≤i<j≤p−1

(mod p3).

So

p−1∑k=1

1

(px+ k)2≡

(p−1∑k=1

1

(px+ k)2

)2

− 2∑

1≤i<j≤p−1

1

(px+ i)(px+ j)

≡ 0− 2∑

1≤i<j≤p−1

1

(px+ i)(px+ j)

≡

(p−1∑k=1

1

(py + k)2

)2

− 2∑

1≤i<j≤p−1

1

(py + i)(py + j)

≡p−1∑k=1

1

(py + k)2(mod p3)

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