Homework 7 solutions M328K
by Mark Lindberg/Marie-Amelie Lawn
Problem 1:
4.4 # 2:x3 + 8x2 − x − 1 ≡ 0 (mod 1331).
a) x3 + 8x2 − x − 1 ≡ 0 (mod 11). This does not break down, so trial and error gives:
x = 0 : f (0) = −1 ≡ 10 6≡ 0 (mod 11).
x = 1 : f (1) = 7 6≡ 0 (mod 11).
x = 2 : f (2) = 8 + 32 − 2 − 1 = 37 ≡ 4 6≡ 0 (mod 11).
x = 3 : f (3) = 27 + 72 − 3 − 1 ≡ 5 + 6 − 3 − 1 = 7 6≡ 0 (mod 11).
Xx = 4 : f (4) = 64 + 128 − 4 − 1 ≡ 9 + 7 − 4 − 1 = 11 ≡ 0 (mod 11).
Xx = 5 : f (5) = 125 + 200 − 5 − 1 ≡ 4 + 2 − 5 − 1 = 0 ≡ 0 (mod 11).
x = 6 : f (6) = 216 + 36 ∗ 8 − 6 − 1 ≡ 7 + 3 ∗ 8 − 6 − 1 = 24 ≡ 2 6≡ 0 (mod 11).
x = 7 : f (7) = 343 + 49 ∗ 8 − 7 − 1 ≡ 2 + 5 ∗ 8 − 6 − 1 = 35 ≡ 2 6≡ 0 (mod 11).
x = 8 : f (8) = 512 + 512 − 8 − 1 ≡ 1015 ≡ 3 6≡ 0 (mod 11).
x = 9 : f (9) = 729 + 81 ∗ 8 − 9 − 1 ≡ 719 + 4 ∗ 8 = 751 ≡ 3 6≡ 0 (mod 11).
x = 10 : f (10) = 1000 + 800 − 10 − 1 ≡ 1789 ≡ 7 6≡ 0 (mod 11).
Solutions to x3 + 8x2 − x − 1 ≡ 0 (mod 11) are x0 ≡ 4, 5 (mod 11)
b) x3 + 8x2 − x − 1 ≡ 0 (mod 11). Note: f 0 (x) = 3x2 + 16x − 1
x0 ≡ 4: f 0 (4) = 48 + 64 − 1 = 111 ≡ 1 6≡ 0 (mod 11). Then, by part 1 of Hensel’s Lemma,
(4)
= − 187
f 0 (4) ≡ 1 ≡ 1 (mod 11). Then t = −(1) f11
11 = −17 ≡ 5 (mod 11). Then
the solution to f (x) ≡ 0 (mod 121) is x1 = 4 + 11 ∗ 5 = 59 (mod 121) .
x0 ≡ 5: f 0 (5) = 75 + 80 − 1 = 134 ≡ 0 (mod 11). Then f (5) = 319 ≡ 77 6≡ 0 (mod 121),
and so there are no solutions mod 121 from this case. (Part 3 of Hensel’s Lemma.)
c) x3 + 8x2 − x − 1 ≡ 0 (mod 1331) x ≡ 59 (mod 121). Then f (59) = 233167. t =
−(1) 233167
121 = −1927 ≡ 9 (mod 11). Then x2 = 59 + 9 ∗ 121 = 1148 mod 1331 .
Problem 2: Since 144 = 24 32 , we can use the Chinese remainder theorem and find simultaneous solutions modulo 24 and modulo 32, then the uniqueness of the Chinese remainder
theorem will give solutions modulo 144. To solve each of these congruences, we use Hensel’s
lemma.
First step: Find solutions modulo 32 :
Step p = 3: As usual, we know that solutions modulo 32 are also solutions modulo 3. Hence
considering the equation modulo 3, the polynomial becomes x5 + x mod 3, since 6 is a multiple of 3.
We see that plugging in 0 gives a solution, whereas plugging in 1 gives 2 and plugging in 2
gives 1 mod 3. Therefore, the only solution modulo 3 is 0.
Step p2 = 9: Now, we use Hensel’s lemma to extend this solution. The derivative of the
1
original polynomial is 5x4 + 1. We see that plugging in 0 gives 1 6≡ 0 mod 3. Therefore, we
are in the first case of Hensel’s lemma.
In this case, the solution modulo 9 = 32 is given by 0 + t3 where
t = −f 0 (0)(6/3) = −1(6/3) ≡ 2
mod 3,
and f 0 (0) is the inverse of f (0) mod 3. Thus, the solution modulo 9 is 0 + 2.3 = 6.
Second step: Find solutions modulo 24 :
On the other hand, reducing this polynomial modulo 2, we get again x5 + x mod 2, since 6
is divisible by 2.
Step p = 2: By inspection we see that there are two solutions to the polynomial congruence
modulo 2 namely x0 ≡ 0, 1 mod 2.
Step p2 = 4: We now want to extend these solutions modulo 4. Take the derivative 5x4 + 1
mod 2.
Plugging in 0 gives 6 ≡ 0 mod 2 and so the roots corresponding to 0 use the first case of
Hensels lemma
Plugging in 1 gives 6 ≡ 0 mod 2. Moreover f (1) = −4 ≡ 0 mod 4 so the roots corresponding
to 1 use the second case of Hensels lemma.
First extend x0 ≡ 0 mod 2, Hensels lemma implies that the solution modulo 4 is given by
0 + t.2 where
f (0)
6
tf 0 (0)
≡ −(1)
mod 2 ≡ −3 mod 2 ≡ 1 mod 2.
2
2
Therefore a solution modulo 4 is 0 + 1.2 ≡ 2 mod 2.
Now extend x0 = 1. We are in the second or third case of Hensel lemma and have that
either 1 and 3 are roots mod 4 or neither of them are roots modulo 4. Plugging in 1 gives
15 + 1 − 6 ≡ −4 ≡ 0 mod 4. Therefore we are in the second case of Hensel lemma and both
1 and 3 are roots modulo 4. Thus, the roots modulo 4 are 1, 2 and 3.
Step p3 = 8: We extend the solutions once more following the same procedure. Since 2
corresponds to the first case of Hensels lemma and plugging in 2 to the original equation
gives 25 + 2 − 6 = 28, Hensels lemma implies that the root of this modulo 8 is given by 2 + t4
where t ≡ −1̄ − 1(28/4) ≡ 1 mod 2.
Therefore, Hensels lemma implies that 2 + 1.4 = 6 is a solution modulo 8. For the other
cases, we are in the second or third case of Hensels lemma so either 1 and 5 are solutions
modulo 8 or neither of them are solutions modulo 8. Since we know that plugging in 1 gives
4 6≡ 0 mod 8, it follows by the third case of Hensel’s lemma that 1 is not a solution modulo
8. Finally, the second case of Hensels lemma implies that 3 and 7 are solutions modulo 8.
Therefore, the solutions modulo 8 are 3, 6, 7.
Step p4 = 16: We now want to extend these solutions modulo 16 by applying Hensels lemma
once again. We get following the same procedure that the solution modulo 16 are 3, 6 and 11.
Now, we must combine these solutions using the Chinese remainder theorem to finish the
problem. The first solution is given by x ≡ 6 mod 9 and x ≡ 3 mod 16. One finds that 51
solves this simultaneous system. The second solution is given by x ≡ 6 mod 9 and x ≡ 6
mod 16. One finds that 6 solves this simultaneous system. Finally, the third solution is given
by x ≡ 6 mod 9 and x ≡ 11 mod 16. One finds that 123 solves this simultaneous system.
Therefore there are three solutions, 6, 51, 123.
Notice that the problem asked only to find how many solutions exist, but this is
a nice example of the use of Hensel’s lemma.
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Problem 3:
1) We can set this problem up as a quadratic congruence. Let 2x + 1 and 2x + 3 be the
two consecutive odd numbers. Then if there product is 11 more than a multiple of 83,
we get
(2x + 1)(2x + 3) ≡ 1 mod 83.
Simplifying, we get 4x2 + 8x − 8 ≡ 0 mod 83. But since 4 and 83 are relatively prime
it is equivalent to
x2 + 2x − 2 ≡ 0 mod 83
√
The roots in R are −1 ± 3 We hence need to find the square roots of 3 modulo 83. By
squaring every number from 0 to 82 (in fact we only need to go half-way), and reducing
modulo 83, we can see that 13 mod 83 and 70 mod 83 are the square roots. (Note that
70 ≡ −13 mod 83, so we could write these as ±13 mod 83, which is why we actually
only need to plug in the numbers 0 to 41). And x ≡ 12, 69 mod 83, which correspond
to the two pairs of consecutive numbers 25, 27 and 139, 141 modulo 83.
2) 4.4 #4 x2 + x + 34 ≡ 0 (mod 81 = 34 ).
x2 + x + 34 ≡ 0 (mod 3)
x = 0 f (0) = 34 ≡ 1 6≡ 0 (mod 3)
Xx = 1 f (1) = 36 ≡ 0 (mod 3)
x = 2 f (2) = 40 ≡ 1 6≡ 0 (mod 3)
x2 + x + 34 ≡ 0 (mod 32 ). f 0 (x) = 2x+1. f 0 (1) = 3 ≡ 0 mod (3). But f (1) = 36 6≡ 0 (mod 81),
so there are no solutions to this or the higher order problems, by part 3 of Hensel’s
Lemma.
Problem 4
Proof by Induction:
Base Case: j = 1. Then we have that f (x) ≡ 0 (mod p) has one solution mod p, namely a1 . Then
we have that f (a1 ) ≡ 0 (mod p), or, by the definition of modulus, that p | f (a1 ). By the
definition of divides, there is an integer n such that f (a1 )−pn = 0. Then, for all x, define
g(x) = f (x) − pn. Because pn is an integer, we have that g(x) is a polynomial. We also
have that pn = f (x) − g(x), or, by the definition of divides, p | (f (x) − g(x)), and by the
definition of modulus, f (x) ≡ g(x) (mod p). Then, we note that g(a1 ) = f (a1 )−pn = 0,
so a1 is a zero of g(x). Then, by some basic theorems of algebra, there must exist some
polynomial q(x) (not necessarily with integer coefficients) such that g(x) = q(x)(x−a1 ).
Then we simply notice that f (x) ≡ g(x) = (x − a1 )q(x), and so we have shown that
such a q(x) exists.
Inductive Step: Note: We assume that p is prime. Inductive Hypothesis: For any f (x) ≡ 0 (mod
p), with solutions a1 , a2 , . . . , aj−1 , there exists a polynomial q(x) such that f (x) ≡
(x−a1 )(x−a2 ) · · · (x−aj−1 )q(x) (mod p). For the inductive step, we have that f (x) ≡ 0
(mod p) has solutions a1 , a2 , . . . , aj−1 , aj . I am assuming that 0 ≤ ai < p, because there
is an equivalent solution (mod p) that is in this range, the least non-negative residue,
by previous theorems, and we would simply use that solution instead. By the inductive
hypothesis, we have that for some polynomial q1 (x), we have that f (x) ≡ (x − a1 )(x −
a2 ) · · · (x − aj−1 )q1 (x) (mod p). Then we know that by the definitions of modulus
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and divides, using x = aj , there is an integer n such that (aj − a1 )(aj − a2 ) · · · (aj −
aj−1 )q1 (aj ) = pn. Then we note that each of (aj − a1 ), (aj − a2 ), . . . , (aj − aj−1 ) is
an integer by the closure of the integers under subtraction, and thus, their product is
an integer by the closure of the integers under multiplication. Then let this integer be
b = (aj − a1 )(aj − a2 ) · · · (aj − aj−1 ), and we have that bq1 (aj ) = pn. Since we have that
each of 0 ≤ a1 , a2 , . . . , aj−1 , aj < p, we have that −p > aj −a1 , aj −a2 , . . . , aj −aj−1 < p.
Then because p is prime, by earlier theorems, we know that, unless aj − ai = ±1,
then it is not a factor of p, and so we have that no factors of b divide p, and so
by previous theorems, b | n. Then there is an integer m such that n = bm by the
definition of divides. Then b · q1 (aj ) = pbm, so q(aj ) = pm, so p | q1 (aj ) by the
definition of divides, and by the definition of modulus, q1 (aj ) ≡ 0 (mod p). Then allow
g(x) = q1 (x) − pm, and note that pm = q(x) − g(x), and so by the definition of divides,
p | (q1 (x) − g(x)), so by the definition of modulus, q1 (x) ≡ g(x) (mod p). Then also
note that g(aj ) = q1 (aj ) − pm = 0, and so aj is a zero of g(x). Then by basic theorems
of algebra, there is some polynomial q2 (x) (Not necessarily with integer coefficients),
such that g(x) = q2 (x)(x − aj ). Then, since q1 (x) ≡ g(x) = (x − aj )q2 (x) (mod p),
and so by congruences, we have that f (x) ≡ (x − a1 )(x − a2 ) · · · (x − aj−1 )q1 (x) ≡
(x − a1 )(x − a2 ) · · · (x − aj−1 )g(x) = (x − a1 )(x − a2 ) · · · (x − aj−1 )(x − aj )q2 (x) (mod p),
as desired. Thus, the inductive step is complete, and the proof by induction is complete.
Problem 5: We use induction over n.
Induction basis: For n = 1 we consider f (x) = c0 + c1 x ≡ 0 mod p, where p does not divide
c1 . Then c1 x ≡ −c0 mod p. Since p does not divide c1 , gcd(c1 , p) = 1. Hence from what
we now about linear congruences, there exists a unique solution x0 modulo p to the equation
c1 x ≡ −c0 mod p, and hence a unique solution to the polynomial congruence, which proves
the case n = 1.
Induction hypothesis: Assume that n > 1, and every polynomial g(x) = b0 + b1 x + . . . bn xn ,
where p does not divide bn and g(x) ≡ 0 mod p has at most n incongruent solutions modulo
p.
Induction step: Let f (x) = a0 + a1 x + . . . an+1 xn+1 , such that p does not divide an+1 . There
are two possibilities: either the equation f (x) ≡ 0 mod p has no solution, then the number
of solutions is 0 < n and there is nothing to check, or there is at least one solution. In that
case we need to check that there are not more than n solutions.
Let x0 be a solution of f (x) ≡ 0 mod p. Then
a0 + a1 x0 + . . . an+1 xn+1
≡0
0
mod p,
and we can factor out a factor (x−x0 ), i.e. there exists a polynomial g(x) = b0 +b1 x+. . . bn xn
of degree n such that
f (x) = (x − x0 )g(x)
mod p = b0 x + b1 x2 + . . . bn xn+1 − (b0 x0 + b1 x0 x + . . . bn x0 xn )
mod p.
Clearly, comparing coefficients of the same power, an+1 = bn , and consequently p does not
divide bn . Hence by induction hypothesis, we have that g(x) ≡ 0 mod p has at most n
solutions modulo p. Together with x0 , f (x) ≡ 0 mod p has then at most n + 1 incongruent
solutions modulo p.
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Problem 6
Part 1:
a) gcd(a, b, c) = 1 ⇒ ax + by + cz = 1 for some integers x, y, z by Bézout’s theorem,
and we have that a2 +b2 = c2 . Then ax+by = 1−cz, or cz = 1−ax−by. Squaring
one of these gives(ax + by)2 = (1 − cz)2 = a2 x2 + 2axby + b2 y 2 = 1 − 2cz − c2 z 2 =
1 − 2cz − a2 z 2 − b2 z 2 . Then a2 x2 + a2 z 2 + b2 y 2 + b2 z 2 + 2axby + 2cz = 1 =
a(a(x2 + z 2 ) + 2xby) + b(b(x2 + z 2 )) + 2(1 − ax − by) = a(a(x2 + y 2 ) + 2xby − 2x) +
b(b(x2 + z 2 ) − 2y) + 2 = 1 ⇒ a(a(x2 + y 2 ) + 2xby − 2x) + b(b(x2 + z 2 ) − 2y) =
−1 ⇒ a(−a(x2 + y 2 ) − 2xby + 2x) + b(−b(x2 + z 2 ) + 2y) = 1. Thus, we have that
−a((x2 + y 2 ) − 2xby + 2x and −b(x2 + z 2 ) + 2y are integers by the closure of the
integers under multiplication and addition. Thus, we have a linear combination of
a and b that equals 1, so by Bézout’s theorem, and the fact that gcd can not be
negative, gcd(a, b) = 1. Similarly, solving for b2 = c2 −a2 , squaring ax+cz = 1−by,
and making similar substitutions to the above leads to a(−a(y 2 +x2 )−2xcz +2x)+
c(c(y 2 − z 2 ) + 2z) = 1, and by the closure of the integers and Bèzout’s theorem,
gcd(a, c) = 1. And, finally, we can see that, with a2 = c2 − b2 , and squaring
by + cz = 1 − ax, we have b(b(y 2 − x2 ) − 2ycz + 2y) + c(−c(y 2 + z 2 ) + 2z) = 1, so
as above, gcd(b, c) = 1. Therefore, they are pairwise relatively prime.
b) Because gcd(a, b) = 1, we have that gcd(a, b, c) =gcd(gcd(a, b), c) =gcd(1, c) = 1
by rules of the gcd. Thus, the numbers are mutually relatively prime, and so the
statements in a) and b) are equivalent.
Part 2:
a) Proof by contradiction: Notice that, when any integer a is even, it can be written
as a = 2x for some integer x, and then a2 = 4x2 +4x+1 = 4(x2 +x)+1 ≡ 1 (mod 4)
by the definitions of divides and modulus. If a is odd, it can be written as a = 2x+1
for some integer x, not necessarily the same. Then, a2 = 4x2 ≡ 0(mod 4), by the
definitions of divides and modulus. Thus, we can conclude that the square of any
odd integer must be ≡ 1 (mod 4), and the square of any even integer must be ≡ 0
(mod 4), and that it is impossible for them to be equivalent to anything else (mod
4). Then, in the context of this problem, a and b can not both be even, because
then gcd(a, b) ≥ 2 by the definition of gcd and the fact that they share a factor
of 2, and as shown in part 1, since gcd(a, b, c) = 1 and pairwise mutual primality
are equivalent, and a and b are not pairwise mutually prime, then gcd(a, b, c) 6= 1,
which is a contradiction. Therefore, at least one of a and b must be odd. If both
are odd, however, then, as noted above, c2 = a2 + b2 ≡ 1 + 1 = 2 (mod 4), but it
is impossible for a perfect square to be equivalent to 2 (mod 4), which is again a
contradiction. Thus, since it is impossible for both a and b to be even or odd, it
must be the case that one of a and b is even, and the other is odd. Also, note that
this means that c2 = a2 + b2 ≡ 1 + 0 = 1 (mod 4), and so c must be odd, since
only odd numbers squared are ≡ 1 (mod 4).
b) If b is even, then a is odd. Also, if b is even, then there is an integer n such
that b = 2n by the definition of even. Then we have that b2 = 4n2 = c2 − a2 =
(c + a)(c − a). Since both c and a are odd, we have that c + a and c − a are even, by
previously shown facts (odd+odd=even, odd-odd=even), and so there are integers
x and y such that c + a = 2x and c − a = 2y. Then 4n2 = 2x · 2y = 4xy.
Then, n2 = xy. Also, note that c + a + c − a = 2c = 2x + 2y ⇒ c = x + y,
and c + a − (c − a) = 2a = 2x − 2y ⇒ a = x − y. If both x and y shared a
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common factor z, then x = pz and y = qz for some integers p and q. But then
c = pz +qz = z(p+q) and a = pz −qz = z(p−q), and so be the definition of divides,
with p + q and p − q in the integers by closure under addition, we have that z | a
and z | b. Since gcd(a, c) = 1 by part 1, we have that z = 1. In essence, this means
that x and y have no common factors. Then, since xy = n2 , both x and y must be
perfect squares with no common factors as well, by previous results. Then there are
integers k and ` such that x = k 2 and y = `2 . So, we have that, by the statement
two sentences ago, k and ` have no common factors, meaning gcd(k, `) = 1, so they
can not both be even. Then p
we have that a = x − y = k 2 − `2 , c = x + y = k 2 + `2 ,
√
√
and then b = c2 − a2 = k 4 + 2k 2 `2 + `4 − k 4 + 2k 2 `2 − `4 = 4k 2 `2 = 2k`
(because b is positive). Here, note that if both k and ` were odd, k = 2f + 1
and ` = 2g + 1 for some integers f and g, we would have that c = k 2 + `2 =
4g 2 + 4g + 4f 2 + 4f + 2 = 2(2g 2 + 2g + 2f 2 + 2f + 1) would be even, which is a
contradiction to the note at the end of step a), so that it impossible. Therefore,
only one of k and ` can be odd, and the other must be even, meaning that k − `
is odd, and so k − ` ≡ 1 (mod 2), or k 6≡ ` (mod 2). Also, note that k > `, or
else a = k 2 − `2 would be negative, which is impossible. Note also that ` 6= 0,
because this would made b = 2k` = 0, which is impossible. In summary, we have
shown that for any primitive Pythagorean triple, there exists a k and ` such that
k > ` > 0, gcd(k, `) = 1, and k 6≡ ` (mod 2), and a = k 2 − `2 , b = 2k`, and
c = k 2 + `2 .
c) If a = k 2 − `2 , b = 2k`, and c = k 2 + `2 , then a2 + b2 = k 4 − 2k 2 `2 + `4 + 4k 2 `2 =
X
k 4 + 2k 2 `2 + `4 = (k 2 + `2 )2 = c2 . Proof by contradiction that (a, b, c) = 1.
Assume that there is an integer d ≥ 2 such that gcd(a, b, c) = d. We know that
d 6= 2, because this would make a, b, and c even, and because one of k and
` is odd and the other even (k 6≡ ` (mod 2)), we that c = k 2 + `2 must be
even2 +odd2 =even+odd=odd, making a contradiction. Then we have that d | a
and d | c by the definition of gcd, so d | a + c and d | a − c by earlier results.
Then d | 2k 2 and d | 2`2 . Since d > 2, d must have a prime factor p such that
p | k 2 and p | `2 . But then p | k and p | `, which contradicts the fact that
gcd(p, `) = 1, and so no such p, and therefore, no such d, can exist. Thus it must
be the case that gcd(a, b, c) = 1, and all conditions are satisfied and we have a
primitive pythagorean triple.
Part 3: Triples:
k
k
k
k
= 2, ` = 1:
= 3, ` = 2:
= 4, ` = 3:
= 5, ` = 4:
(a, b, c) = (3, 4, 5)
(a, b, c) = (5, 12, 13)
(a, b, c) = (7, 24, 25)
(a, b, c) = (9, 40, 41)
Problem 7
Part 1: Proof:
(⇒) Let f (x) ≡ 0 (mod p) have n solutions. We know that n ≤ p because there are
a maximum of p solutions to any congruence (mod p) (0, 1, . . . , p − 1). Divide
xp − x by f (x) using Euclidean division to get xp − x = f (x)q(x) + r(x), and we
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have that deg(r) < deg(f ) = n. If a is a root of f (x) (mod p), then we have that
ap − a = f (a)q(a) + r(a) ≡ 0q(a) + r(a) ≡ r(a) (mod p), so a is a solution to
r(x) ≡ 0 (mod p). Then for each solution to f (x) ≡ 0 (mod p), there is a distinct
solution to r(x) ≡ 0 (mod p), so there are n distinct such solutions. However,
since deg(r) < n, we have that r(x) ≡ 0 (mod p). This is the case because if the
leading coefficient rm is not divisible by p, we have that by problem 5 there are
at most deg r solutions. But this is not the case, so it must be true that p | rm ,
or by the definition of modulus, rm ≡ 0 (mod p). Then, for the the highest term,
we have that rm xm ≡ 0xm = 0 (mod p), and so we can essentially discard our
highest term, then move on to the next and apply the same reasoning. Applying
this logic recursively (or inductively), we get that r(x) ≡ 0 (mod p) (Note: I just
showed that if deg r(x) < n has n solutions (mod p), then r(x) ≡ 0 (mod p), or
essentially that a polynomial of degree n can have at most n (mod p) roots. I’ll
use this fact in the next part as well.). Then we have that b(x) = f (x)q(x)+r(x) ≡
f (x)q(x) + 0 = f (x)q(x) (mod p). This meets the definition of divisors (mod p),
so we have that if we have n solutions to f (x) ≡ 0 (mod p), then f (x) divides
b(x) = xp − x (mod p).
(⇐) By Fermat’s little theorem (proved a few weeks ago in the homework), xp ≡ x
(mod p), or in our case, xp −x ≡ 0 (mod p). We have that f (x) | (xp −x) (mod p),
so by the definition of divisors, there is a polynomial q(x) such that f (x)q(x) ≡
xp − x ≡ 0 (mod p). Note that deg(f (x)) = n and thus deg(q(x)) = p − n. I know
from my aside in the previous part that this means that q(x) can have at most
p − n solutions, and f (x) can have at most n solutions, but since f (x)q(x) has
p = n + p − n solutions, we can see that we must have the maximum number of
solutions for both q(x) and f (x), namely p − n and n. We have just shown that
f (x) has n solutions (mod p).
Part 2: The previous proposition says that it is true if xn − 1 it divides xn − x. Now we have
(xn − 1)(1 + xn + (xn )2 + · · · + (xn )k−1 = xnk − 1.
Multiply both sides by x:
x(xn − 1)(1 + xn + (xn )2 + · · · + (xn )k−1 = xnk+1 − x.
We want the right side to be xn − x, so just take nk + 1 = p. Of course it is only possible
if n divides p − 1. Hence, if n|p − 1, then the polynomial xn − 1 has exactly n roots
modulo p.
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