MA156 – MidTerm Exam – Wednesday, March 10, 2010 – Solutions
MA156 – MidTerm Exam – Wednesday, March 10, 2010 –
Solutions
1. (9 points)
Compute the following Legendre and Jacobi symbols.
µ ¶
µ ¶
µ ¶
−1
2
17
(a)
(b)
(c)
51
143
41
¡−1¢
Solution. (a) 51 ≡ 3 (mod
¡ 24),
¢ so 51 = −1.
(b) 143 ≡ 7 (mod 8), so 143 = 1.
¡¢
¡¢
¡ ¢ ¡41¢ ¡ 7 ¢ ¡17¢ ¡3¢
= 17 = 17 = 7 = 7 = − 73 = − 31 = −1.
(c) 17
41
2. (10 points)
that
Let p be a prime and let a, m, n be positive integers. Suppose
am ≡ 1 (mod p) and an ≡ 1
(mod p) and
gcd(m, n) = 2.
Prove that a ≡ ±1 (mod p).
Solution. We can find integers u and v so that mu + nv = gcd(m, n) = 2.
Then
a2 ≡ amu+nv ≡ (am )u · (an )v ≡ 1u · 1v ≡ 1 (mod p).
Since p is prime, the polynomial x2 − 1 has a most two roots modulo p, so
its roots are exactly ±1. Therefore a ≡ ±1 (mod p).
3. (10 points) (a) Compute ϕ(3175). (Hint: 3175 = 25 · 127 and 127 is
prime.)
(b) Suppose that f is an arithmetic function that satisfies
X
f (d) =
d|n
1
n
for all integers n ≥ 1.
What is the value of f (12)?
Solution. (a) ϕ(3175) = ϕ(52 · 127) = ϕ(52 )ϕ(127) = (52 − 5)(127 − 1) =
20 · 126 = 2520.
(b) Möbius inversion gives
f (n) =
X
µ(d)f (n/d) =
d|n
X
d|n
–1–
d
µ(d) .
n
MA156 – MidTerm Exam – Wednesday, March 10, 2010 – Solutions
So
1
2
3
4
6
12
+ µ(2) + µ(3) + µ(4) + µ(6) + µ(12)
12
12
12
12
12
12
2
3
6
−
−
+0+
+0
12 12
12
1
= .
6
f (12) = µ(1)
1
12
2
=
12
=
4. (10 points)
Let K be a finite field with q elements, say
K = {a1 , a2 , . . . , aq }.
What is the value of
a41 + a42 + · · · + a4q ?
(Be sure to explain why your answer is correct, i.e., give a proof. No credit
will be given for the answer with no justification.)
Solution. We know K ∗ is cyclic. Let g be a generator of K ∗ . Since we don’t
have to include 04 in the sum, it’s equal to the sum of the fourth powers of
g i for 0 ≤ i < q − 1. Thus
X
a∈K
4
a =
q−2
X
g 4i =
i=0
1 − g 4(q−1)
= 0,
1 − g4
provided g 4 6= 1,
since g q−1 = 1.
The exceptional case g 4 = 1 occurs when q − 1 | 4, since g has exact
order q − 1. Thus the exceptional cases are q = 2, q = 3, and q = 5. In each
of these cases, the value of the sum is −1, since there are q − 1 terms that
are equal to 1.
Alternative Solution: If a4 = 1 for all a ∈ K ∗ , then q − 1 | 4 and we get
the exceptional cases as above. Otherwise there exists some b ∈ K ∗ satisfying
b4 6= 1. Then
!
Ã
X
X
X
(ba)4 =
a4 .
a4 =
b4
a∈K
Since b4 6= 1, it follows that
P
a∈K
a∈K
a∈K
a4 = 0.
5. (10 points) Suppose that p ≡ 3 (mod 4) and that a is a quadratic
residue modulo p. Let
p+1
b≡a 4
(mod p).
–2–
MA156 – MidTerm Exam – Wednesday, March 10, 2010 – Solutions
Prove that
b2 ≡ a
(mod p).
(This gives an explicit way to compute square roots modulo p for primes
p ≡ 3 (mod 4).)
¡¢
Solution. We are given that ap = 1, so Euler’s criterion says that a(p−1)/2 ≡ 1
(mod p). Hence b2 ≡ a(p+1)/2 ≡ a · a(p−1)/2 ≡ a (mod p).
–3–