MATH 313: SOLUTIONS ASSIGNMENT 1.
Problem 1
Let p and q be odd prime numbers such that p = q + 4a for some nonzero integer a. Then
p ≡ q (mod 4), and p - a, otherwise p = q. By the law of quadratic reciprocity (Theorem 11.7),
we have:
(
p q
1
=
q
p
−1
if either p ≡ 1 (mod 4) or q ≡ 1 (mod 4),
if p ≡ q ≡ 3 (mod 4)
Case 1: p ≡ q ≡ 1 (mod 4):
p q
p
q
−1
−1
=1⇒
=
and
=
= 1.
q
p
q
p
p
q
Since p = q + 4a, we have q = p − 4a, hence:
a
4 a
4a
q + 4a
p
=
=
=
=
q
q
q
q
q
q
and
a
4 a
−1 4a
−4a
p − 4a
q
=
=
=
=
=
p
p p
p
p
p
p
p
Case 2: p ≡ q ≡ 3 (mod 4):
p
q
−1
−1
p q
= −1 ⇒
=−
and
=
= −1.
q
p
q
p
p
q
Since p = q + 4a, we have q = p − 4a, hence:
a
4 a
4a
q + 4a
p
q
=
=
=
=
=−
q
q
q
q
q
q
p
and
a
4 a
−1 4a
−4a
p − 4a
q
=
=−
=−
=−
=−
.
p
p p
p
p
p
p
p
Problem 2
First, note that gcd(p, 10) = 1, so p 6= 2, 5. By the law of quadratic reciprocity, since 5 ≡
1 (mod 4), we have
5
p
=
.
p
5
Moreover, from class we know that:
(
5
1
=
p
−1
if p ≡ ±1 (mod 5)
if p ≡ 2, 3 (mod 5).
Further, by Theorem 11.5,
(
−1
1
=
p
−1
if p ≡ 1 (mod 4)
if p ≡ 3 (mod 4).
Combining these two results, we conclude that :


1

1
−5
−1 5
=
=

p
p
p
−1



−1
if p ≡ 1 (mod 4) and p ≡ ±1 (mod 5)
if p ≡ 3 (mod 4) and p ≡ 2, 3 (mod 5)
if p ≡ 1 (mod 4) and p ≡ 2, 3 (mod 5)
if p ≡ 3 (mod 4) and p ≡ ±1 (mod 5).
We have 8 systems of congruences to solve, from which we obtain the following result:
(
−5
1
=
p
−1
if p ≡ 1, 3, 7, 9 (mod 20)
if p ≡ 11, 13, 17, 19 (mod 20).
Problem 3
We will use the Chinese Remainder Theorem (CRT) : if gcd(p, q) = 1, then the system
(
x ≡ b (mod p)
x ≡ c (mod q)
has exactly 1 solution modulo pq.
Now consider x2 ≡ b0 (mod p). We know that it has at most 2 solutions modulo p, xp , and x0p .
Case 1: If p | a, then x2 ≡ a ≡ 0 (mod p) and hence there is only one solution.
Case 2: If p = 2 and p - a, then x2 ≡ a ≡ 1 (mod 2) so that there is one solution.
Case 3: If p odd and p - a, then x2 ≡ a (mod p) has either 0 or 2 solutions.
Keeping those cases in mind, we can find the number of solutions for each of the two parts of the
system: x2 ≡ b0 (mod p) and x2 ≡ c0 (mod q). Then depending on the number of solutions for
each we find 0, 1, 2, or 4 linear systems to solve, each leading to a unique solution.
For instance, if we get 2 for each, such that we have 4 distinct solutions x0p , x00p , x0q , x00q , creating
four different system of congruences of the form :
(
x ≡ xp (mod p)
x ≡ xq (mod q)
Note that if p = q, we get different results since we are looking at modulo p2 , and we cannot use
the CRT.
Case 1:
Case 2:
Case 3:
Case 4:
If p2 | a, then x2 ≡ a ≡ 0 (mod p2 ). There is only one solution.
If p | a but p2 - a, then there is no solution.
If p = 2 and p - a, then x ≡ ±1 (mod p2 ) only if p ≡ 1 (mod 4).
If p odd and p - a, then x2 ≡ a (mod p2 ) has either 0 or 2 solutions.
Here the possible number of solutions when p = q is 0, 1, or 2 solutions.
Problem 4
To solve
x2 − 3x − 1 ≡ 0 (mod 31957),
we begin by multiplying by 4, so that we can complete the square to find that
(2x − 3)2 − 13 ≡ 0 (mod 31957).
We thus want to find when 13 is a square modulo 31957. Using the Legendre symbol and the fact
that 31957 ≡ 1 (mod 4), we thus have
13
31957
3
13
1
=
=
=
=
=1
31957
13
13
3
3
and hence there is a solution to our initial congruence.
Problem 5
Claim 1: There are infinitely many primes congruent to 2 modulo 3.
Proof (by mimicking Euclid’s proof) :
Suppose that there are finitely many primes congruent to 2 modulo 3 such that they can be listed
p0 , p1 , ..., pn , with p0 = 2. Now let N = 3 · p1 · · · pn + 2. Observe that N is odd since the product
of odd numbers is odd; thus 2 - N . Note also that N ≡ 2 (mod 3); thus 3 - N . Moreover, no odd
prime congruent to 2 modulo 3 (none of the p1 , ..., pn ) divides N , otherwise they would have to divide 2 which is not possible. But if 3 doesn’t divide N , and no prime congruent to 2 mod 3 divides
N , then all prime divisors of N must be congruent to 1 mod 3. We have reach a contradiction,
since if all primes divisors of N are 1 mod 3, then N can be written as the product of those primes
and thus would be 1 mod 3 as well. But we have shown that N is congruent to 2 mod 3. The resulting contradiction impose that there are necessarily infinitely many primes congruent to 2 mod 3.
Claim 2: There are infinitely many primes congruent to 1 modulo 3.
Proof (by considering N = (2p1 p2 ...pk )2 + 3 and using Legendre symbols) :
Suppose that there are finitely many primes congruent to 1 mod 3 such that they can be listed
p1 , p2 , ..., pk . Now, let N = (2p1 p2 ...pk )2 + 3, and let q be any prime divisor of N . Note that q
cannot be 2 since N is 3 mod 4, and cannot be 3 either, since N is 1 mod 3. We have
(2p1 p2 ...pk )2 ≡ −3 (mod q),
and hence
−3
q
= 1 whereby either
−1
3
−1
3
=
= 1 or
=
= −1.
q
q
q
q
In the first case, we have q ≡ 1 (mod 4) and so
3
q
1=
=
q
3
implies that q ≡ 1 (mod 3). In the second case, necessarily q ≡ 3 (mod 4) and so
3
q
−1 =
=−
,
q
3
and we have that, again, q ≡ 1 (mod 3).
Therefore, in either case, we need q ≡ 1 (mod 3); that is any prime divisors of N must be 1 mod 3.
This is a contradiction since by construction, any prime congruent to 1 mod 3 cannot divide N . It
follows that our initial assumption must be false and hence that there exist infinitely many primes
congruent to 1 mod 3.
Problem 6
Observe that 15841 = 7 · 31 · 73.
Let’s first show that 15841 is a Carmichael number. Let a be an arbitrary integer such that
gcd(a, 15841) = 1, so that gcd(a, 7) = gcd(a, 31) = gcd(a, 73) = 1. Now consider a15840
modulo 7, 31 and 73. We can check that 6, 30, and 72 all divide 15840. Thus, by Fermat’s little
theorem, we know that a15840 is 1 mod 7, mod 31, and mod 73. Finally, by Chinese remainder
Theorem, we find that a15840 ≡ 1 (mod 15841). Therefore, 15840 is a Carmichael number.
Now let’s show
that 15841 is an euler pseudoprime to base 2. We want to compute the Jacobi
2
symbol 15841 , which is:
2
2
2
2
=
=1
15841
7 31 73
since all 3 Legendre symbols are equal to 1. Now we need to show that
2
15841−1
2
= 27920 ≡ 1 (mod 15841).
Since 7920 is divisible by 6,30, and 72, by Fermat’s little theorem, we get that a7920 is 1 mod 7,
mod 31, and mod 73. Finally, by the Chinese Remainder Theorem, a7920 ≡ 1 (mod 15841).
Finally, we need to show that 15841 is a strong pseudoprime to the base 2. So we write 15840 = 25 ·
r
495. We need to check that either a495 ≡ 1 (mod 15841) or a495·2 ≡ −1 (mod 15841). Let’s start
by computing 2495 mod 15841. Note that by definition of a pseudoprime, if 2495 ≡ 1 (mod 15841)
or 2495 ≡ −1 (mod 15841) we are done. We can do the computation as follows:
2495 ≡ (23 )165 ≡ 1 (mod 7)
2495 ≡ (25 )99 ≡ 1 (mod 31)
2495 ≡ (29 )55 ≡ 1 (mod 73)
By Chinese Remiander Theorem, we find that 2495 ≡ 1 (mod 15841), showing that 15841 is a
strong pseudoprime to the base 2.
Problem 7
Suppose n is an Euler pseudoprime to the base 3, whereby we have 3(n−1)/2 ≡ n3 (mod n).
Moreover, if n ≡ 5 (mod 12), we have that n ≡ 1 (mod 4) and n ≡ 2 (mod 3), and hence
3
n
2
=
=
= −1,
n
3
3
and so 3(n−1)/2 ≡ −1 (mod n). It thus follows form the definition that n is a strong pseudoprime
modulo 3.