Math 437/537—Group Work #6
Friday, October 23, 2015
1. Determine whether the following congruences have solutions. You don’t have to find the
solutions—just decide whether solutions exist. You may use the fact that 41, 227, and 367 are
all prime.
(a)
(b)
(c)
(d)
(a)
(b)
(c)
(d)
x2 ≡ 21 (mod 41)
5x2 − x − 1 ≡ 0 (mod 41)
x2 ≡ 55 (mod 367)
x2 ≡ 137 (mod 227). Hint: 137 − 227 factors more nicely than 137.
7
21
3
By multiplicativity, 41
= 41
. Since
41 ≡ 1 (mod 4), quadratic
results in
41
reciprocity
3
41
7
41
41
2
no sign changes: 41 = 3 and 41 = 7 . By periodicity,
= 3 = −1 (by brute
3
41
−1
force, or since 3 ≡ ±3 (mod 8), for example),
while
7 = 7 = −1 (by brute force,
21
2 −1
or since 7 ≡ 3 (mod 4)). Therefore 41 = 3 7 = (−1)(−1) = 1. We conclude that
x2 ≡ 21 (mod 41) does have solutions.
The discriminant of 5x2 − x − 1 equals (−1)2 − 4 · 5 · (−1) = 21, which is a square modulo
41 by part (a); therefore 5x2 − x − 1 ≡ 0 (mod 41) has solutions. (More concretely: since
(20, 41) = 1, the congruence 5x2 − x − 1 ≡ 0 (mod 41) is equivalent to 20(5x2 − x − 1) ≡
0 (mod 41), or (10x − 1)2 − 21 ≡ 0 (mod 41). Therefore the solutions to y 2 ≡ 21 (mod 41)
can be transformed, via 10x − 1 ≡ y (mod 41) or equivalently x ≡ 10−1 (y + 1) (mod 41),
into solutions to the original
congruence.)
11 55
5
5
367
By multiplicativity, 367 = 367
.
By
quadratic
reciprocity,
we
have
=
367 367
5
11
55
since 5 ≡ 1 (mod 4), but 367
= − 367
since
11
≡
367
≡
3
(mod
4);
therefore
11
367 =
367 367
367
2
2
− 5 11 . By periodicity, we have 5 = 5 = −1 by the formula for p , while
367
4
=
= 1 since 4 = 22 is obviously a square modulo any prime. Therefore
11 11 367 367
55
= − 5 11 = −(−1)1 = 1, and we conclude that x2 ≡ 55 (mod 367) does have
367
solutions.
By periodicity and then multiplicativity,
2 −90
−1
2
3
5
137
=
=
.
227
227
227 227 227 227
−1
2
Since 227 ≡ 3 (mod 4) and 227 ≡ 3 (mod 8), we have 227
= −1 and 227
= −1; we
32
137
5
5
also have p = 1 for any odd prime p. Therefore 227 = (−1)(−1)1 227 = 227
.
5
227
Since
; periodicity then gives
5 ≡21 (mod 4), quadratic reciprocity
gives 227 = 5 137
227
2
= 5 = −1 by the formula for p . We conclude that 227 = −1, and so x2 ≡
5
137 (mod 227) has no solutions.
(continued on next page)
2.
(a) For which primes p does there exist an integer x such that x2 ≡ 5 (mod p)? State your
answer in terms of the last digit of p.
(b) For which primes p does there exist an integer x such that x2 ≡ −5 (mod p)? State your
answer in terms of the last two digits of p.
We note that when p = 2, both congruences have the solution x = 1, while when p = 5, both
congruences have the solution x = 0. During the proofs, therefore, we may assume p 6= 2, 5.
(a) Since 5 ≡ 1 (mod 4), quadratic reciprocity tells us that p5 = p5 for any odd prime p 6= 5.
2
3
4
1
=
1,
=
−1,
=
−1,
and
= 1.
It’s easy to establish
(by
brute
force
even)
that
5
5
5
5
5
5
Therefore p = 1 when p ≡ ±1 (mod 5), while p = −1 when p ≡ ±2 (mod 5). Note
that these two cases correspond to the last digit being 1 or 9, and 3 or 7, respectively.
Therefore x2 ≡ 5 (mod p) has a solution when the last digit of p is 1, 2, 5, or 9 but not
when the last digitof p is 3 or 7.
(b) We know that −1
= 1 when p ≡ 1 (mod 4) and −1
= −1 when p ≡ 3 (mod 4). Therep
p
−5
−1 5
fore p = p p = 1 in precisely two cases: either p ≡ ±1 (mod 5) and p ≡ 1 (mod 4),
or else p ≡ ±2 (mod 5) and p ≡ 3 (mod 4). Either by doing a bunch of little Chinese remainder calculations, or else just going through all the reduced residue classes modulo 20
by hand, we see that these cases correspond to the residue classes p ≡ 1, 3, 7, 9 (mod 20).
We conclude, amazingly, that the congruence x2 ≡ −5 (mod p) has solutions if and only if
the second-to-last digit of p is even! (This works for single digit primes too, as long as we
call the second-to-last digit 0.)
3. Let p be an odd prime, and let g be a primitive root modulo p, so that any
a that is not a multiple
a
k
of p can be written as a ≡ g (mod p) for some integer k. Prove that p = 1 if k is even while
a
= −1 if k is odd.
p
There are multiple ways to see this. If we set k = 2j + ε with ε ∈ {0, 1}, we can use Euler’s
criterion to write
ε
a
≡ a(p−1)/2 ≡ (g k )(p−1)/2 = (g p−1 )j g (p−1)/2 ≡ 1j (−1)ε (mod p).
p
(Note that g (p−1)/2 is a solution to x2 ≡ 1 (mod p) that is not congruent to 1 (mod p), which by
Lemma 2.10 forces g (p−1)/2 ≡ −1 (mod p).) Both integers ap and (−1)ε are either 1 or −1, and
so they differ by at most 2; consequently,
prime can divide their difference unless that
no odd
a
ε
difference equals 0. We conclude that p = (−1) , which is what we want to prove.
Alternatively, if k = 2j is even, then clearly x2 ≡ g 2j (mod p) has solutions, namely x ≡
±g j (mod p); therefore ap = 1 when a ≡ g k (mod p) with k even. And the p−1
even integers
2
2 4
p−1
{2, 4, . . . , p − 1} give rise to distinct residue classes g , g , . . . , g
(mod p) (or else the quotient
of two of them would be 1 (mod p), contradicting the fact that the order of g is p − 1) which are
all quadratic residues; but we know that there are only p−1
quadratic residues (mod p). Therefore
2
p−1
the other 2 integers {1, 3, . . . , p − 2}, which are all odd, must give rise to quadratic nonresidues
g 1 , g 3 , . . . , g p−2 (mod p).