Number Theory ArielU 2016
Practical Session 5
Practical session 5
Ÿ1. Basic properties of Congruence
Exercise 1.
For n ≥ 1, use congruence theory to establish each of the following divisibility statement:
1. 7|52n + 3 · 25n−2
2. 27|25n+1 + 55n+2
Answer.
1. As 52 = 25 ≡ 4 (mod 7) it follows that 52n ≡ 4n (mod 7) by Theorem 2.13 in the Congruences
lecture. Next, note that 25 ≡ 4 (mod 7) it follows 25n ≡ 4n (mod 7)
For n ≥ 1, 25n · 4−1 ≡ 4n · 4−1 (mod 7) so that 25n−2 ≡ 4n−1 (mod 7). By Theorem 2.1 in the
Congruences lecture we get 3 · 25n−2 ≡ 3 · 4n−1 (mod 7)
52n + 3 · 25n−2 ≡ 4n + 3 · 4n−1 (mod 7)
≡ 4 · 4n−1 + 3 · 4n−1 (mod 7)
≡ 7 · 4n−1 (mod 7)
≡ 0 (mod 7)
and the claim follows.
2. As 25 ≡ 5 (mod 27) it follows that 25n ≡ 5n (mod 27) by Theorem 2.13 in the Congruences lecture.
By Theorem 2.1 in the Congruences lecture we get 25n · 2 ≡ 5n · 2 (mod 27)
25n+1 + 5n+2 ≡ 2 · 5n + 5n+2 (mod 27)
≡ 5n · (2 + 25) (mod 27)
≡ 5n · 27 (mod 27)
≡ 0 (mod 27)
and the claim follows.
Exercise 2.
Prove or disprove the following claims.
1. If a is an odd integer, then a2 ≡ 1 (mod 8)
2. For any a ∈ Z: a3 is congruent to 0, 1, or 6 modulo 7.
Answer.
1. As a is an odd integer, let us a = 4k + 1 or a = 4k + 3, for some k ∈ Z.
a2 = 16k 2 + 8k + 1 or a2 = 16k 2 + 24k + 9
a2 − 1 = 8(2k 2 + k) or a2 − 1 = 8(2k 2 + 3k + 1)
The claim follows a2 ≡ 1 (mod 8)
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Number Theory ArielU 2016
Practical Session 5
2. By the Division Algorithm a = 7q + r for some 0 ≤ r < 7. For each of the seven options for r one can
show that (7q + r)3 is congruent to 0, 1, or 6 modulo 7. Let us exemplify one case. The treatment
of all other cases is similar in spirit (though the details are dierent).
Suppose r = 4, i.e., a is of the form 7q + 4. Then
(7q + 4)3 = (7q)3 + 3 · (7q)2 · 4 + 3 · (7q) · 42 + 43
= (7q)3 + 3 · (7q)2 · 4 + 3 · (7q) · 42 + 64.
Note that 64 = 7 · 9 + 1 so that
= (7q)3 + 3 · (7q)2 · 4 + 3 · (7q) · 42 + 7 · 9 + 1.
Then
(7q + 4)3 − 1 = (7q)3 + 3 · (7q)2 · 4 + 3 · (7q) · 42 + 7 · 9
= 7(72 q 3 + 3 · 72 · q 2 · 4 + 3q · 42 + 9)
implying that in this case a3 − 1 ≡ 0 (mod 7) so that a3 ≡ 1 (mod 7).
Repeating this type of analysis for each of the possible values of r one can prove the statement.
Exercise 3.
Use the theory of congruences to verify that
89|244 − 1.
Answer.
Here
28
Note that 244 = 232 · 28 · 24 .
≡ 78 (mod 89) and 24 ≡ 16 (mod 89). Let us calculate 232 (mod 89)
232 ≡ 216·2 ≡ (216 )2 ≡ 322 ≡ 45 (mod 89).
Then by Theorem 2.7 in the Congruences lecture we can multiply all the results together and take the
remainder modulo 89. We have
244 ≡ 232 · 28 · 24 ≡ 45 · 78 · 16 ≡ 1 (mod 89).
It follows that 89|244 − 1.
Ÿ2. Linear congruences
Exercise 1.
Find all solutions to the equations:
1. 36x ≡ 8 (mod 102).
2. 5x ≡ 2 (mod 26)
Answer.
1. Here we have that (36, 102) = 6 but 6 - 8 so there are no solutions to this equation.
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Number Theory ArielU 2016
Practical Session 5
2. As (5, 26) = 1 there is a unique solution modulo 26 (See Corollary 5.8 in the Congruences lecture).
Viable value of x then satisfy x = x0 +(26/1)·t, t ∈ Z (see the Congruences lecture for an explanation
about this). We set out to nd x0 .
Applying the extended Euclid's algorithm for (5, 26) we rst generate the equations
26 = 5 · 5 + 1
5=5·1+0
and arrive at
1 = 26 − 5 · 5 = 1 · 26 − 5 · 5
Then 1 · 2 = (1 · 26 − 5 · 5) · 2 so that 2 = 2 · 26 + 5 · (−10). We may then set x0 = −10 leading to
the values of x to be given by
x = −10 + 26t, t ∈ Z.
Note now that
x = 16 − 26 + 26t, t ∈ Z.
Set t0 = t − 1 and write
x = 16 + 26t0 , t0 ∈ Z.
It follows that x ≡ 16 (mod 26).
Ÿ3. System of two linear congruences with two variables
Now we see how to solve a system of two linear congruences in two variables with the same modulus.
Theorem 3.1. Let
a, b, c, d, r
and
s ∈ Z.
Let
n ∈ Z+ .
The system of linear congruences
ax + by ≡ r (mod n)
cx + dy ≡ s (mod n)
has a unique solution modulo
n
whenever
gcd(ad − bc, n) = 1.
Let us multiply the rst congruence of the system by d, the second congruence by b, and subtract
the lower result from the upper. These calculations yield
Proof.
(ad − bc)x ≡ dr − bs (mod n)
(3.2)
The assumption gcd(ad − bc, n) = 1 ensures that the congruence
(ad − bc)z ≡ 1 (mod n)
posseses a unique solution; denote the solution by t. When congruence (3.2) is multiplied by t, we obtain
x ≡ t(dr − bs) (mod n)
A value for y is found by a similar elimination process. That is, multilply the rst congruence of the
system by c, the second one by a, and subtract to end up with
(ad − bc)y ≡ as − cr (mod n)
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(3.3)
Number Theory ArielU 2016
Practical Session 5
Multiplication of this congruence by t leads to
y ≡ t(as − cr) (mod n)
A solution of the system is now established.
Exercise 1.
Find the solution of the systems of congruences:
3x + 4y ≡ 5 (mod 13)
2x + 5y ≡ 7 (mod 13)
Answer.
As gcd(3 · 5 − 2 · 4, 13) = 1, a solution exists. First we multiply the rst congruence by 5 and get
15x + 20y ≡ 25 (mod 13).
(3.4)
Then we multiply the second congruence by 4 and get
8x + 20y ≡ 28 (mod 13).
(3.5)
Now by Theorem 2.7 in the Congruences lecture we subtract (3.4)-(3.5):
7x ≡ −3 (mod 13)
or by Theorem 2.1
14x ≡ −6 (mod 13).
Therefore x ≡ 7 (mod 13).
From the rst congruence 3x + 4y ≡ 5 (mod 13) follows
3x ≡ 5 − 4y (mod 13)
(3.6)
If x ≡ 7 (mod 13) so 3x ≡ 21 (mod 13). Then substitute the result x ≡ 7 (mod 13) to the (3.6) and get
5 − 4y ≡ 21 (mod 13)
−4y ≡ 16 (mod 13)
−12y ≡ 48 (mod 13)
The solution of the systems of congruences is y ≡ 9 (mod 13) and x ≡ 7 (mod 13)
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