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MA 261 — Worksheet
Monday, April 21, 2014
1. Show that 211,213 − 1 is not divisible by 11.
Solution: To say that 211,213 − 1 is divisible by 11 is equivalent to saying that
211,213 ≡ 1 (mod 11).
Thus, to show our statement is equivalent to showing that 211,213 is congruent to something
different from 1 modulo 11.
Let us take (small) consecutive powers of 2 and see if there is a pattern that we can use to
simplify our calculations. Note that
20 ≡ 1 (mod 11)
21 ≡ 2 (mod 11)
22 ≡ 4 (mod 11)
23 ≡ 8 (mod 11)
24 ≡ 5 (mod 11)
25 ≡ 10 (mod 11)
26 ≡ 9 (mod 11)
27 ≡ 7 (mod 11)
28 ≡ 3 (mod 11)
29 ≡ 6 (mod 11)
210 ≡ 1 (mod 11).
Observe that we could have simplified are calculation as follows:
25 ≡ 10 (mod 11) ; 25 ≡ −1 (mod 11) ; 210 = (25 )2 ≡ (−1)2 = 1 (mod 11).
No matter what, we have shown that 2 raised to the tenth power is congruent to 1 modulo
11. If we divide the exponent 11, 213 by 10 we obtain 11, 213 = 10 · 1, 121 + 3. Thus
211,213 = 210·1,121+3 = (210 )1,121 · 23 ≡ (1)1,121 · 23 ≡ 1 · 8 ≡ 8 (mod 11).
Thus, we have shown that
211,213 ≡ 8 (mod 11),
which implies that 211,213 − 1 cannot be divisible by 11. The above calculations shows that
211,213 − 8 is divisible by 11.
2. (1) Describe all solutions of the congruence 22x ≡ 5 (mod 15).
Solution 1 (classic method): By Theorem 3.19 this congruence has a solution if and only
if there exist integers x and y such that 22x + 15y = 5. We know that this Diophantine
equation has a solution if and only if gcd(22, 15) divides 5, by Theorem 1.48 (see also
Theorem 3.20). As gcd(22, 15) = 1, we do have a solution. Moreover, all the solutions are
described by Theorem 1.53. Namely,
x = x0 +
15k
gcd(22, 15)
y = y0 −
22k
gcd(22, 15)
k ∈ Z,
where x0 and y0 are a particular solution of 22x + 15y = 5.
Do not memorize those formulas! For your instruction, we repeat next the proof in this
specific example. To find a particular solution we use the Euclidean Algorithm (Exercise
1.35) applied to a = 22 and b = 15. We have:
22 = 15 · 1 + 7
15 = 7 · 2 + 1
1
7 = 7 · 1 + 0.
2
This says that the last nonzero remainder is 1; this is the gcd between 22 and 15. We can
also use the above equations to express 1, the gcd, as a linear combination of 22 and 15, as
follows:
1 = 15 − 7 · 2
;
1 = 15 − (22 − 15) · 2
;
1 = 15 · 3 + 22 · (−2).
If we multiply the last equation by 5 we obtain
22 · (−10) + 15 · (15) = 5.
(1)
Thus x0 = −10 and y0 = 15.
Consider now, another set of integer solutions x and y of 22x + 15y = 5. If we subtract
equation (1) from 22x + 15y = 5 we obtain
22 · (x + 10) + 15 · (y − 15) = 0
⇐⇒
15 · (15 − y) = 22 · (x + 10).
(2)
This means that 15 divides the product 22 · (x + 10). However, by Theorem 1.41 we have
that 15 divides x + 10, as gcd(15, 22) = 1. Thus
x + 10 = 15 · k
⇐⇒
x = −10 + 15 · k,
k ∈ Z.
Substituting back in equation (2) we obtain that y = 15 − 22 · k, even if we don’t need the
values for y.
Thus the solutions of 22x ≡ 5 (mod 15) are x = −10 + 15 · k where k is any integer. The
solution set consists of the numbers {. . . , −25, −10, 5, 20, 35, . . . }. Observe, tough, that
the only solution that satisfies 0 ≤ x ≤ 14 is x = 5 . This confirms the results in Theorem
3.24. There is exactly 1 = gcd(22, 15) solution in the range 0 ≤ x ≤ 14.
Solution 2 (easier, but requires some tricks): Notice that, by reducing the coefficient 22
modulo 15, the original congruence is equivalent to
22x ≡ 5 (mod 15)
⇐⇒
7x ≡ 5 (mod 15).
Now, multiply both sides of the latter congruence by 2. We obtain
14x ≡ 10
(mod 15)
⇐⇒
−x ≡ 10 (mod 15),
as 14 ≡ −1 (mod 15). Multiply both sides by −1 and observe that −10 ≡ 5 (mod 15):
x ≡ −10
(mod 15)
⇐⇒
x≡5
(mod 15).
Hence our solutions are of the form x = 5 + 15ℓ, where ℓ ∈ Z.
(2) Describe all solutions of the congruence 45x ≡ 15 (mod 24).
Solution 1 (classic method): By Theorem 3.19 this congruence has a solution if and only
if there exist integers x and y such that 45x + 24y = 15. We know that this Diophantine
equation has a solution if and only if gcd(45, 24) divides 15, by Theorem 1.48 (see also
Theorem 3.20). As gcd(45, 24) = 3, we do have a solution. Moreover, all the solutions are
described by Theorem 1.53. Namely,
x = x0 +
24k
gcd(45, 24)
y = y0 −
45k
gcd(45, 24)
where x0 and y0 are a particular solution of 45x + 24y = 15.
k ∈ Z,
3
Do not memorize those formulas! For your instruction, we repeat next the proof in this
specific example. To find a particular solution we use the Euclidean Algorithm (Exercise
1.35) applied to a = 45 and b = 24. We have:
45 = 24 · 1 + 21
24 = 21 · 1 + 3
21 = 3 · 7 + 0.
This says that the last nonzero remainder is 3; this is the gcd between 45 and 24. We can
also use the above equations to express 3, the gcd, as a linear combination of 45 and 24, as
follows:
3 = 24 − 21
;
3 = 24 − (45 − 24)
;
3 = 24 · 2 + 45 · (−1).
If we multiply the last equation by 5 we obtain
45 · (−5) + 24 · (10) = 15.
(3)
Thus x0 = −5 and y0 = 10.
Consider now, another set of integer solutions x and y of 45x + 24y = 15. If we subtract
equation (3) from 45x + 24y = 15 we obtain
45 · (x + 5) + 24 · (y − 10) = 0
⇐⇒
8 · (10 − y) = 15 · (x + 5),
(4)
after we divided both sides by 3. This means that 8 divides the product 15·(x+5). However,
by Theorem 1.41 we have that 8 divides x + 5, as gcd(8, 15) = 1. Thus
x+5=8·k
⇐⇒
x = −5 + 8 · k,
k ∈ Z.
Substituting back in equation (4) we obtain that y = 10 − 8 · k, even if we don’t need the
values for y.
Thus the solutions of 45x ≡ 15 (mod 24) are x = −5 + 8 · k, where k is any integer.
The solution set consists of the numbers {. . . , −21, −13, −5, 3, 11, 19, 27, . . . }. Observe,
tough, that the only solutions that satisfies 0 ≤ x ≤ 23 are x = 3, 11, 19 . This confirms
the results in Theorem 3.24. There are exactly 3 = gcd(45, 24) distinct solutions in the
range 0 ≤ x ≤ 23.
Solution 2 (easier, but requires even more tricks): Notice that, by reducing the coefficient 45 modulo 24, the original congruence is equivalent to
45x ≡ 15 (mod 24)
⇐⇒
−3x ≡ 15
(mod 24).
Now, multiply both sides of the latter congruence by −1. We obtain
3x ≡ −15
(mod 24)
⇐⇒
3x ≡ 9
(mod 24),
as −15 ≡ 9 (mod 24). By Theorem 3.19, the congruence 3x ≡ 9 (mod 24) has a solution
if and only if there exist integers x and y such that 3x + 24y = 9. All terms are divisibile
by 3, so that x + 8y = 3. Hence, by using Theorem 3.19 again, we conclude that x is a
solution of the simpler congruence x ≡ 3 (mod 8)
3x ≡ 9
(mod 24)
⇐⇒
x ≡ 3 (mod 8).
Hence our solutions are of the form x = 3 + 8ℓ, where ℓ ∈ Z. The only solutions that
satisfy 0 ≤ x ≤ 23 are again x = 3, 11, 19.
4
3. (1) Solve the simultaneous system of congruences
x ≡ 1 (mod 8)
x≡2
x ≡ 3 (mod 81).
(mod 25)
Solution: Notice that the numbers 8, 25, and 81 are pairwise relatively prime. Using the
proof that we gave for the Chinese Remainder Theorem (see Theorem 3.29), we are looking
for a solution of the form
8 · 25 · 81
8 · 25 · 81
8 · 25 · 81
x1 +
x2 +
x3 ,
x=
8
25
81
that is,
x = 25 · 81 · x1 + 8 · 81 · x2 + 8 · 25 · x3 ,
(5)
for suitable x1 , x2 , and x3 .
If we now substitute the proposed solution described in (5) into the three given congruences
we realize that we need to solve the following independent (from each other) congruences
25 · 81 · x1 ≡ 1 (mod 8)
8 · 81 · x2 ≡ 2
(mod 25)
8 · 25 · x3 ≡ 3 (mod 81).
The first congruence reduces to
25 · 81 · x1 ≡ 1 (mod 8)
⇐⇒
1 · 1 · x1 ≡ 1 (mod 8),
because both 25 and 81 are congruent to 1 modulo 8. Thus we can choose x1 = 1 as a
solution.
The second congruence reduces to
8 · 81 · x2 ≡ 2 (mod 25)
⇐⇒
8 · 6 · x2 ≡ 2 (mod 25),
because 81 is congruent to 6 modulo 25. Again notice that we can simplify the congruence
as follows
48 · x2 ≡ 2 (mod 25)
⇐⇒
−2 · x2 ≡ 2
(mod 25),
as 48 is congruent to −2 modulo 25. If we multiply the latter congruence by −13 we obtain
−2 · x2 ≡ 2 (mod 25)
⇐⇒
26 · x2 ≡ −26
(mod 25).
Since 26 ≡ 1 modulo 25 we have that
x2 ≡ −1
(mod 25).
Thus we can choose x2 = −1 as a solution.
The third congruence reduces to
8 · 25 · x3 ≡ 3
(mod 81)
⇐⇒
38 · x3 ≡ 3
(mod 81),
because 8 · 25 = 200 is congruent to 38 modulo 81. If we multiply the latter congruence
by 32 we obtain
32 · 38 · x3 ≡ 32 · 3
(mod 81)
⇐⇒
x3 ≡ 15
(mod 81).
since 32 · 38 = 1216 ≡ 1 and 32 · 3 = 96 ≡ 15 modulo 81. Thus we can choose x3 = 15
as a solution.
I am sure you will ask: how did you know that multiplying by 32 would have given such
an easy value? It has to do with the fact that gcd(38, 81) = 1. Hence if you apply the
Euclidean Algorithm to the pair 81 and 38, you will see that 1 = 38 · 32 + 81 · (−15). Thus
38 · 32 ≡ 1 (mod 81).
5
Summing up what we did so far, we have x = 25 · 81 · 1 + 8 · 81 · (−1) + 8 · 25 · 15 = 4, 377
satisfies the given congruences. By Theorem 3.29 any other solution x′ is such that
x′ ≡ 4, 377
(mod 8 · 25 · 81)
x′ ≡ 4, 377 (mod 16, 200).
⇐⇒
In other words, the only solution x of the three given congruences such that 0 ≤ x ≤
16, 199 is x = 4, 377 .
(2) Solve the simultaneous system of congruences
y ≡ 5 (mod 8)
y ≡ 12
y ≡ 47
(mod 25)
(mod 81).
Solution: Modify the previous calculations on your own. You will look for a solution of
the form
y = 25 · 81 · y1 + 8 · 81 · y2 + 8 · 25 · y3 ,
(6)
for suitable y1 , y2 , and y3 . You will obtain that
y2 = −6
y1 = 5
y3 = 43
do work. Thus y = 15, 437. Again by Theorem 3.29 any other solution y ′ is such that
y ′ ≡ 15, 437 (mod 16, 200).
In other words, the only solution y of the three given congruences such that 0 ≤ y ≤
16, 199 is x = 15, 437 .
4. (Challenge): Show that for every integer n, the number n33 − n is divisible by 15.
Sketch of the solution: Notice that 15 = 3 · 5 and gcd(3, 5) = 1. Thus by Theorem 1.42
(or Theorem 2.25) it is enough to show that n33 − n is divisible by 3 and by 5, respectively.
In other words we need to show
n33 ≡ n (mod 3)
and
n33 ≡ n (mod 5).
The above relations will follow once we prove that for any n one has
n3 ≡ n (mod 3)
and
n5 ≡ n
(mod 5).
(7)
Indeed,
n33 = (n3 )11 ≡ n11 = (n3 )3 · n2 ≡ n3 · n2 ≡ n · n2 = n3 ≡ n
(mod 3),
and
n33 = (n5 )6 · n3 ≡ n6 · n3 = n9 = n5 · n4 ≡ n · n4 = n5 ≡ n (mod 5).
Now, to prove the congruences in equation (7) we have to distinguish various cases.
When we consider the congruences modulo 3 we have the following three cases
n ≡ 0 (mod 3)
n ≡ 1 (mod 3)
n ≡ 2 (mod 3)
that we need to analyze.
When we consider the congruences modulo 5 we have the following five cases
n ≡ 0 (mod 5)
n ≡ 1 (mod 5)
that we need to analyze.
n≡2
(mod 5)
n≡3
(mod 5)
n ≡ 4 (mod 5)
6
Let me verify for you (7) in one of these cases so that you understand the ‘trick’. For
instance, suppose that n ≡ 2 (mod 5). Then by Theorem 1.18 we have that n5 ≡ 25 (mod
5). But 25 = 32 which is congruent to 2 modulo 5. Hence
n ≡ 2 (mod 5)
;
n5 ≡ 2 (mod 5).
Now, by the symmetric (Theorem 1.10) and the transitive (Theorem 1.11) properties of
congruences, the two congruences above yield
n5 ≡ n (mod 5).
The remaining cases can be verified similarly.