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MATH10101, Exam January 2010
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A3. Use the method of successive squaring to calculate
17184 mod 37.
Find the smallest non-negative solution to
1790 x ≡ 4 mod 37.
[6 marks] Solution
mod 37 172 30 174 ≡ 302 12 178 ≡ 122 33 1716 ≡ 332 16 1732 ≡ 162 34 1764 ≡ 342 9 17128 ≡ 92 7 Then, because 184 = 128 + 32 + 16 + 8, we have
17184 = 17128 × 1732 × 1716 × 178 ≡ 7 × 34 × 16 × 33 ≡ 12 mod 37.
[4 marks] Because 90 = 64 + 16 + 8 + 2 we have
1790 = 1764 × 1716 × 178 × 172 ≡ 9 × 16 × 33 × 30 ≡ 36 ≡ −1 mod 37.
So the equation becomes −x ≡ 4 mod 37. Multiply both sides by −1 to get the least non-negative solution x = 37 − 4 = 33. Standard example [2 marks]
Comments on A
Some students gave mod 37 172 − 7 174 ≡ (−7)^2 178 ≡ 122 − 4 1716 ≡ (−4)^2 1732 ≡ 162 − 3 1764 ≡ (−5)^2 17128 ≡ 92 7
which is helpful if you don’t have a calculator and want to keep the numbers smaller in magnitude. You can go further and say
17184 = 17128 × 1732 × 1716 × 178 ≡ (7 × −3) × (16 × −4) = − 21 × − 64 ≡ 16 × 10 ≡ 12 mod 37.
Some students used Fermat’s Little Theorem, so 17^36 ≡ 1 mod 37. This does ease calculation since
17184 =
174 ≡ 174 mod 37,
but the question had asked that you use the Method of Successive Squaring. Always do what the question asks.
Note that 17^90 ≡ 36 mod 37 means that the congruence 17^90 x ≡ 4 mod 37 becomes 36x ≡ 4 mod 37, i.e. 9x ≡ 1 mod 37. This can be solved by Euclid’s algorithm.
Some students tried to use the results from the first part of the question to solve the second part. For instance, you might notice that
17184 ≡ 12 ≡ 174 mod 37.
Hence 17^180 ≡ 1 mod 37. So multiply both sides of 17^90 x ≡ 4 mod 37 by 17^90 to get 17^180 x ≡ 4 × 1790 mod 37, i.e. x ≡ 4 × 1790 mod 37. You can then use the results from the table to simplify the right hand side.
Finally a few students squared both sides of 17^90 x ≡ 4 mod 37 to get 17180 x^2 ≡ 42 mod 37, i.e. x^2 ≡ 42 mod 37. Unfortunately this has two solu- tions x ≡ 4 and −4 mod 37. Only one of these is a solution to the original congruence and you can’t find which one by this method. Conclusion? Don’t use this method.
A5 Consider σ and ρ to be the following permutations in S 11 :
σ =
ρ =
Calculate σ−^1 ◦ ρ and σ ◦ ρ^2. Calculate the orders of σ, ρ, σ−^1 ◦ ρ and σ ◦ ρ^2. [6 marks] Solution
σ−^1 ◦ ρ =
[1 mark]
σ ◦ ρ^2 =
[1 mark] In cycle form
σ = (1, 2 , 4 , 8 , 5) ◦ (3, 11) ◦ (6, 10 , 7) , ρ = (1, 9) ◦ (2, 5 , 10 , 3 , 11 , 6 , 7) ◦ (4, 8) , σ−^1 ◦ ρ = (1, 9 , 5 , 6 , 10 , 11 , 7) ◦ (2, 8) , σ ◦ ρ^2 = (1, 2 , 7) ◦ (3, 10) ◦ (4, 8 , 5 , 11 , 6).
Hence
order of σ = lcm (5, 2 , 3) = 30, order of ρ = lcm (2, 7 , 2) = 14, order of σ−^1 ◦ ρ = lcm (7, 2) = 14, order of σ ◦ ρ^2 = lcm (3, 2 , 5) = 30.
Standard type of example [4 marks]
- a). Binomial Theorem. For any a, b ∈ R and all n ≥ 1 ,
(a + b)n^ = an^ +
n 1
an−^1 b +
n 2
an−^2 b^2 +
n 3
an−^3 b^3 + ...
n n − 1
abn−^1 +
n n
bn,
∑^ n
r=
n r
an−rbr. (1)
Bookwork [1 mark] b) When n = 7 we get, for any a ∈ Z,
(a + 1)^7 = a^7 + 7a^6 + 21a^5 + 35a^4 + 35a^3 + 21a^2 + 7a + 1 = a^7 + 7
a^6 + 3a^5 + 5a^4 + 5a^3 + 3a^2 + a
≡ a^7 + 1 mod 7,
because all the terms that have vanished had a coefficient divisible by 7. [2 marks] c) Base case n = 1, 1^7 = 1 so result holds with equality. Assume the result holds with n = k, so k^7 ≡ k mod 7. Look at the k + 1 case by considering
(k + 1)^7 ≡ k^7 + 1 mod 7, by part 2.b, ≡ k + 1, by the inductive hypothesis.
So the result holds for k + 1.
Hence, by induction, the result holds for all n ∈ N. Seen for 5, not 7, in question sheet [3 marks] d)
x^14 + 3x^7 + 2 ≡ x^2 + 3x + 2 mod 7, by part 2.c, = (x + 2) (x + 1) mod 7.
Because 7 is prime the only way (x + 2) (x + 1) ≡ 0 mod 7 is if either 7| (x + 2) or 7| (x + 1). So possible solutions are x ≡ −2 mod 7 or x ≡ −1 mod 7, i.e. either x ≡ 5 mod 7 or x ≡ 6 mod 7. [3 marks]
Comments on B
- The general solution to 7 = 252m + 889n is
(m, n) =
t, − 17 −
t
= (60 + 127t, − 17 − 36 t)
for t ∈ Z. You were not asked for this general solution but when t = − 1 we get a solution (m, n) = (− 67 , 19), and some students gave this as their answer.
- a). Binomial Theorem. It is possible to write down the Binomial expansion in the reverse order:
(a + b)n^ = bn^ +
n 1
bn−^1 a +
n 2
bn−^2 a^2 +
n 3
bn−^3 a^3 + ...
n n − 1
ban−^1 +
n n
an,
∑^ n
r=
n r
arbn−r. (2)
Alternatively, write r = n − s in (1). Then as r varies from 0 to n, s varies from n to 0. Thus
(a + b)n^ =
∑^ n
r=
n r
an−rbr^ =
∑^0
s=n
n n − s
an−(n−s)bn−s
∑^ n
s=
n s
asbn−s^ (3)
since, as we have seen on a Question Sheet, ( n n − s
n s
for all 0 ≤ s ≤ n. In the last sum in (3), we have a summation over s, but this is simply a label and can be renamed as we like, to r for example. But when we do that we get the summation seen in (2).
If we put a = 1 and b = x in the answer given above, (1) , we get
(1 + x)n^ =
∑^ n
r=
n r
xr, (4)
You can start the solution differently as
x^14 + 3x^7 + 2 =
x^7 + 2
x^7 + 1
Thus, since 7 is prime we get x^14 +3x^7 +2 ≡ 0 mod 7 iff either x^7 +2 ≡ 0 mod 7 or x^7 + 1 ≡ 0 mod 7. You could then either use Fermat’s Little Theorem or just search 0 ≤ x ≤ 6 to find solutions.
You should note immediately, though, that x^7 + 1 ≡ 0 mod 7, i.e. x^7 ≡ −1 mod 7, has the solution x = −1.
B
- Suppose that ∼ is equivalence relation on a non-empty set S.
a) For a ∈ S define the equivalence class [a]. b) If a, b ∈ S and a ∼ b prove that [a] = [b]. c) If a, b ∈ S and a b prove that [a] ∩ [b] = ∅.
- Define a relation ∼ on R^2 by
(x 1 , y 1 ) ∼ (x 2 , y 2 ) if, and only if y 2 − y 1 = 2 (x 1 − x 2 ).
Prove that this is an equivalence relation.
- Describe geometrically the equivalence classes of R^2. [15 marks]
Solutions
- a) [a] = {x ∈ S : x ∼ a}. Bookwork [1 mark] b) Assume that a ∼ b. Let k ∈ [a]. Then by definition of the class, k ∼ a. Thus we have both k ∼ a and a ∼ b. So by transitivity k ∼ b. Hence k ∈ [b] by definition of the class. True for all k ∈ [a] means that [a] ⊆ [b].
Let now ∈ [b]. Then by definition of the class, ∼ b. Use symmetry to write a ∼ b as b ∼ a. Thus we have both ∼ b and b ∼ a. So by transitivity ∼ a. Hence ∈ [a] by definition of the class. True for all ∈ [b] means [b] ⊆ [a].
Combining [a] ⊆ [b] and [b] ⊆ [a] gives [a] = [b]. Bookwork [4 marks]
c) Assume a b. Assume, for a contradiction [a] ∩ [b] 6 = ∅. So there exists c ∈ [a] ∩ [b]. Here c ∈ [a] means c ∼ a while c ∈ [b] means c ∼ b. Use symmetry to write c ∼ a as a ∼ c. Thus we have both a ∼ c and c ∼ b. So by transitivity a ∼ b. But this contradicts a b. Hence our last assumption is false, thus [a] ∩ [b] = ∅ as required. Bookwork [4 marks]
- For any (a, b) ∈ R^2 we have b − b = 0 and 2 (a − a) = 0 in which case b − b = 2 (a − a), the condition for (a, b) ∼ (a, b). Hence ∼ is reflexive. [1 mark]
Comments on B
- a) You were not asked to give the definition of an equivalence relation, though a lot of students did (for no marks).
Instead, you were given an equivalence relation ∼, and asked to define the equivalence class [a]. Because ∼ is symmetric, the condition x ∼ a in the definition is identical to a ∼ x and so the definition can be written as
[a] = {x ∈ S : a ∼ x}.
b) and c) Note that these statements are both of the form “If something holds then prove something else”, they are not “if, and only if” statements, and so you do not have to prove the implications in the reverse direction.
Lots of students proved that ∼ is reflexive by saying
“ (a, b) ∼ (a, b) b − b = 2 (a − a) 0 = 2 × 0
which is true, hence ∼ is reflexive”. This proof is “going the wrong way”, you should really start with what you know, i.e. 0 = 2 × 0, and end with what you are trying to prove, i.e. (a, b) ∼ (a, b).
B
- Prove that there are infinitely many prime numbers.
(You may assume without proof that integers greater than 1 have a prime divisor.)
- a) Prove that if a ∈ Z and p is prime with p - a then gcd (p, a) = 1. b) Prove that if a, b ∈ Z, p is prime and p|ab then either p|a or p|b.
(You may assume without proof that if gcd (m, n) = 1 then there exist s, t ∈ Z such that sm + tn = 1.)
- State Fermat’s Little Theorem.
Show that 73739797 + 9797^7373 is divisible by 11. [15 marks]
Solutions
- Assume for contradiction that there are only finitely many primes, p 1 , p 2 , ..., pr say. Consider the integer which is the product of these primes plus 1:
N = p 1 p 2 ...pr + 1.
Then N is not divisible by any of the primes pi, 1 ≤ i ≤ N, for the remainder is 1 on dividing by any of these. Yet N > 1, because this is a non-empty list of primes, containing as it does 2. So, by the assumption in the question, N is divisible by a prime q say. Since q divides N it is not any of the pi, ..., pr, which contradicts the assumption that this list contains all primes. Hence the assumption is false and there are infinitely many primes. Bookwork [4 marks]
- a) The common divisors of a and p must be divisors of p, namely −p, − 1 , 1 or p. But we are told p - a so −p and p are not common divisors. So the only common divisors are −1 and +1. Since +1 is the greatest of these we have gcd (p, a) = 1. Bookwork [2 marks]
b) Assume p is prime and p|ab. If p|a we are finished.
Comments on B
- a) Too many students thought that
N = p 1 p 2 ...pr + 1
is prime. This is not always the case.
Examples
It might by that N is a prime, as in
i) 2 × 3 × 5 × 7 + 1 = 211, and 211 is prime.
ii) 2 × 3 × 5 × 7 × 11 + 1 = 2311 and 2311 is prime.
But often N is not a prime.
iii) If all the primes in the world were { 2 , 3 , 5 , 7 , 11 , 13 } , then
2 × 3 × 5 × 7 × 11 × 13 + 1 = 30031,
but 30031 is not prime, since 30031 = 59 × 509. The important fact is that
59 , 509 ∈ {/ 2 , 3 , 5 , 7 , 11 , 13 } ,
and so we have found some primes not in the original list.
iv) If all the primes in the world were { 2 , 3 , 5 , 7 , 11 , 13 , 17 } , then
2 × 3 × 5 × 7 × 11 × 13 × 17 + 1 = 510511,
but 510511 is not prime, since 510511 = 19 × 97 × 277. The important fact is that again 19 , 97 , 277 ∈ {/ 2 , 3 , 5 , 7 , 11 , 13 , 17 } ,
and so we have found some primes not in the original list.
v) If all the primes in the world were { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 } , then
2 × 3 × 5 × 7 × 11 × 13 × 17 × 19 + 1 = 9699691,
but 9699691is not prime, since 9699691 = 347 × 27953. The important fact is that 347 , 27953 ∈ {/ 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 } ,
and so we have found some primes not in the original list.
b) One student claimed that if p is prime then 2p^ − 1 is prime (which would give an infinitude of primes). This is FALSE.
Claim if n ≥ 3 is odd then 3 divides 2n^ − 1 (and so since all but one prime is odd, all but one of the 2p^ − 1 is composite).
Proof of Claim n odd implies n = 2m + 1 for some m ∈ N. Thus
2 n^ + 1 = 22 m+1^ + 1 = 2 (4)m^ + 1 ≡ 2 (1)m^ + 1 mod 3 ≡ 2 + 1 ≡ 0 mod 3.
Hence 3| 2 n^ + 1.
- a) Too many students forgot that the divisors of p include −p and − 1 as well as p and 1.
We can rewrite the proof using notation from the course, namely that D (n) is the set of all divisors of n, while D (m, n) = D (m) ∩ D (n) is the set of all common divisors of m and n. Then
D (a, p) ⊆ D (p) = {−p, − 1 , 1 , p}.
But p - a implies −p, p /∈ D (a) and so −p, p /∈ D (a, p). Hence
D (a, p) ⊆ {− 1 , 1 }.
Yet − 1 , 1 ∈ D (a, p) and so D (a, p) = {− 1 , 1 }. Therefore
gcd (a, p) = max D (a, p) = max {− 1 , 1 } = 1.
- In Fermat’s Little Theorem too many students forgot the condition that p - a (which, because p is prime, is the same as gcd (p, a) = 1). Luckily, since I was only giving 1 mark for the statement of this Theorem I had to be generous.
I also accepted ap^ ≡ a mod p which does not require the condition p - a.
