Question 2:
2 | (a) | (i) | 136 | B1
[1] | 1.1 | May see math.factorial(136!) % 137
2 | (a) | (ii) | 137 is prime by part 2(a)(i) and Wilson’s
theorem. | B1
[1] | 1.1 | Must include clear statement regarding primality of 137.
2 | (b) | (i) | Appropriate structured program
Loop for 𝑗 up to 𝑛+1.
Check for solution with if statement and print
result.
Fully correct programme. | M1
A1
A1
[3] | 3.3
2.1
2.5 | Pseudo code accepted, condone lack of syntax, give reasonable
BOD on possible transcription errors based on correct answer to
2(b)(ii).
Program must include code to implement Wilson’s theorem. This
may include use of math.factorial or direct computation of
factorial.
Example Python code
def wlistprimes(n):
for j in range (2, n +1):
if (math.factorial(j - 1) + 1) % j == 0:
print(j)
Must not produce 1 as a result.
2 | (b) | (ii) | 263, 269, 271, 277, 281, 283 and 293. | B1
[1] | 1.1
2 | (c) | (i) | 2 is the only even prime number. | B1
[1] | 2.2a
2 | (c) | (ii) | A number congruent to 0 (mod 4) is divisible
by 4 and hence not prime. | B1
[1] | 2.2a
2 | (c) | (iii) | 263, 271 and 283. | B1
[1] | 1.1
2 | (c) | (iv) | 𝑁 leaves a remainder 3 when divided by
𝑐 ,𝑐 or 𝑐 .
1 2 3 | B1
[1] | 2.2a
2 | (c) | (v) | 3 | B1
[1] | 1.1
2 | (d) | (i) | 3 | B1
[1] | 2.2a | Accept −1.
2 | (d) | (ii) | 𝑀 is a positive integer so by the fundamental
theorem of arithmetic 𝑀 can be written
uniquely as the product of powers of prime
numbers.
Expressing each power of a prime as a
product gives the required form. | B1
B1
[2] | 1.2
2.4 | It is sufficient to state that the 𝑞 s are not necessarily distinct.
𝑖
2 | (d) | (iii) | For a contradiction suppose 𝑞 ≡ 1(mod 4)
𝑖
for all integers 𝑖 where 1 ≤ 𝑖 ≤ 𝑟.
Then
𝑀 ≡ 𝑞 ⋯𝑞 (mod 4)
1 𝑟
≡ 1×⋯×1 (mod 4)
≡ 1 (mod 4).
But 𝑀 (mod 4) = 3.
This is a contradiction so 𝑞 ≢ 1(mod 4) for
𝑖
all integers 𝑖 where 1 ≤ 𝑖 ≤ 𝑟. | M1
A1
A1
A1
[4] | 3.1a
2.1
1.1
2.2a | Discussion of 𝑀 > 2 not required.
May also see consideration of 𝑞 = 2. Since 𝑀 is odd, the case
𝑖
𝑞 ≡ 2(mod 4) is not possible. Condone no discussion of this
𝑖
point.
Or equivalently, this is a contradiction, so 𝑞 ≡ 3(mod 4) for
𝑖
some integer 𝑖 where 1 ≤ 𝑖 ≤ 𝑟.
2 | (d) | (iv) | By (d)(iii) at least one 𝑞 satisfies 𝑞 ≡
𝑖 𝑖
3(mod 4) and (d)(ii) implies 𝑀 ≡
0 (mod 𝑞 ).
𝑖
By assumption 𝑝 ,…,𝑝 is a complete list of
1 𝑛
primes congruent to 3 (mod 4).
[Since 𝑀 is not divisible by 3,] this 𝑞 must be
𝑖
one of the 𝑝 ,…,𝑝 .
2 𝑛
So 𝑀 ≡ 4𝑝 …𝑝 + 3 (mod 𝑞 )
2 𝑛 𝑖
≡ 3(mod 𝑞 ).
𝑖
This is a contradiction to 𝑀 (mod 𝑞 ) ≡ 0 so
𝑖
there are infinitely many primes congruent to
3(mod 4) | B1
B1
B1
B1
B1
[5] | 3.1a
1.1
2.2a
2.1
2.1 | Seen or implied use of this assumption from the question stem.
May see an equivalent argument that none of the 𝑞 s is in the list
𝑖
of the 𝑝 s.
𝑗
If making an argument using the fact that none of 𝑞 s is in the list
𝑖
of the 𝑝 s then will be able to assert the existence of a new
𝑗
unlisted prime congruent to 3(mod 4).