Question 2:
2 | (a) | Suppose x2 – 2y2 = 1 where x and y are
integers.
Then x2 = 2y2 + 1 and so x2 is odd and
therefore x is odd.
So x = 2t + 1 for some integer t and
2y2 = (2t + 1)2 – 1 = 4t2 + 4t.
Thefore y2 = 2t2 + 2t is even and so y is even.
Alternative version
Suppose x2 – 2y2 = 1 where x and y are
integers.
Then x2 = 2y2 + 1 and so x2 is odd and
therefore x is odd.
So
x1 (mod 4) or x3 (mod 4)
x2 1 (mod 4)
x210 (mod 4)
2y2 0 (mod 4)
 y2 is even
 y is even | B1
M1
E1
[3]
B1
M1
E1
[3] | 2.1
2.1
2.1
2.1
2.1
2.1 | For x2 is odd, conclusion (allow
suitable mod 2 argument)
For 2y2 is a multiple of 4
For y2 is even and conclusion.
For x2 is odd, conclusion (allow
suitable mod 2 argument)
For 2y2 ≡ 0 (mod 4).
For y2 is even and conclusion.
(b) | Example code for Python
for x in range(1,s+1):
for y in range(1,s+1):
if x*x-2*y*y==1:
print(x,y) | M1
M1
A1
[3] | 3.3
2.1
2.5 | Pseudo code accepted, condone
lack of syntax, give reasonable
BOD on possible transcription
errors and consider a correct
answer to 2(c) as evidence of a
correct prog.
At least one correct loop
Appropriate conditional statement
Complete prog
(c) | x = 3 y = 2
x = 17 y = 12
x = 99 y = 70
x = 577 y = 408
cao | B1
[1] | 1.1b | Set s to 600 in the above. Ignore
x = 1, y = 0 if seen.
(d) | For the smallest solution
32222 [1]
  
 32 2 32 2 [1]
 2 2
 32 2 32 2 1
  
 1712 2 1712 2 1
So the next solution is x = 17, y = 12
By cubing and taking the fourth power of the second line
above the other solutions are found. | M1
M1
A1
E1
[4] | 2.1a
2.1a
2.1a
2.4 | Substituting x = 3, y = 2 (or
square and simplify the given
algebraic expression)
Squaring and
expanding/simplifying
(or substitute x = 3, y = 2 if
previous step done algebraically)
Stating that this means x = 17, y =
12 is a solution or rearranging to
172 – 2 × 122 = 1
(e) |  n n
32 2 32 2 1 simplified gives a solution for any
positive integer n. So the equations has infinitely many
positive integer solutions x,y | B1
[1] | 2.2a | Infinitely many solutions.
(f) | Example code for Python
for m in range(1,s+1):
for n in range(1,s+1):
if m*(m+1)==2*n*n: (oe)
print(m,n) | M1
A1
[2] | 2.1
2.5 | Pseudo code accepted, condone
lack of syntax, give reasonable
BOD on possible transcription
errors and consider a correct
answer to 2(g) as evidence of a
correct prog.
At least one correct loop and
conditional statement
Complete program
(g) | m = 1 n = 1
m = 8 n = 6
m = 49 n = 35
m = 288 n = 204
cao | B1
[1] | 1.1b | Set s = 300 in the above. Ignore
m = 0, n = 0 if seen.
(h) | Comparing answers to (c) and (g) suggests that, if x, y is a
solution of x22y2 1 in which x, y are positive integers,
x1 y
then m ,n satisfy T = n2
m .
2 2
Since x is odd, m is an integer.
Since y is even, n is an integer.
x1x1
  
 2  2  x21
T  
m
2 8
y2
n2 
4
Since x22y2 1, we have 8T = 8n2 and so
m
T = n2 as required.
m
Since there are infinitely many positive integer solutions to
x22y2 1 (part (v)) there are infinitely many triangular
numbers which are square. | M1*
M1
M1
M1
B1(de
p*)
[5] | 3.1a
2.1
2.4
2.1
3.2a | Relationship between pairs in (c)
and pairs in (g).
Allow x = 2m + 1, y = 2n
Seen in part (a).
Substitutions into T and n2.
m
Allow substitution of x = 2m + 1,
y = 2n into x22y2 1.
Verification that x22y2 1
leads to T = n2.
m
Reference to part (e). Do not
allow if only converse argument
seen (i.e. a solution of T = n2
m
gives a solution of
x2 – 2y2 =1).