Question 2:
2 | (a) | Suppose c=b+1 and substitute into 𝑎2+𝑏2 =
𝑐2.
Simplify to a 2 = b + c .
Suppose a 2 = b + c . Substitute into 𝑎2+𝑏2 =
𝑐2. Simplify to b + c + b 2 = c 2 .
Factorise and rearrange,
( b + c ) ( 1 + b − c ) = 0 .
Since b+c0, c=b+1. | M1
A1
M1
A1
[4] | 1.1a
2.2a
1.1
2.2a | If argument is present as an “if and only if” must see
valid reasoning to support this assertion.
1±√(2𝑏+1)2
Or 𝑐 = = 𝑏+1 or 𝑐 = −𝑏.
2
Since 𝑏,𝑐 ≥ 0,𝑐 = 𝑏+1
Accept CAS for finding 𝑐.
(b) | Appropriate program structure
Loops with correct range
Check required condition on a and c with if
statement.
Fully correct program | M1
M1
A1
[3] | 3.3
2.1
2.5 | Pseudo code accepted, condone lack of syntax, give
reasonable BOD on possible transcription errors.
For example, use of loops, if statement(s) to check
condition(s) and print final output.
Condone one incorrect loop with one wrong value for
M1 mark.
Example program
For a in range(1,501):
for c in range(a,1001):
if a*a + (c-1)*(c-1)==c*c and c-1!=0:
print(a,c-1,c)
(c) | 21 | B1
[1] | 1.1a
Alternative method 1
If b 2 = a + c then, by a similar argument to that | M1 | M1 | 2.1 | 2.1
in (a), c = a + 1.
So 𝑏2 = 2𝑎+1. | M1 | 2.1
For 𝑎 ≥ 3, 𝑏 = √2𝑎+1 < 𝑎. Which is a | A1 | A1 | 3.2a | 3.2a
contradiction to 0  a  b  c .
When 𝑎 = 1, 𝑏 = √3 and 𝑎 = 2, 𝑏 = √5.
Which is impossible since 𝑏 is an integer.
[3]
Alternative method 2
If b 2 = a + c then, by a similar argument to that | M1 | M1 | 2.1 | 2.1 | Accept any correct method leading to deduction. | Accept any correct method leading to deduction.
in (a), c = a + 1.
Since a, b and c are integers and 𝑐 = 𝑎 + 1 | M1 | 2.1
0  a  b  c , this implies that 𝑏 = 𝑎 = 𝑐−
1.
Since b 2 = a + c , 𝑏2−2𝑏−1 = 0. | A1 | 3.2a | Accept any correct method leading to deduction, for
This equation has no integer solutions. | example CAS to find roots which are not integers.
[3]