Date | May 2022 | Marks available | 6 | Reference code | 22M.1.AHL.TZ1.8 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 1 |
Command term | Prove | Question number | 8 | Adapted from | N/A |
Question
Consider integers a and b such that a2+b2 is exactly divisible by 4. Prove by contradiction that a and b cannot both be odd.
Markscheme
Assume that a and b are both odd. M1
Note: Award M0 for statements such as “let a and b be both odd”.
Note: Subsequent marks after this M1 are independent of this mark and can be awarded.
Then a=2m+1 and b=2n+1 A1
a2+b2≡(2m+1)2+(2n+1)2
=4m2+4m+1+4n2+4n+1 A1
=4(m2+m+n2+n)+2 (A1)
(4(m2+m+n2+n) is always divisible by 4) but 2 is not divisible by 4. (or equivalent) R1
⇒a2+b2 is not divisible by 4, a contradiction. (or equivalent) R1
hence a and b cannot both be odd. AG
Note: Award a maximum of M1A0A0(A0)R1R1 for considering identical or two consecutive odd numbers for a and b.
[6 marks]
Examiners report
Most candidates did not present their proof in a formal manner and merely relied on an algebraic approach rendering the proof incomplete. Very few candidates earned the first mark for making a clear assumption that a and b are both odd. A significant number of candidates only considered consecutive or identical odd numbers. The required reasoning to complete the proof were often poorly expressed or missing altogether. Only a small number of candidates were awarded all the available marks for this question.