Zadanie 8, Paper 1, maj 2022, TZ1, Matematyka AA IB HL

Assume both are odd, so a=2m+1,b=2n+1 for integers m,n. Then a² + 3b² = 4m(m + 1) + 12n(n + 1) + 4. Each consecutive product is even; write m(m+1)=2u,n(n+1)=2v. The sum becomes 8(u+3v)+4, which has remainder 4 on division by 8. This contradicts divisibility by 8, so a,b cannot both be odd.

M1:state the contradictory assumption. A1:general odd forms 2m+1,2n+1. A1:correct expansion/factorization. A1:use evenness of consecutive products to obtain 8N+4. R1:this is not divisible by 8. R1:explicit contradiction and conclusion. Subsequent marks are independent of the initial assumption mark. Considering only equal or consecutive a,b cannot establish the general claim. If only equal or consecutive odd a,b are considered, cap at M1 A0 A0 A0 R1 R1 (3/6), preserving the source restricted-case tolerance. An equivalent general odd-square modulo-8 argument earns corresponding marks.