Prove that if \(a\) is any odd integer, then \(a^2-1\) is divisible by 8. (2 marks) --- 8 WORK AREA LINES (style=lined) --- \(\text {Prove if \(a\) is odd, \(a^2-1\) is divisible by 8.}\) \(\text {Let}\ \ a=2 n+1, \ \ n \in \mathbb{Z}\) \(a^2-1=4(2 k+1)(2 k+2)=8(2 k+1)(k+1) /8\) \(\text{If \(n\) is even \((n=2 k)\):}\) \(a^2-1=4(2 k)(2 k+1)=8 k(2 k+1) /8\) \(\therefore \text{ If \(a\) is odd, \(a^2-1\) is divisible by 8}\) \(\text {Prove if \(a\) is odd, \(a^2-1\) is divisible by 8.}\) \(\text {Let}\ \ a=2 n+1, \ \ n \in \mathbb{Z}\) \(a^2-1=4(2 k+1)(2 k+2)=8(2 k+1)(k+1) /8\) \(\text{If \(n\) is even \((n=2 k)\):}\) \(a^2-1=4(2 k)(2 k+1)=8 k(2 k+1) /8\) \(\therefore \text{ If \(a\) is odd, \(a^2-1\) is divisible by 8}\)
\(a^2-1\)
\(=(2 n+1)^2-1\)
\(=4 n^2+4 n+1-1\)
\(=4 n(n+1)\)
\(\text{If \(n\) is odd (\(n=2 k+1, k \in Z)\):}\)
\(a^2-1\)
\(=(2 n+1)^2-1\)
\(=4 n^2+4 n+1-1\)
\(=4 n(n+1)\)
\(\text{If \(n\) is odd (\(n=2 k+1, k \in Z)\):}\)
Proof, EXT2 P1 2024 HSC 14a
Prove that if \(a\) is any odd integer, then \(a^2-1\) is divisible by 8. (2 marks) --- 8 WORK AREA LINES (style=lined) ---