Question

Prove algebraically that the square of any odd number is always 1 more than a multiple of 8.

Note: Let n stand for any integer in your working

Answer 1

Answer: For odd n, [4n(n+1) + 1] = [4(n+1)n + 1

Step-by-step explanation: the term 4*(n+1) is always a multiple of 8, and so is [4(n+1) * n] . Thus the square of odd is yet again 1 more than some multiple of 8.