Prove algebraically that the square of any odd number is always 1 more than a multiple of 8. Let n stand for any integer in your working

If n is "any integer", then 2n+1 is "any odd number."

The square of any odd number is then ...

(2n+1) ² = 4n² + 4n + 1 = 4n (n+1) + 1

Since n is any integer, one of n and n+1 will be an even integer, so the product 4n (n+1) will be divisible by 8.

Then the sum 4n (n+1) + 1 is one more than a number divisible by 8, hence ...

the square of an odd number is 1 more than a multiple of 8.