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

Explanation:

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.

step-by-step explanation:

no solution.

the functions have the same rate of change i think. i narrowed it down to this answer and "the rate of change of tour 1 is greater" but i got it wrong using that answer so im assuming its the answer above. (this is a question i also answered from the iready site if anyone needs a reference)