I got 2/4 for this question
Remember that any odd number can be repersented as 4n+1 so your proof is pretty solid...
CPhill will explain im sure!
Since the number you start with is odd you can write it 2n+1, that is, it is one more than an even number. Squaring 2n+1 gives
Look at it like this:
Thus the square of 2n+1 is one more than 4n(n+1). Since n and n+1 are consecutive integers one of them must be even, and hence 2 divides n(n+1). Thus 8 divides 4n(n+1) and hence 4n(n+1)+1 is one more than a multiple of 8. That is dividing 4n(n+1)+1 by 8 leaves a remainder of 1.
Does this help?
I think there may be better ways to do this....but here's my attempt
Any odd can be written as (2n + 1) where n is an integer .....and any multiple of 8 can be written as 8m....where m is some integer
(2n + 1)^2 = 8m + 1 (1) expand the left side
4n^2 + 4n + 1 = 8m + 1 subtract 1 from both sides
4n^2 + 4n = 8m divide both sides by 4
n^2 + n = 2m factor the left side
n ( n + 1) = 2m (2)
Note that n, n + 1 are consecutive integers....and the product of any two consecutive integers is even....and the right side - 2m - is always even...so....since (2) is always true.....then (1) is always true