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#1

#2**0 **

Remember that any odd number can be repersented as 4n+1 so your proof is pretty solid...

CPhill will explain im sure!

StarStrike
Feb 14, 2019

#3**+2 **

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:

(2n+1)^2=

4n^2+4n+1=

4(n^2+n)+1=

4n(n+1)+1

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?

StarStrike
Feb 14, 2019

#4**+1 **

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

So

(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

CPhill Feb 14, 2019