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I need help with this question

 Feb 14, 2019

I got 2/4 for this question


 Feb 14, 2019

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

CPhill will explain im sure!cheeky

StarStrike  Feb 14, 2019

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?frown

StarStrike  Feb 14, 2019
edited by StarStrike  Feb 14, 2019

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



cool cool cool

 Feb 14, 2019

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