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i have no idea about this?????????????

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2 Answers

If n = 4k + 3 does 8 divide n^2-1 [evenly; i.e., no remainder]?
 
(n^2 - 1) / 8 = (n + 1)(n - 1) / 8
 
= (4k + 3 + 1)(4k + 3 - 1) / 8
 
= (4k + 4)(4k + 2) / 8
 
= 4(k + 1)(2)(2k + 1) / 8
 
= 8(k + 1)(2k + 1) / 8
 
= (k + 1)(2k + 1) with no remainder.
 
So the answer is YES.
Hi Daniel;
n=4k+3
I think the second equation is...
n2-1
(4k+3)2-1
Let's FOIL the (4k+3)(4K+3)
FIRST...(4K)(4K)=16K2
OUTER...(4k)(3)=12k
INNER...(3)(4k)=12k
LAST...(3)(3)=9
16k2+12k+12k+9
16k2+24k+9
The original equation includes -1...
16k2+24k+9-1
16k2+24k+8
Can we factor out 8?
8(2k2+3k+1)
YES!!!!