Question 1191208: Prove that the sum of the cubes of any two consecutive odd integers is divisible by four Found 2 solutions by Edwin McCravy, math_tutor2020:Answer by Edwin McCravy(20056) (Show Source):
2n-1 is an odd integer for any integer n
The next larger odd integer is 2 more than that or
2n-1+2 or 2n+1
The sum of their cubes is
(2n-1)3+(2n+1)3
Factor this as the sum of two cubes:
[(2n-1)+(2n+1)][(2n-1)2-(2n-1)(2n+1)+(2n+1)2]
[2n-1+2n+1][(2n-1)2-(2n-1)(2n+1)+(2n+1)2]
[4n][(2n-1)2-(2n-1)(2n+1)+(2n+1)2]
There is no need to simplify the factor on the right
because the factor on the left is divisible by 4, so
the whole thing is divisible by 4.
Edwin
You can put this solution on YOUR website!
k = any integer
2k = an even integer
2k+1 = an odd integer
(2k+1)+2 = 2k+3 = the next odd integer right after 2k+1
Their respective cubes are (2k+1)^3 and (2k+3)^3
Let's see what happens if we expand things out for the first cubic expression
(2k+1)^3 = (2k+1)(2k+1)^2
(2k+1)^3 = (2k+1)(4k^2+4k+1)
(2k+1)^3 = 2k(4k^2+4k+1)+1(4k^2+4k+1)
(2k+1)^3 = 8k^3+8k^2+2k+4k^2+4k+1
(2k+1)^3 = 8k^3+12k^2+6k+1
Do the same for the other cubic expression
(2k+3)^3 = (2k+3)(2k+3)^2
(2k+3)^3 = (2k+3)(4k^2+12k+9)
(2k+3)^3 = 2k(4k^2+12k+9)+3(4k^2+12k+9)
(2k+3)^3 = 8k^3+24k^2+18k+12k^2+36k+27
(2k+3)^3 = 8k^3+36k^2+54k+27
Then add up the results
(2k+1)^3+(2k+3)^3 = (8k^3+12k^2+6k+1) + (8k^3+36k^2+54k+27)
(2k+1)^3+(2k+3)^3 = (8k^3+8k^3)+(12k^2+36k^2)+(6k+54k)+(1+27)
(2k+1)^3+(2k+3)^3 = 16k^3+48k^2+60k+28
(2k+1)^3+(2k+3)^3 = 4*4k^3+4*12k^2+4*15k+4*7
(2k+1)^3+(2k+3)^3 = 4*(4k^3+12k^2+15k+7)
(2k+1)^3+(2k+3)^3 = 4*(some integer)
The fact we're able to factor a 4 out front means that the sum of any two consecutive odd perfect cubes is divisible by 4.
4 is a factor of (2k+1)^3+(2k+3)^3 for any integer k.
In other words, (2k+1)^3+(2k+3)^3 is a multiple of 4.
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