Question 1191208
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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.


This concludes the proof.


Examples:
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	1^3 + 3^3 = 1 + 27 = 28 = <font color=blue>4</font>*7
	3^3 + 5^3 = 27 + 125 = 152 = <font color=blue>4</font>*38
	5^3 + 7^3 = 125 + 343 = 468 = <font color=blue>4</font>*117
	7^3 + 9^3 = 343 + 729 = 1072 = <font color=blue>4</font>*268
	9^3 + 11^3 = 729 + 1331 = 2060 = <font color=blue>4</font>*515
	11^3 + 13^3 = 1331 + 2197 = 3528 = <font color=blue>4</font>*882
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