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Fun math question. Let's see who is truly elite.?
Given two arbitrary odd integers a and b, prove that a^3 - b^3 is divisible by 2^n iff a-b is divisible by 2^n.
Note: I have half the work done. I have one way proved. In other words I have found that 2^n | a-b -> 2^n | a^3 - b^3. I have yet to show that 2^n | a^3- b^3 -> 2^n | a-b.
Any help just on that second half would be greatly appreciated!
Question:
How do you figure that a^2 + ab + b^2 will always be odd? For instance... a and b are both even... then a^2 + ab + b^2 is even, right?
3 Answers
- Anonymous1 decade agoFavorite Answer
Let a = 2n + 1 and b = 2m + 1
a^3 - b^3 = (a - b)(a^2 + ab + b^2) = 2(n - m)(4n^2 + 4m^2 + 4mn + 6n + 6m + 3)
If 2^n divides a - b then it must divide a^3 - b^3 which has a - b as factor.
If 2^n divides a^3 - b^3 then it must divide 2(n - m) because the long factor is odd, and so divides a - b = 2(n - m).
Edit. You defined a, b as odd in the question.
- ToddLv 61 decade ago
Suppose 2^n | a^3- b^3. a^3-b^3 = (a-b)*(a^2+ab+b^2), but (a^2+ab+b^2) will always be odd (odd + odd + odd = odd). Thus, 2^n | a-b.
