• Dec 15th 2012, 08:30 PM
DirectorRico
Help!:

Prove by contradiction. If a,b and c are consecutive integers s.t. a<b<c then a^3 + b^3 /=/ c^3 (Cannot equal to c^3).

Here are my steps... let me know if I am on the right track:
For the sake of contradiction: a^3 + b^3 = c^3

(at this point, a<b<c would still hold right?
It is the problem to find a^3 + b^3 = c^3 and find it contradicting the consecutive integer a<b<c?)

I would go on with stating: b = a+1 ; c = b+1 <=> c = (a+1)+1

From here I am somewhat stuck.

Thank you!
• Dec 15th 2012, 09:02 PM
jakncoke
if a, b, c are consecutive integers, then if a is an even number, b is an odd number, and c is an even number.
$a = 2k \text{ } b = 2k + 1 \text{ } c = 2k + 2$

(even number)^3 = even number
$a^2 = (2k)^3 = 8k^3 = 2(4k^3)$

(odd number)^3 = odd number
[tex] (2k+1)^3 = (4k^2 + 4k + 1)(2k+1) = 8k^3 + 4k^2 + 8k^2 + 4k + 2k + 1 = 8k^3 + 12k^2 + 6k + 1 = 2(4k^3 +6k^2 + 3k) + 1

even number + odd number = odd number

2k + 2m + 1 = 2(k+m) + 1

but if c is an even number, (even)^3 = even, so then this cant hold.

then just calculate,
$(2k+1)^3 + (2k+2)^3 = 16k^3 + 36k^2 + 30k + 9$
$(2k+3)^3 = 8k^3 + 36k^2 + 54k + 27$
So $(2k+1)^3 + (2k+2)^3 - (2k+3)^3 = 8k^3 - 24k - 18 = 0$
So if you calculate the discriminat, this eqn. has only 1 real root. Then you can use another test to see the one real root isn't an integer (remember k must be an integer)

The calculation is tedious, use wolfram alpha, just type in (2k+1)^3 and all is done for you.
• Dec 16th 2012, 12:03 AM
DirectorRico