Thread: Prove by Contradiction

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!

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.


(even number)^3 = even number


(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.

if u start with an odd, even, odd
then just calculate,


So
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.

That's great, thanks

+rep

Could you also help me with these questions?:

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