## Tuesday, April 29, 2014

### An easier challenge?

NOTE: Don't forget to submit solutions and/or questions via the NEW Contact form at top of right sidebar...

And the winner is...
I'm happy to report that no one submitted an incorrect proof for my Challenge. Then again no one submitted anything! Disappointed but not surprised. Of course I've learned not to take this personally since it was a last minute announcement but I will leave the Challenge open for now for anyone who still wants to submit their proof or a counterexample to show I'm wrong!
A comment from anyone who tried the Challenge would be appreciated.  I saw the large number of views so it's hard to determine why no responses...
I don't give up easily so here's another Algebra Challenge which might also elicit zero responses!
Note: These are really designed for students...
1^3 + 2^3 < 3^3
2^3 + 3^3 < 4^3
Show algebraically that if a,b,c are consecutive positive integers with a<b<c,  then
a^3 + b^3 > c^3 for a> 5.

#### 1 comment:

Andrew Ciszewski said...

I am working on the harder proof, I just need some Pascal Triangle Numbers... or to recall the formula, I should say.

On this one, I used a purely algebraic, rather than inductive proof:

a=5+k
b=6+k
c=7+k

1) a^3+b^3?c^3

2) (5+k)^3 + (6+k)^3 ? (7+k)^3

3] (125 + 75k + 15k^2 + k^3)
+
(216 + 108k + 18k^2 + k^3)
?
343 + 147k + 21k^2 + k^3

4] 341 + 183k + 33k^2 + 2k^3
?
343 + 147k +21k^2 + k^3

5] 36k+12k^2 + k^3 ? 2

if k>1 then 36k+ 12k^2 +k^3 > 2

If k>1 then a>5.
QED