SEN POTW #2Weekly, I will post a mathematical question to be solved by us amazing SEN'ers! The problems may vary wildly, and most will be rather difficult to solve. Depending on difficulty, I will award more or less points than previous POTW's! A scoreboard will be kept in a collapsible box at the bottom of each POTW! Hints may or may not be given depending on the question, and will never affect the points to be awarded! The points will be awarded to the first person to answer the POTW correctly, guessing is allowed ~at your own risk! Work must be shown upon giving your answer, just so that other participants can see your thoughts and possibly learn from them. If sufficient work isn't shown then you may be awarded less points than stated for that particular POTW!
Without further ado, here is POTW #2
Points to be awarded upon a correct answer: 3
Note: For this problem sufficient work MUST be shown in order to obtain points whatsoever!
" Prove that three consecutive integers can not satisfy the equation: A
3+B
3 = C
3 "
Solution (Spoiler)
2x^3 + x3^2 + 3x + 1 != x^3 + 6x^2 + 12x + 8;
a^3 + (a+1)^3 = (a+2)^3
a^3 + a^3+3a^2+3a+1 = a^3+6a^2+12a+8
a^3 - 3a^2 - 9a - 7 = 0
This cubic has three roots, they can be found using the cubic equation, i'll just skip that step and give approximations of the answers: 5.05456 and -1.02728+0.574095i and -1.02728-0.574095i
None of these are integers.
EDIT: For those of you who want the cubic equation:
http://en.wikipedia.org/wiki/Cubic_function#General_formula_of_rootsPOTW Score Board
1. 5 points - azala
2. 3 points - Doodle77
Post has been edited 3 time(s), last time on Oct 4 2009, 8:59 pm by CecilSunkure.
None.
>be faceless void >mfw I have no face
OK cecil wants more steps...
A = A
A + 1 = B
A + 2 = C
Because A B and C are consecutive.
A^3 + B^3 = C^3
We substitute B and C with A+1 and A+2, respectively.
A^3 + (A + 1)^3
This comes out as 2A^3 + 3A^2 + 3A + 1
Whereas C^3, which after substitution is (A + 2)^3, equals A^3 + 6A^2 + 12A + 8.
Clearly the two sides cannot equal each other, so
A^3 + B^3 != C^3
Post has been edited 2 time(s), last time on Oct 4 2009, 8:31 pm by zany_001. Reason: Happy now? XD
Red classic.
"In short, their absurdities are so extreme that it is painful even to quote them."
Relatively ancient and inactive
A^3 + (A + 1)^3 is NOT A^3 + 3A^2 + 3A + 1. It's 2A^3 + 3A^2 + 3A +1.
It comes down to proving that 2A^3 + 3A^2 + 3A +1 != x^3 + 6x^2 + 12x + 8 for any integer x. It's not very difficult.
None.
>be faceless void >mfw I have no face
Ah yeah, forgot to add A^3. Comes from my being on a cellphone.
Red classic.
"In short, their absurdities are so extreme that it is painful even to quote them."
Sorry zany, your work is invalid
Still looks invalid to me. You need to show all your steps anyways to recieve points.
Post has been edited 1 time(s), last time on Oct 4 2009, 8:25 pm by CecilSunkure.
None.
a^3 + (a+1)^3 = (a+2)^3
a^3 + a^3+3a^2+3a+1 = a^3+6a^2+12a+8
a^3 - 3a^2 - 9a - 7 = 0
This cubic has three roots, they can be found using the cubic equation, i'll just skip that step and give approximations of the answers: 5.05456 and -1.02728+0.574095i and -1.02728-0.574095i
None of these are integers.
EDIT: For those of you who want the cubic equation:
http://en.wikipedia.org/wiki/Cubic_function#General_formula_of_roots
None.
a^3 + (a+1)^3 = (a+2)^3
a^3 + a^3+3a^2+3a+1 = a^3+6a^2+12a+8
Just this would have been sufficient, though zany did correct his mistake before this was posted.
Although I really did want all the steps you guys took to get from where doodle went in line 1 to line 2, and then combine like terms between each side.. I'll award zany 3 points since he showed work, just remember -next time I add a note in the OP asking for work to be shown, show more than what was shown this time around, or I might knock off some points
Congratz zany!
None.
>be faceless void >mfw I have no face
Thanks! Now, don't make the next one trig and I should get a lead O.o
Red classic.
"In short, their absurdities are so extreme that it is painful even to quote them."
Relatively ancient and inactive
a^3 + a^3+3a^2+3a+1 = a^3+6a^2+12a+8
How is this sufficient? It goes down to, as Doodle showed, 'a^3 - 3a^2 - 9a - 7 = 0', which is an equation in a format which can easily have integer solutions.
None.
Although I really did want all the steps you guys took to get from where doodle went in line 1 to line 2, and then combine like terms between each side.
Huh? How does it take more than one step to get between those? Don't you know your binomial expansions?
Just this would have been sufficient, though zany did correct his mistake before this was posted.
No, that's not, you need to prove that the cubic has no integer solutions.
Also, zany's work is incorrect. It is possible for A^3+B^3 to equal C^3, just not for an integer.
None.
Relatively ancient and inactive
I think for the next PotW, we need a 10+ AIME problem.
None.
>be faceless void >mfw I have no face
Also, zany's work is incorrect. It is possible for A^3+B^3 to equal C^3, just not for an integer.
That's what we are trying to find, no?
And I see what you mean. I haven't done cubic yet, just quadratic, so doodle can have the points.
Red classic.
"In short, their absurdities are so extreme that it is painful even to quote them."
These are too easy? Want more difficult questions?
[Edit]@zany if you want to let Doodle have the points, that's fine. Though you did give the answer I was looking for. I'll modify the OP.
None.
>be faceless void >mfw I have no face
But not a total proof, which is what was asked for.
Red classic.
"In short, their absurdities are so extreme that it is painful even to quote them."
Grand Moderator of the Games Forum
A^3+B^3 = AB^9
I found math after sixth grade rather useless =P
Next question is already up:
http://www.staredit.net/180621/
None.
