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Topic: Two cubes with Total Volume 17 (Read 4606 times) |
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Barukh
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Two cubes with Total Volume 17
« on: Jul 6th, 2007, 5:47am » |
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Two equal cubes with side length 2 units have total volume of 16 cubic units.
Can you find the exact sizes of 2 cubes that have total size of 17 cubic units?
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| « Last Edit: Jul 6th, 2007, 5:48am by Barukh » |
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SMQ
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Re: Two cubes with Total Volume 17
« Reply #1 on: Jul 6th, 2007, 6:07am » |
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Huh? 
Is the cube root of 17/2 not an exact size? Is there some further restriction? Is there some significance I'm missing to your using "length" and "volume" in the first line, but "size" (apparently to mean two different things) in the second?
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towr
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Re: Two cubes with Total Volume 17
« Reply #2 on: Jul 6th, 2007, 6:40am » |
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Do the two cubes that total 17 have to be the same size? Must the side lengths be integer? Can they be negative?
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Barukh
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Re: Two cubes with Total Volume 17
« Reply #3 on: Jul 6th, 2007, 7:45am » |
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Well, well... I should state it more carefully.
What was meant is: the side lengths should be positive integers or rationals.
Sorry for inconveniece.
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Sir Col
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Re: Two cubes with Total Volume 17
« Reply #4 on: Jul 7th, 2007, 3:48pm » |
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[edit]Duh!  [/edit]
I've not solved it, butand I've made a littleno progress...
[edit]Ignore fumbling nonsense below[/edit]
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(a/c)3 + (b/d)3 = 17
a3 + b3 = 17(bd)3 = 17e3
(a+b)(a2-ab+b2) = 17e3
As 17 is prime, a+b=17 or a2-ab+b2 = 17
If a+b=17 then a2-ab+b2 = e3.
(a+b)2 = a2+2ab+b2 = 289 => a2+b2 = 289-2ab
So 289-3ab = e3 => 289-e3 = 3ab.
Only 289-64=225 and 289-1=288 are divisible by 3 and with 3ab=225 or 288 it can be verified that there are no rational solutions in a.
So if a solution exists then a2-ab+b2 = 17 and a+b = e3.
(a+b)2 = a2+b2+2ab = e6
a2+b2 = e6-2ab
So e6-3ab = 17 or e6-17 = 3ab.
As 17==2 mod 3 we are looking for e==2 mod 3.
But that's as far as I can get!
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| « Last Edit: Jul 8th, 2007, 2:07am by Sir Col » |
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Obob
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Re: Two cubes with Total Volume 17
« Reply #5 on: Jul 7th, 2007, 3:56pm » |
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on Jul 7th, 2007, 3:48pm, Sir Col wrote:
As 17 is prime, a+b=17 or a2-ab+b2 = 17
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Doesn't it only follow that 17 divides a+b or 17 divides a2-ab+b2?
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FiBsTeR
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Re: Two cubes with Total Volume 17
« Reply #6 on: Jul 7th, 2007, 7:34pm » |
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on Jul 7th, 2007, 3:48pm, Sir Col wrote:(a/c)3 + (b/d)3 = 17
a3 + b3 = 17(bd)3 |
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Can you explain how you got from the first line to the second line?
EDIT: I assumed that maybe you meant a3 + b3 = 17(cd)3, but even then I still don't follow how you can arrive at this, as I get a3d3 + b3c3 = 17(cd)3
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| « Last Edit: Jul 7th, 2007, 7:36pm by FiBsTeR » |
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SWF
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Re: Two cubes with Total Volume 17
« Reply #7 on: Jul 7th, 2007, 9:37pm » |
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We need to find a3+b3=17c3 with a,b, and c integers. Then a/c and b/c are the sizes of the two cubes. Assume a and b are relatively prime since can multiply a,b,c by any integer for other solutions.
If you look at the values of integers cubed mod 17, it turns out that the form of and and b must be a=17*A+n and b=17*B-n where A and B are integers and n is an integer from 1 through 16. The original equation then becomes:
c3=(A+B)( 289*(A2-AB+B2) + 51*n*(A-B) +3*n2)
That doesn't look much better, but it did eliminate many possibilities.
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towr
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Re: Two cubes with Total Volume 17
« Reply #8 on: Jul 8th, 2007, 10:48am » |
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Using a3 +b3 =17c3, I think c has to be a multiple of 7
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Barukh
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Re: Two cubes with Total Volume 17
« Reply #9 on: Jul 8th, 2007, 10:59am » |
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on Jul 8th, 2007, 10:48am, towr wrote:| Using a3 +b3 =17c3, I think c has to be a multiple of 7 |
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Why?
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towr
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Re: Two cubes with Total Volume 17
« Reply #10 on: Jul 8th, 2007, 11:13am » |
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on Jul 8th, 2007, 10:59am, Barukh wrote:Why not?
I tried sums of two cubes modulo 7 that are also 17 times a cube mod 7. And the latter turns out to 0 mod 7, hence a multiple of 7.
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| « Last Edit: Jul 8th, 2007, 11:41am by towr » |
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towr
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Re: Two cubes with Total Volume 17
« Reply #11 on: Jul 9th, 2007, 2:41am » |
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Maybe it would work to combine the results when you use results modulo primes.
For example with p=19 we have only 7 of the possible 19 sums of cubes (mod 19) that can be 17 times a cube (mod 19). With
337 it's 113, for 2953 it's 985. It seems to be about a third each time.
Now if these results were somewhat independant, you could quickly narrow down our a,b,c.
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Barukh
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Re: Two cubes with Total Volume 17
« Reply #13 on: Jul 9th, 2007, 5:18am » |
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on Jul 9th, 2007, 4:28am, towr wrote:| Seems the maths required to deduce an answer is a bit beyond what I know |
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I don't think so.
Now, when the answer is known: how difficult it is to find the solution with negative sizes?
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revenge
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Re: Two cubes with Total Volume 17
« Reply #14 on: Jul 9th, 2007, 3:42pm » |
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Would this help?
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SWF
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Re: Two cubes with Total Volume 17
« Reply #15 on: Jul 9th, 2007, 9:36pm » |
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Looks like people have managed to look up the answer already, but let me give the progress I have made so far. Continuing from where I left off last time, make the substitution A=x+e and B=x-e, and the equation becomes:
c3=2x*(289*x2+3*q2) where q=17*e+n
The problem is reduced to finding integers x and q that give a cube for this formula, and e is found from floor(q/17), and n=mod(q,17). I gave up on an elegant way to solve the equation, but did find some solutions.
Solution 1: From the formula, c must be a multiple of 2, and from towr's observation also a multiple of 7. Try for the simpliest possible solution: c=2*7=14. A simple thing to try is x=1, and fortunately solving for q gives an integer: q=19. So e=1 and n=2. A=x+e=2, B=x-e=0, a=17*A+n=36, b=17*B-n=-2. A solution is a=36, b=-2, c=-14, or can factor out the 2 for a=18, b=-1, c=7. Unfortunately this has a negative number.
Solution 2: Another easy thing to try is x a perfect cube: x=d3 and c must be of the form c=14*d*f, and
(14*f)3=2*(289*d6+3*q2)
Try various values of f and d, and solve for q until an integer q is found. A simple computer program quickly finds d=19 and f=307 as a solution, and it gives x=6859, q=93277, a=209880, b=23326, c=81662. A common factor of 2 can be removed giving: a=104940, b=11663, c=40831. I could find no other positive solution except multiples of this.
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| « Last Edit: Jul 9th, 2007, 9:38pm by SWF » |
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towr
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Re: Two cubes with Total Volume 17
« Reply #16 on: Jul 10th, 2007, 12:40am » |
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on Jul 9th, 2007, 9:36pm, SWF wrote:| I could find no other positive solution except multiples of this. |
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From as much as I understood from the link I found, once you have one solution, you can, somehow, find infinitely many others (alternating negative and positive, and not merely multiples).
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Barukh
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Re: Two cubes with Total Volume 17
« Reply #17 on: Jul 10th, 2007, 12:40am » |
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Well done, SWF! 
Your negative solution is what I meant in my previous post by negative sizes. Fortunately, it's easy to find in this particular case (but not in general).
Once you've got this solution, an infinite number of others may be generated by a simple transformation formula. Can you find it? As a reference, we've got 2 solutions:
a1 = 18, b1 = -1, c1 = 7.
a2 = 104940, b2 = 11663, c2 = 40831.
What's the relation between the two?
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Hippo
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Re: Two cubes with Total Volume 17
« Reply #18 on: Jul 10th, 2007, 3:49am » |
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The answer to the last question was given in already mentioned link ... the squaring in elliptic curves ... btw: this connection with elliptic curves was unknown to me.
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towr
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Re: Two cubes with Total Volume 17
« Reply #19 on: Jul 10th, 2007, 4:16am » |
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on Jul 10th, 2007, 3:49am, Hippo wrote:| The answer to the last question was given in already mentioned link ... the squaring in elliptic curves ... |
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But what does that mean, that's what I'm wondering.
An actual procedure to find the next triple can be found halfway down on http://www.math.niu.edu/~rusin/known-math/95/numthy.cub
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Hippo
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Re: Two cubes with Total Volume 17
« Reply #20 on: Jul 10th, 2007, 7:28am » |
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For description of the group operations you can also google the description of Lenstra's Elliptic Curve Factorizacion Algorithm.
But may be there is a difference in the curve shape ... so I should think about it a little bit ...
... I can see I know very little about such curves ... if I have read it well in wikipedia ... this was the base for the proof of the Last Fermat Theorem ...
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| « Last Edit: Jul 10th, 2007, 4:21pm by Hippo » |
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Barukh
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Re: Two cubes with Total Volume 17
« Reply #21 on: Jul 10th, 2007, 8:37am » |
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Quote:
Exactly! Looks like Dave knows everything...
on Jul 10th, 2007, 4:16am, towr wrote:
But what does that mean, that's what I'm wondering. |
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Maybe, it's the same as the above? I will need to find out...
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SWF
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Re: Two cubes with Total Volume 17
« Reply #22 on: Jul 10th, 2007, 6:27pm » |
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No wonder I couldn't find any more positive solutions, they get big fast! The formulas given in towr's link seem to have an error. I think it should say that new values of a,b,c can be generated from those of a previous step with:
anew= a*(a3+2*b3)
bnew= -b*(2a3+b3)
cnew= c*(a3-b3)
In the x and q of the formula I posted earlier, the next x of a solution is found from the previous x and q with the simple formula, xnew=x*q3. The next q is qnew=(3*(17x)4 + 6*(17xq)2 - q4)/8. Values of a and b in terms of x and q are: a=(17x+q)/2 b=(17x-q)/2.
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| « Last Edit: Jul 15th, 2007, 11:48am by SWF » |
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Hippo
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Re: Two cubes with Total Volume 17
« Reply #23 on: Jul 25th, 2007, 2:43am » |
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I don't know how to continue ... using elliptic curves requires other curve than x^3+y^3=n.
One can choose v=(x+y),w=xy and got nice elliptic curve, but rational v,w solution seems do not help finding rational solution for x,y.
So I look forward for anyone's explanation of the citations.
SWF: Can you give more details of your computations? (may be subresults?)
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srn437
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Re: Two cubes with Total Volume 17
« Reply #24 on: Aug 29th, 2007, 10:05am » |
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17 expressed as two cubes. Can the sides have irrational or negative length. If no to both, it's impossible.
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