Ed,
> 3370318
> 2716057
> We're both expressible as the sum of two cubes!
>
> I invite the zealous viewer to check that claim.
> If you'd like to try trickier numbers, here a a few more
> numbers that are the sum of two cubes: 6669, 7222,
> 119041, 2716343, 2741256, 3370087, 6017193,
> 6742008, 9016488, 16776487, 24375176.
I found solutions for all but 7222 and 119041. Since I think I used an exhaustive search, I'm looking forward to seeing the solutions for these two. The ones I found are:
3370318 = 119³ + 119³
2716057 = 952³ + (-951)³
6669 = 20³ + (-11)³
2716343 = 191³ + (-162)³
2741256 = 207³ + (-183)³
2741256 = 168³ + (-126)³
2741256 = 140³ + (-14)³
2741256 = 114³ + 108³
3370087 = 150³ + (-17)³
6017193 = 246³ + (-207)³
6017193 = 209³ + (-146)³
6017193 = 185³ + (-68)³
6017193 = 180³ + 57³
6017193 = 166³ + 113³
6742008 = 252³ + (-210)³
6742008 = 189³ + (-21)³
6742008 = 188³ + 46³
6742008 = 168³ + 126³
9016488 = 234³ + (-156)³
9016488 = 216³ + (-102)³
9016488 = 208³ + 26³
9016488 = 195³ + 117³
16776487 = 292³ + (-201)³
16776487 = 256³ + (-9)³
16776487 = 255³ + 58³
16776487 = 220³ + 183³
24375176 = 303³ + (-151)³
24375176 = 290³ + (-24)³
24375176 = 272³ + 162³
I also found the following sums of rational, non-integral cubes. In each case, the two numbers being cubed add up to an integer.
119041 = (45 4/15)³ + (29 11/15)³
6017193 = (181 7/8)³ + (10 1/8)³
16776487 = (485 2/5)³ + (-460 2/5)³
16776487 = (245 1/6)³ + (126 5/6)³
24375176 = (460 2/3)³ + (-418 2/3)³
Regards,
Al Zimmermann
-----------------------------------------------------
Dear Ed,
I suppose you might not have used complex cubes, but then you'd have to have
made some typos. I can't figure out what gave you 7222, but 119041 could be
119401, giving (200, -199). For the others:
2716343 191, -162
2741256 114, 108
3370087 150, -17
6017193 166, 113
6742008 168, 126
9016488 195, 117
16776487 220, 183
24375176 272, 162
Thank goodness for my Voyage 200. Without it, I may not have found the factor
(79 + 54i) of 119041, nor the quadratic form of (79 + 54i - x)^3 - x^3. Despite
the difficulty of keeping the coefficients in the quadratic straight on my
calculator, I've finally calculated the cubes that make up this last number.
(62 + 6i)^3 + (17 + 48i)^3 = 119041
Bryce Herdt
-------------------------------------------------------
Another trick to finding the cubes:
Since the smallest number that can be obtained between two cubed numbers when subtracted is always going to be from sequential numbers, then this can be used as a limit.
So if cubed numbers subtracted from the next lower cubed number is graphed and a curve is fitted to the graph this can be used to determine a limit to search. A perfectly fitted line for this is y=3x^2+3x+1 where y is the desired number and the two x's are the sequentially cubed numbers (ie y=37 so x=4 and -3 (it actually comes out with the signs opposite on the x values but is easily seen the desired result)).
Therefore the search for a number like 6669 is limited to 48 and there is no need to search above 48 since 48^3-47^3=6769 and that's the smallest number between cubes.
Which leads me to believe that 7222 is not expressible by two cubes since the limit for a negative and positive cube is 50
(50^3-49^3=7351) and the limit for two positives is 20 (7222^(1/3)=19.33) which is only 450 possibilities to check and none of them are equal to 7222.
There are also only 2700 numbers to check for 119041 and I get nothing.
Thanks
Colin Sturm
PS - I have attached my original email to you below
So is the challenge to find all the tricks or all the numbers?
I have only thought of a few tricks (more just observations than tricks):
There can only be two positive numbers or one positive and one negative to check for. Two negatives would be the same as checking for a positive answer, the signs would just be switched.
If searching for two positive cubed numbers equal to the desired value (x) then neither number can be greater than x^(1/3) so if writing a program the for loop is much shorter only going from 1 to x^(1/3).
Checking for one negative and one positive is trickier. My method was to take the number in question (6669) and add increasing cubed numbers and checking if the result^(1/3) is an integer. There's probably a better trick for this part.
I wrote a small Matlab program to find the numbers.
The numbers:
Futurama numbers:
119 119 3370318
952 -951 2716057
Your numbers: (There were a few with two results)
20 -11 6669
191 -162 2716343
114 108 2741256
140 -14 2741256
150 -17 3370087
180 57 6017193
185 -68 6017193
188 46 6742008
189 -21 6742008
208 26 9016488
216 -102 9016488
255 58 16776487
256 -9 16776487
272 162 24375176
290 -24 24375176
I was unable to find numbers for 7222 or 119041
I checked all the numbers from 1 to 10000 (there are 506 under 10000) and the closest numbers I found to 7222 were
7202 and 7254. I didn't check around 119041 because it was taking too long.
Please get back with me on some hints.
Thanks
Colin Sturm
-----------------------------------------------------
Aw, that's cheating! Using negative cubes? That eliminates the lower bound!
Anyway, the first two are 2 * 119^3 and 952^3 + (-951)^3. I'll find the others
later.
The only other one I've found is 6669 = 20^3 - 11^3. What, are you using
complex cubes, too? I can't handle that! (in all honesty im using a voyage 200 for
this and it doesnt do complex factoring) Yeeauggh....
119041 = (62, 6)^3 + (17, 48)^3
Bryce Herdt (MathIdentity)
---------------------------------------------------
Ok, I'm (at least for now) stumped.
3370318 = 119^3 + 119^3
For 2716057 exhaustive search shows that a^3 and b^3
cannot both be positive (it was so easy to set up in
The Mathematical Explorer, it was irrestible, I'm
still trying to convince my employer to spring for
full-blown Mathematica 5.0). Ok, on to more advanced
ideas:
2716057 = 499 * 5443
where 499 and 5443 are both prime. Therefore since:
a^3 + b^3 factors as (a+b)(a^2-ab+b^2) then the
solution must satisfy either:
a + b = 499
a^2 - a b + b^2 = 5443
or
a + b = 5443
a^2 - a b + b^2 = 499
But according to TME neither of these sets of
equations is solvable over the integers.
Can you give at least a hint what I am missing
(probably something obvious).
[I gave the hint Gaussian Integers]
Ok, that I had taken into account (I remembered
negatives, I hadn't attempted to solve over Gaussian
integers however I would have recognized them if the
solution had taken that form!). What I didn't take
into account is that the list of factorizations
included:
2716057 = 499 * 5443 = 5443 * 499
= (-499) * (-5443) = (-5443) * (-499)
and more importantly I left out!
2716057 = 1 * 2716057 = 2716057 * 1
= (-1) * (-2716057 ) = (-2716057 ) * (-1)
Therefore:
2716057 = 952^3 - 951^3
My daughter is demanding the computer so I don't have
time to work on the rest yet. I did do a
FactorInteger on each to see that only one of the
examples has two factors! My plan would be to map all
possible factorizations x*y and then attempt to solve
the equations as below.
a + b = x
a^2 - a b + b^2 = y
********
20^3 - 11^3 = 6669
My general method is to write down all possible products of the number expressed as integers (or Gaussian integers) N = xy and then perform:
Solve[{a+b==x, a^2-a b+b^2 == y},{x,y}]
with the reasoning being that a^3 + b^3 = (a+b)(a^2-ab+b^2) and if a, b are integers (Gaussian integers), so are the two parts of the product.
and look for cases where (a,b) are integers (or Gaussian integers).
However, I haven't had time to write a function that efficiently goes from the prime factors to all possible products, or to test a predicate that tests if something's an integer. I did try a few cases to show that there is no integral solution. Hence I leave the search as an exercise to the reader.
Cheers,
Lyman Hurd
---------------------------------------------------
-11^3 + 20^3 = 6669
(too lazy to do the others!)
Denis Borris
---------------------------------------------------
The problem:
a. Each of these numbers is expressible as the sum of two integer cubes.
What are the cubes?
3370318
2716057
b. If you'd like to try trickier numbers, here a a few more numbers that
are the
sum of two cubes:
6669, 7222, 119041, 2716343, 2741256, 3370087, 6017193, 6742008,
9016488, 16776487, 24375176. Can you figure out all the tricks?
Notation:
<> is "not equal"
~= is "congruent"
Solution:
In order to solve these by a means other than brute force,
we first need to note some properties of x^3 + y^3.
We can assume that x+y <> 0, because in that case
x = -y and x^3 = -y^3 and x^3 + y^3 = 0, which is
not the case.
x^3 + y^3 = (x+y)^3 - 3xy(x+y), so
x^3 + y^3 ~= (x+y)^3 mod 3
But n^3 ~= n mod 3, so
x^3 + y^3 ~= x+y mod 3
Observe also that x^3 + y^3 ~= x+y mod 2, so
x^3 + y^3 ~= x+y mod 6
The next step is to factor all of the target numbers, and see what
falls out. For each of the factorizations there is an implied
(-1 * -1), with the -1's to be distributed where appropriate.
3370318 = 2 * 7^3 * 17^3
2716057 = 499 * 5443
6669 = 3^3 * 13 * 19
7222 = 2 * 23 * 157
119041 = 13 * 9157
2716343 = 7 * 29 * 13381
2741256 = 2^3 * 3^3 * 7^3 * 37
3370087 = 7 * 19 * 25339
6017193 = 3^3 * 7 * 13 * 31 * 79
6742008 = 2^3 * 3^3 * 7^4 * 13
9016488 = 2^3 * 3^3 * 13^3 * 19
16776487 = 7 * 13 * 19 * 31 * 313
24375176 = 2^3 * 7 * 19 * 739
Some of these are "almost solved" by factoring. With these,
only a little manipulation is required:
3370318 = 2 * 7^3 * 17^3
= 2 * 119^3
= 119^3 + 119^3
2741256 = 2^3 * 3^3 * 7^3 * 37
= (2*3*7)^3 * 37
= (2*3*7)^3 * (64 - 27)
= (2*3*7)^3 * (4^3 - 3^3)
= (2*3*7*4)^3 - (2*3*7*3)^3
= 168^3 + (-126)^3
6742008 = 2^3 * 3^3 * 7^4 * 13
= (2*3*7)^3 * 91
= (2*3*7)^3 * (64 + 27)
= 168^3 + 126^3
9016488 = 2^3 * 3^3 * 13^3 * 19
= (2*3*13)^3 * (27 - 8)
= 234^3 + (-156)^3
One of these is close enough to a well-known cube that we can
save some time by checking a guess (if it doesn't work out,
we can try the final method):
16776487 = 16777216 - 729
= 256^3 + (-9)^3
Most of the rest can be solved by using the congruence above
and the factorizations. I was able to solve all but two in
this manner.
For example:
2716343 = 7 * 29 * 13381
(x^3 + y^3) = 2716343 ~= 5 mod 6, so (x+y) ~= 5 mod 6
x^3 + y^3 = (x+y) * [(x+y)^2 - 3xy], so (x+y) is a factor of x^3 + y^3
Factors of 2716343 that are ~= 5 mod 6 are -7, +29, and -13381.
Trying (x+y) = 29, we have [(x+y)^2 - 3xy] = 93667.
So xy = -30942. Since (x+y) = 29, we have
x = 191, y = -162 (or vice versa).
So 2716343 = 191^3 + (-162)^3
Most of the remainder of the target numbers may be solved
in this manner. The only trial and error involved is in
selecting among the factors of equal congruence mod 6.
There were two targets that I was unable to solve this way.
Final results:
3370318 = 119^3 + 119^3
2716057 = 952^3 + (-951)^3
6669 = 20^3 + (-11)^3
7222 = ???
119041 = ???
2716343 = 191^3 + (-162)^3
2741256 = 168^3 + (-126)^3
3370087 = 150^3 + (-17)^3
6017193 = 180^3 + 57^3
6742008 = 168^3 + 126^3
9016488 = 234^3 + (-156)^3
16776487 = 256^3 + (-9)^3
24375176 = 272^3 + 162^3
-- Susan Hoover
---------------------------------------------------------
Colin Sturm (2+11i)^3+(20+1i)^3 = 7222
--------------------------------------------------------
1^3 = 1
28115742^3 = 2225347273222227224
7^3 = 343
13464^3 = 2440744441344
8121^3 = 535585155561
18821^3 = 6666963601661
426859^3 = 77777383297757779
2^3 = 8
999...999^3 = 999...99700...0029999...999
Juha Saukkola