1729, and related questions

Question

I just read this paragraph: (written by G. H. Hardy, on Ramanujan)

I remember once going to see him when he was lying ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. ‘No,’ he replied, ‘it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways.’

Was Ramanujan right?

What are other numbers having such property (expressible as the sum of two cubes in two different ways)?

Are there infinite number of them?

And, on the other hand:

What if the word "cubes" is replaced by "5-degree power"? Would such numbers exist? If yes, what would be the smallest?

Another SO question related to 1729: Proof that 1729 is the smallest taxicab number

See also: https://math.stackexchange.com/questions/67406/special-numbers — , Jul 31 '14 at 22:43
Not as widely known, soon after this Ramanujan was interrupted by a man from Porlock, and completely lost his train of thought. — Will Jagy, Jul 31 '14 at 23:03
@copper.hat, not hemlock, $$\begin{array}{l}\text{For he on honey-dew hath fed,}\cr \text{And drunk the milk of Paradise.}\end{array}$$ — Will Jagy, Aug 01 '14 at 01:15
@copper.hat, there you go, or perhaps a damsel with a dulcimer. http://www.poetryfoundation.org/poem/173247 — Will Jagy, Aug 01 '14 at 01:26
@Will Jagy +1 for the Kubla Khan reference on math.SE of all places :) — user_of_math, Aug 01 '14 at 01:36
@copper.hat, he was happy, had all those ponies, yak butter for his tea — Will Jagy, Aug 01 '14 at 02:43
see also http://www.amazon.com/Managing-Dental-Practice-Genghis-Khan/dp/1846193966 — Will Jagy, Aug 01 '14 at 02:48
See http://math.stackexchange.com/questions/226333/generalised-hardy-ramanujan-numbers for a discussion of 5th powers. Summary: no nontrivial solution of $a^5+b^5=c^5+d^5$ is known, no solution is expected, no proof of nonexistence is known. — Gerry Myerson, Aug 01 '14 at 03:26
For symmetric solutions formula there. http://math.stackexchange.com/questions/469151/find-all-integer-solutions-to-diophantine-equation-x3y3z3-w3/776918#776918 — individ, Aug 01 '14 at 05:23
@GerryMyerson All answers are good and interesting. Something to learn from each of them. I find discovering generalizations or different aspects of a given problem one of the most exciting experiences related to mathematics. I will mark the best answer as "the answer" in few days. — VividD, Aug 02 '14 at 13:32

score 5 · Answer 1 · answered Aug 01 '14 at 02:06

5

If negative numbers are allowed, then $91=3^3+4^3=6^3+(-5)^3$.

answered Aug 01 '14 at 02:06

Senex Ægypti Parvi

2,419

Ross Millikan · Answer 2 · 2014-08-01T03:00:07.010

4

There are many more: they are given in OEIS

1729, 4104, 13832, 20683, 32832, 39312, 40033, 46683, 64232, 65728, 110656, 110808, 134379, 149389, 165464, 171288, 195841, 216027, 216125, 262656, 314496, 320264, 327763, 373464, 402597, 439101, 443889, 513000, 513856, 515375, 525824, 558441, 593047, 684019, 704977

There are references and a link to a list of the first 10000, but no formula is given that guarantees there are infinitely many.

Somewhere I read that if you sum the inverses of the highest exponents of the variables in a Diophantine equation and get more than $1$, you should expect infinitely many solutions. In this case it is $a^3+b^3=c^3+d^3$. These sum to $4/3$ so there should be infinitely many. For fifth powers it is less than $1$, so by this test there should be finitely many. I do not remember the justification (name of the theorem?)

edited Aug 01 '14 at 03:00

answered Aug 01 '14 at 02:52

Ross Millikan

374,822

1

Woah, that rule actually works for Fermat's Last Theorem! Was this actually a theorem, or just a rule of thumb with exceptions? For example, it doesn't apply to $2x+4y=1$, but maybe such trivial cases are to be excluded? – Nishant Aug 01 '14 at 03:15
1

@Nishant: it certainly needs there not to be some "trivial" reason for there to be or not be infinitely many. In your example, the GCD of the coefficients is that trivial reason. In other examples, there can be (say) a recurrence that proves there are infinitely many, even though the rule says there won't be. – Ross Millikan Aug 01 '14 at 03:18
Yes, there is infinitely many in positive integers, as shown by Choudhry. Pls see answer below. – Tito Piezas III Feb 13 '18 at 08:54

score 3 · Accepted Answer · answered Aug 01 '14 at 03:43

Theorem 412 of Hardy and Wright, An Introduction to the Theory of Numbers, 6th edition, page 442, says,

"Whatever $r$, there are numbers that are representable as sums of two positive cubes in at least $r$ different ways."

The proof is quite elementary, but involves a bit more typing than I am keen to do. The notes say the proof was found by Fermat, but there was one place in the argument where he just assumed something that actually needed proof; Mordell was the first to write a complete proof, but didn't publish it. So I suppose the hardy & Wright book was the first place a full proof was published.

For the question about 5th powers, see my comment on the original question.

score 1 · Answer 4 · edited Feb 24 '23 at 17:01

Very late for this party, but yes, there is an infinite number of taxicab numbers. The complete solution in positive integers to,

$$x_1^3+x_2^3 = x_3^3+x_4^3$$

was given by Choudhry's On Equal Sums of Cubes (1998). For positive integers $a,b,c$,

$$\begin{aligned} d\,x_1 &= (a^2 + a b + b^2)^2 + (2a + b)c^3\\ d\,x_2 &= (-a^3 + b^3 + c^3)c\\ d\,x_3 &= (a^2 + a b + b^2)^2 - (a - b)c^3\\ d\,x_4 &= (a^3 + (a + b)^3 + c^3)c\end{aligned}$$

where, $$(a^3-b^3)^{1/3}<\,c\,<\frac{(a^3-b^3)^{2/3}}{a-b}$$

and $d=1$, or chosen such that $\text{GCD}(a,b,c)=1$.

P.S. For Choudhry's complete solution in positive integers to $x_1^3+x_2^3+ x_3^3=x_4^3$, see this post.

1729, and related questions

4 Answers4