Ramanujan and sum of four cubes

Every multiple of 6 can be written as a sum of four cubes

The proof of the theorem is elementary as well as elegant.

Consider $$(n+1)^3 + (n-1)^3 = 2n^3 + 6n$$

Thus,

$$6n = (n+1)^3 + (n-1)^3 + (-n)^3 + (-n)^3$$

Effectively proving the theorem and also giving the required four numbers. The professor also made a remark that a proof is due to Ramanujan. I recently found this scribbled in my notebook and have since not been able to find any reference to this on the web. As far as I know, Kennigel's biography, 'The Man Who Knew the Infinity' does not mention it. There are tons of references to Taxicab numbers, Sum of four square proofs etc..

Does anyone know of any reference to the theorem and the proof? Is it a part of a more general theory?

• You might want to look at mathworld.wolfram.com/CubicNumber.html it mentions this identity along with some more about writing whole numbers as the sum of 4 signed cubes. Jun 27, 2013 at 16:04
• Cool, I missed the Wolfram page. The page does provide a general set of theorems for representing numbers as sums of cubes. The identity in question does appear in on the page, so does the Taxicab number. But it does not mention the source of the identity. Maybe it did not originate from Ramanujan and my notes were erroneous. Jun 27, 2013 at 16:32
• the book Hardy wright theory of numbers has the equation you stated.author is probably Mordell and Hammond.
– user242371
May 21, 2015 at 10:20
• It follows that all primes $\ge5$ are the sum of five cubes. Sep 20, 2016 at 12:24
• +1 , for " Ramanujan"
– user926846
Dec 7, 2021 at 4:12

Similar to what happens with a famous Lagrange's theorem (that of the four squares), likewise an old conjecture still unproved says that on $$\mathbb Z$$ every number is a sum of four cubes. We have proven that this is so on $$\mathbb Q$$ (*).

This old conjecture is quite plausible and has been partially proven. It is known to hold for all numbers that are not congruent to $$±4$$ modulo $$9$$ (in particular for numbers of the form $$6n$$); that is to say, for its full proof it would suffice to limit itself to the integers of the form $$9m ±4$$.

The proof in $$\mathbb Q$$, concisely given, is as follow: for rational arbitrary $$a$$, putting $$A_1 =a^5 +10a^4 -8a^3 +16a^2 +64a-32\\ A_2 =a^5 -10a^4 -8a^3 -16a^2 +64a+32\\ A_3 =-a^5 +8a^4 +8a^3 -16a^2 +80a+32\\ A_4 =-a^5 -8a^4 +8a^3 + 16a^2 +80a-32$$ we have the equality $$a(6a^4 +24a^2 +96)^3 =A_1^3 +A_2^3+A_3^3+A_4^3.$$

Details in (*) Pro-Mathematica, Vol 13, nº25-26 (1999)

Indeed, it was who discovered the sum of four squares, it was G.Xylander that tried to solve what Diophantus had pointed much earlier. Also, it was discovered that, by Fermat that every number is the sum of four squares. The final proof came with the Lagrange's Four-Square Theorem that follows from Euler's Four-Square Identity.

And the proof uses quarternions and can be found here: https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem

However, the sum of cubes for a multiple of 6 was generalized by Edward Waring, Hibert Theorem (which as to find $$g(k)$$, this is, for every $${\displaystyle k}$$ , let $${\displaystyle g(k)}$$ denote the minimum number $${\displaystyle s}$$ of $${\displaystyle k}$$th powers of naturals needed to represent all positive integers). And the source is: https://en.wikipedia.org/wiki/Waring%27s_problem