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April 10, 2012

Integers equal to the sum of three cubes

Filed under: mathematics — ckrao @ 11:32 am

One easy-to-state open problem in number theory is whether every integer not of the form 9k \pm 4 is the sum of three cubes. Some solutions are easy to find, while others are notoriously difficult.

For example 29 can be written as 3^3 + 1^3 + 1^3, or even 4^3 - 3^3 - 2^3.

It is clear that 9k \pm 4 cannot be written as the sum of three cubes since x^3 \equiv -1, 0 \text{ or } 1 \mod 9.

There also exist infinite families of solutions in certain cases (though these need not represent all solutions). For example,

\displaystyle (9t^3 + 1)^3 + (9t^4)^3 + (-9t^4 - 3t)^3 = 1

\displaystyle (6t^3 + 1)^3 - (6t^3 - 1)^3 - (6t^2)^3 = 2

For a long time it was not known whether 30 can be written as the sum of cubes. The following is the “simplest” way this can be done, found as recently as 1999.

\begin{array}{lcl} 30 &=& (2,220,422,932)^3 + (-2,218,888,517)^3 + (-283,059,965)^3\\ &=& 10,947,302,325,566,084,787,191,541,568 - 10,924,622,727,902,378,924,946,084,413 - 22,679,597,663,705,862,245,457,125 \end{array}

It is still not known whether 33, 42 or 74 can be written as the sum of three cubes (edit: it is now known that 74 and 33 can be – see [5] and [6])! In these cases no solution has been found when any of the cubes is less than (10^{14})^3 in magnitude [4].

One of the computational algorithms to find solutions to a^3 + b^3 + c^3 = n where n is relatively small, proposed by Elkies, involves converting the equation to (-a/c)^3 + (-b/c)^3 \approx 1 and then considering rational points close to the curve y = \sqrt[3]{1-x^3}, x \in [0, 1/\sqrt[3]{2}]. This curve is covered by small parallelograms and the problem is converted to finding lattice points in a pyramid using basis reduction followed by the Fincke-Pohst algorithm [4].



[2] Hisanori Mishima, Chapter 4. n=x^3+y^3+z^3

[3] D.J. Bernstein,

[4] A.-S. Elsenhans and J. Jahnel, New sums of three cubes, Math. Comp. 78 (2009), 1227–1230, available here.

[5] Sander G. Huisman, Newer sums of three cubes,, April 2016.

[6] Andrew R. Booker, Cracking the Problem With 33.



  1. Hmm, wondering what the name for this equation is:

    Or the differences between the differences between consecutive cubes. Or for that matter the proof that if the differences process is recursively used on any give power of x (y=x^n, x is within positive integers), that the result is always factorial based on n. In other words, it equals ‘n!’. This HAS to have been looked into during the days when machines designed to do that automatically were being used around the time after Colin Maclaurin died.

    Comment by Joe — November 1, 2013 @ 12:07 pm | Reply

    • Not sure, but it is referred to a 3.1.3 equation here:

      To understand why the n! result is true, refer to the final equation of this earlier blog post and let p(x) = x^n. Match the coefficients of x^n of both sides of that equation to conclude that a_n = n! where a_n is the bottom number of the finite difference table.

      Comment by ckrao — November 2, 2013 @ 11:24 am | Reply

      • Well I kind of figured it was related to the n’th derivative function for any given n’th power of x. Forgive my attempt at doing this in text: Dx^5 of x^5=5*4*3*2*1
        This is true since Dx x^5=5x^4 and you can repeat the process for all powers until you get x^0 (in other words, a constant).

        What would be the name for a generalized equation family like this?
        a_1^n + … + a_n^n = b^n (where _ means subscript which is a way of representing an array member)
        So sum of 4 hypercubes having a ‘content’ of exactly 1 hypercube. That is the geometric description of a^4+b^4+c^4+d^4=z^4 obviously. Change n to 2 and you get good ol’ Pythagoras’ Theorem. Which of course has a geometric description exactly matching a circle (constant radius) or triangle (hypotenuse changes). 😉

        Thanks for answering earlier. Diophantine equations are… a bit broad for something that specific. It’s sort of the other side of Fermat’s (big) Last Theorem. It could be fun (or frustrating – haven’t checked if anyone’s succeeded for this specific case) to prove there’s infinite solutions for some families of this nature. Many arbitrary polynomials have non-algebraic solutions…

        Comment by Joe — November 2, 2013 @ 11:26 pm

      • Sorry, should have read those first before replying. Reading them right now!

        Comment by Joe — November 2, 2013 @ 11:33 pm

  2. sum of three different cube of positive integers is equal to a another cube of positive integers

    Comment by Subhankar Mondal — February 28, 2014 @ 7:19 am | Reply

  3. 74 = 66229832190556³ + 283450105697727³ – 284650292555885³


    Comment by John Doe — June 1, 2016 @ 12:35 pm | Reply

    • Many thanks for this! The post has now been updated with the link you have provided.

      Comment by ckrao — June 19, 2016 @ 5:39 am | Reply

  4. 21^3 + 28^3 + 35^3 = 42^3 , which is the series 3 4 5 6 multiplied by 7.

    by a program I wrote.

    There are so many sums of cubes; perhaps they follow a rule similar to the Pythagorean theorem! (multiples work)

    Comment by — August 2, 2016 @ 8:38 pm | Reply

    • But we talk about $a^3+b^3+c^3=42$, not about $a^3+b^3+c^3=42^3$.

      Comment by guest — July 8, 2018 @ 5:07 am | Reply

  5. and k stands for…

    Comment by mos — August 19, 2018 @ 3:06 am | Reply

  6. Could somebody please explain in further detail why the values that can be represented as 9k+-4 can’t be solved. I’m struggling to wrap my head around why.

    Comment by sunny — October 15, 2018 @ 1:51 pm | Reply

    • Yes, a perfect cube has remainder -1, 0 or 1 when divided by 9 so the sum of three of these cannot be of the form 9k+-4.

      Comment by ckrao — October 16, 2018 @ 1:35 am | Reply

  7. (8866128975287528)**3 + (-8778405442862239)**3 + (-2736111468807040)**3 = 33

    Comment by Sualeh — March 8, 2019 @ 8:51 pm | Reply

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