42 is the new 33 - Numberphile
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange users cite it as a reference on the sums-of-three-cubes problem, especially the cracking of 33 and 42 and why such searches are hard.
Recommended by 0 people across 0 different months, 2011 to 2025.
What it is
A Numberphile video about the problem of writing numbers as sums of three cubes, covering how Andrew Booker found a solution for 33 and why 42 was the last unsolved number below 100. It features number theory discussion and references to computational search methods and related papers.
Cited in 3 Wikipedia articles, including Numberphile, Sums of three cubes, Andrew Booker (mathematician).
What people said
Cites KCd · 2011-12 · Mathematics Stack Exchange · on “Can you give an example of a complex math problem that is easy to solve?”
For an integer $n$ , let's seek integral solutions of $x^3 + y^3 + z^3 = n$ . 1) When $n = 29$ a solution is easy to find: $(x,y,z) = (3,1,1)$ . 2) When $n = 33$ it is harder to find a solution, but one is known: $$ (x,y,z) = (8866128975287528, -8778405442862239, -2736111468807040). $$ This was found in 2019 by Andrew Booker. See [LINK] and [this video] . 3) Here I had earlier used $n = 42$ , saying we expect it is a sum of three cubes in $\mathbf Z$ but that no representation of $42$ in that form was currently known at the time I wrote that. Now (Sept. 2019) a representation is known: $42 = (-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3$ . I don't want to keep updating this a
Cites corey979 · 2018-08 · Mathematics Stack Exchange · on “What problems have been frequently computationally verified for large values?”
Searching of solutions of the Diophantine equation $$x^3+y^3+z^3=k$$ for small $k$ ( $k<1000$ ) has been performed [LINK] for $|x|,|y|,|z|$ up to $10^{15}$ . For $k<100$ , solutions were not yet found for $k=33$ and $k=42$ . For $k<1000$ , there are 14 values without solution. Edit: 33 has been cracked. [this video] Edit2: 42 has been cracked. [LINK]
Cites D.Matthew · 2019-03 · Mathematics Stack Exchange · on “Integer solutions to the diophantine equation $a^3+b^3+c^3=6$”
I once thought that this equation $x^3+y^3+z^3=6$ has only a few smaller integer solutions. Until somebody told me this $6=192722201207819^3+162765491944499^3+(-225522344776678)^3$ existed. I don't know how it came about. Question : How to find more integer solutions to the equation $a^3+b^3+c^3={\color{red}{6}}$ ? a few smaller integer solutions : \begin{align*} (-637)^3+(-205)^3+644^3&=6\\ (-235)^3+(-55)^3+236^3&=6\\ (-58)^3+(-43)^3+65^3&=6\\ (-1)^3+(-1)^3+2^3&=6\\ \end{align*} Something happened recently I've also heard that mathematicians have recently solved this problem: The least integer solutions to the equation $a^3+b^3+ c^3=33$ Triple cubic Diophantine equations [LINK] 42 is the ne
Cites Eric Towers · 2019-05 · Mathematics Stack Exchange · on “Solution to a Third degree diophantine equation”
Do you have any structural reason to believe such a method to find suitable $k$s exists? Even the question of which integers can be represented as sums of three cubes is unresolved for the equivalent of $k = 42$... 42 is the new 33 - Numberphile [this video] .
Cites Arthur · 2025-12 · Mathematics Stack Exchange · on “Can $2025$ and $2026$ be written as sums of five fifth powers?”
Considering how the sum of three third powers [LINK] it apparently quite difficult [this video] aside from where modular arithmetic immediately excludes a solution, I don't think five fifth powers is easier. I could be wrong, though. So I suspect your best hope is to find some modulus that makes it obviously impossible.