42

On April 1st 2019, my blog post was titled "42 is the new 33" and that time a solution to the problem of expressing 42 as a sum of three cubes had not been found. However, very recently, a solution was found as this Numberphile Video explains:

The solution is visible there in the thumbnail for the video but I'll repeat it below: $$X^3+Y^3+Z^3=42 \text{ when }$$ $X = -80538738812075974$ 

$Y = 80435758145817515$ 

$Z = 12602123297335631$

Certain numbers cannot be the sum of three cubes. The reason is that any cubic number must be equal to 0 or ±1 mod 9. If we add three cubic numbers together, the sum must therefore lie between -3 and +3 and so any number that is equal to 4 or 5 mod 9 (or equivalently ±4) cannot be expressed as a sum of three cubes. So between 1 and 99, the following numbers fall into this category: 4, 5, 13, 14, 22, 23, 31, 32, 40, 41, 49, 50, 58, 59, 67, 68, 76, 77, 85, 86, 94, 95.

It wasn't obvious to me why "any cubic number must be equal to 0 or ±1 mod 9". However, I finally realised why this was the case. Here is what I discovered:

Consider a number $x$ that can be divided, by 9, $n$ times leaving a remainder of $r$. This means that $x=9n +r$ and so $x^3=(9n+r)^3$. This latter expression becomes: $$(9n)^3+3 \times (9n)^2 r+ 3 \times (9n)r^2+r^3$$ Clearly 9 divides each term of the above expression, except perhaps the $r^3$ term. Because we are considering modulus 9, the remainder $r$ must lie between -4 and +4 (-4 is the same as +5, -3 is the same as +6 etc.). Let's consider each possible remainder:

• $±4^3=±64$ and $±64 \equiv ±1 \bmod 9$
• $±3^3=±27$ and $±27 \equiv 0 \bmod 9$
• $±2^3=±8$ and $±8 \equiv ±1 \bmod 9$
• $±1^3=±1$ and $±1 \equiv ±1 \bmod 9$
• $0^3=0$ and $0 \equiv 0 \bmod 9$

So there you have it, any cubic number must be equal to ±1 or 0 and so the sum of three cubes must lie between ±3. This excludes any number that is ±4 or ±5 mod 9.

Here are the numbers from 1 to 99 expressed as a sum of three cubes (zeroes allowed). The table was taken from an old site and I had to manually enter the values for 33, 42 and 74 because they were still empty:

Figure 1: numbers between 1 and 29

Figure 2: numbers between 30 and 64

Figure 3: numbers between 65 and 99

As the YouTube video makes clear, the search is on for a solution to 114, the next "unknown" after 42, but also for a third solution to 3. The two current solutions are $1^3+1^3+1^3=4^3+4^3+(-5)^3=3$ but is there another solution? As yet, we don't know.

ADDENDUM: LINK

Well actually now we do know: $$3=569936821221962380720^{3}+(-569936821113563493509)^{3}+(-472715493453327032)^{3}$$ The only remaining unsolved cases up to 1,000 are 114, 165, 390, 579, 627, 633, 732, 906, 921, and 975.