# A curious phenomenon for the cubic diophantine $x^3+y^3+z^3 = 1$?

Given positive integers $a,b,c$,

$$a^3+b^3 = c^3\pm1\tag{1}$$

then there are exactly 133 solutions with $c<10^6$. (Wroblewski has tables for the general $x_1^3+x_2^3+x_3^3+x_4^3 = 0$ with all terms $<10^6$.) The first twenty-four for $(1)$ are,

$$\begin{array}{cccc} |t|&a&b&c\\ 1&6&8&\color{red}{9}\\ 1&10&\color{red}{9}&12\\ -&64&94&103\\ 2&71&\color{blue}{138}&\color{red}{144}\\ 2&73&\color{red}{144}&150\\ -&\color{blue}{135}&\color{blue}{138}&172\\ -&\color{blue}{135}&235&249\\ -&334&438&495\\ -&372&\color{blue}{426}&505\\ -&\color{blue}{426}&486&\color{blue}{577}\\ 3&242&720&\color{red}{729}\\ 3&244&\color{red}{729}&738\\ -&566&823&904\\ -&791&812&\color{blue}{1010}\\ -&236&1207&1210\\ -&368&1537&1544\\ -&1033&1738&1852\\ -&\color{blue}{1010}&1897&1988\\ 4&575&2292&\color{red}{2304}\\ 4&\color{blue}{577}&\color{red}{2304}&2316\\ -&1938&2820&\color{blue}{3097}\\ 1&2676&3230&\color{red}{3753}\\ -&\color{blue}{3097}&3518&4184\\ 1&\color{red}{3753}&4528&5262\\ \end{array}$$

The pairs of red numbers are caused by two identities, mostly by,

$$(1 - 9t^3)^3 + \color{red}{(9t^4)}^3 + (3t - 9t^4)^3=1$$

for $|t|=1-4$, and the last pair by,

$$(1+9t^3+648t^6-3888t^9)^3 + \color{red}{(-135t^4+3888t^{10})}^3 + (-3t-81t^4+1296t^7-3888t^{10})^3 = 1$$

The absolute value of $t$ is given in the table above, with one sign giving one solution, and the opposite sign for the other. (The middle term is unchanged.) There are an infinite more polynomial identities, but their degrees and coefficients get larger and larger, and their first manifestations have $c$ beyond this range.

Question: So what causes the pairs of blue numbers?

If you extend this table to cover all 133 solutions, there are many more blue pairs. I want to know how others will approach this problem, or if they originally will dismiss it as mere coincidence. (It's not, really.)

P.S. I found this "pairing" while considering the asymptotics of the solutions to $(1)$, discussed in this post.

Since bounty has ended, might as well answer this question for those who are curious. Some of the blue pairs of numbers belong to the identity,

$$(1 - a c + b c)^3+(a + c^2 - a c^3)^3+(a c^3 - b - c^2)^3 =1\tag{1}$$

where,

$$a^2+ab+b^2 = 3c(ac-1)^2\tag{2}$$

or since $(2)$ is a quadratic in $b$,

$$b=\frac{1}{2}\Big(-a\pm\sqrt{a^2-4\big(a^2-3c(ac-1)^2\big)}\Big)\tag{3}$$

Since the middle term of $(1)$ does not contain $b$, then it remains unchanged regardless of which sign of $(3)$ is chosen. For example, let $a,c = -570,\, 12/19$ and using the - then + sign of $(3)$ yields,

$$505^3+(\color{blue}{-426})^3+(-372)^3 = 1$$

$$577^3+(\color{blue}{-426})^3+(-486)^3 = 1$$

Making the discriminant of $(3)$ a square,

$$a^2-4\big(a^2-3c(ac-1)^2\big) = y^2\tag{4}$$

in the integers can also be done. Given an initial solution, one can then use a Pell equation to generate an infinite more.

A lot then, but not all, of the blue pairs can be explained by both signs of $(3)$ yielding integer terms to $(1)$. Maybe the rest are coincidence. Then again, maybe it's not.

• I'd look at the factorization $a^3+b^3=(a+b)\cdot {a^3+b^3\over a+b}$ and $c^3+1=(c+1)\cdot {c^3+1\over c+1}$ because the primefactorization of the factors are disjunct (except if 3 is involved as a primefactor), so the factors of the one-equation must match exactly the factors of the other equation - maybe that leads just to your equations above... – Gottfried Helms Oct 9 '13 at 5:14