# Sums of three cubes in arithmetic progression equal to a cube $x^3+(x+y)^3+(x+2y)^3 = z^3$

Using exhaustive search, small positive and primitive integer solutions to,

$$x^3+(x+y)^3+(x+2y)^3 = 3 x^3 + 9 x^2 y + 15 x y^2 + 9 y^3= z^3\tag1$$

are,

$$x,y = 3,1,\quad x+y =2^2$$

$$x,y = 149,107,\quad x+y =2^8$$

$$x,y = 317,808,\quad x+y =3^2\cdot 5^3$$

P.S. The equation,

$$ax^3 +bx^2 y + c x y^2 + d y^3= z^3$$

with initial rational solution $x_0, y_0$ can be transformed into an elliptic curve. Hence $(1)$ has an infinite number of primitive integer solutions. (Edit: I just recalled I asked something similar two years ago, but without the positivity requirement. See this post.)

Question 1: What are the others with six digits or less?

$\color{brown}{Update:}$ Zander found,

$$x,y = 243800,249239,\quad x+y =79^3$$

Question 2: Why does $x+y$ have interesting factorizations?

For non-positive $x,y$ we have,

$$x,y = −1839,1871,\quad x+y =2^5$$

$$x,y = 13898941449153,-12222218425537 ,\quad x+y =2^8\cdot1871^3$$

• If you choose $x-y,\,x,\,x+y$ instead, we get a simpler equation $3x^3 + 6xy^2 = z^3$. Since $z^3$ must be a multiple of 3, we can set $z=3m$ to turn it into $x^3 + 2xy^2 = 9m^3$. From mathworld.wolfram.com/DiophantineEquation3rdPowers.html it seems there is no general solutions like the Pythagoras triples? – kennytm Apr 14 '15 at 13:00
• I have evaluated the table in math.uni.wroc.pl/~jwr/eslp/tables.htm and only got OP's 3 pairs. @Tito With $x-y,x,x+y$ we also require $0<y<x$. The two forms are equivalent so it should take the same number of brute-force entries. – kennytm Apr 14 '15 at 13:19
• @kennytm: You know about Jarek's tables? I know it has $z<10^6$, and you found there's nothing other than the 3 pairs? Hm, curious. – Tito Piezas III Apr 14 '15 at 13:23
• Here's one: $x=243800, y=249239$. Found with nearly-brute-force search. – Zander Apr 20 '15 at 22:17
• @kennytm: Why did your code miss Zander's solution? – Tito Piezas III Apr 21 '15 at 1:31

Denote $s=x+y$. Then consider equation $$(s-y)^3+s^3+(s+y)^3=z^3,\tag{1}$$ for positive $s,y,z$. First, consider any such integer solutions: for $s>y$ and for $s<y$.

Eq. $(1)$ is equivalent to $$3s^3+6sy^2=z^3.\tag{2}$$ Denote $z=3c$, then $(2)$ is equivalent to

$$s^3+2sy^2=9c^3.\tag{3}$$

To check all up-to-6-digital $x,y$ (or $s,y$), one can use ineq. $s^3<9c^3$, and consider $c<10^7/\sqrt[3]{9}$, i.e. $c<480750$.

Exhaustive search (up to $c<1\,000\,000$) gives us this table of solutions:

\begin{array}{|l|l|l|l|} \hline x & y & s=x+y & z\\ \hline -1 &2 &1 &3\cdot 1 \\ \bf{+3} &\bf{1} &\bf{2^2} &\bf{3\cdot 2} \\ -10 &11 &1 &3\cdot 3 \\ -16 &25 &3^2 &3\cdot 11 \\ \bf{+149} &\bf{107} &\bf{2^8} &\bf{3\cdot 136} \\ -919 &955 &6^2 &3\cdot 194 \\ -1839 &1871 &2^5 &3\cdot 292 \\ -8545 &8549 &2^2 &3\cdot 402 \\ +\bf{317} &\bf{808} &\bf{3^2\cdot5^3} &\bf{3\cdot 685} \\ -12759 &14956 &13^3 &3\cdot 4797 \\ -11589 &54181 &2^5\cdot11^3 &3\cdot 33132 \\ -560239 &590614 &3^5\cdot5^3 &3\cdot 133095 \\ \bf{+243800} &\bf{249239} &\bf{79^3} &\bf{3\cdot 271997} \\ \hline \end{array} Other solutions have $x+y>10^6$ (even $x+y>2\cdot 10^6$).

On Question 2.

Denote $$a=18\frac{c}{s}=6\frac{z}{s},\qquad b=36\frac{y}{s},\tag{4}$$ then eq. $(3)$ is equivalent to $$b^2=a^3-3\times 6^3\tag{5}$$

"Walking" on rational points of Elliptic Curve described by eq. $(5)$, one can find more primitive solutions of $(1)$ (with $x>0$ too). And I guess they cover all such solutions...

Note that all $s=x+y$ has one of $4$ forms here (why?): $$s=q^3,\quad s=4q^3, \quad s=9q^3, \quad s=36q^3.\tag{6}$$ This form is true for $s>y$ and for $s<y$.

Here is table with positive primitive $x,y$ (sorted by $s$):

\begin{array}{|r|r|rl|} \hline x & y & s=x+y & \\ \hline 3 &1 &4&=4\cdot 1^3 \\ 149 &107 &256&=4(2^2)^3 \\ 317 &808 &1125&=9\cdot 5^3 \\ 243800 &249239 &493039&=79^3 \\ 4062853 &437147 &4500000&=36(2 \cdot 5^2)^3 \\ 720469 &11105855 &11826324&=36(3\cdot 23)^3 \\ 2957658 &181262351 &184220009&=569^3 \\ 87092698624 &2477541935 &89570240559&=9(3^2\cdot 239)^3 \\ 456326994059 &722215431505 &1178542425564&=36(7\cdot 457)^3 \\ 3170389673427 &431591782862 &3601981456289&=15329^3 \\ 10460723603247633 &5072768120572198 &15533491723819831&=(59\cdot4229)^3 \\ 132098636066470851 &28361108004240976 &160459744070711827&=(7\cdot 149\cdot 521)^3 \\ 74556823768778731 &160637266326925333 &235194090095704064&=4(2^3\cdot 13\cdot 3739)^3 \\ \cdots & \cdots & \cdots\\ \hline \end{array}

• Thanks for this comprehensive data! (I factorized the $s$ column of your first table.) I hope you don't mind. – Tito Piezas III May 3 '15 at 12:10