Solving the diophantine equation $x^3+y^3 = z^6+3$

Question

I've the following problem:

Show that the congruence $x^3+y^3 \equiv z^6+3\pmod{7}$ has no solutions. Hence find all integer solutions if any to $x^3+y^3-z^6-3 = 0.$

We can rearrange the first equation to $z^6 \equiv (x^3+y^3-3) \mod{7}$. But $z^6\equiv 1\mod{7}$ so this is only possible when $x^3+y^3=4$ which no $x,y \in \mathbb{Z}$ satisfy.

Now I don't know if that helps at all solve the second bit. We have $z^6 = x^3+y^3-3$. I know that $x^3+y^3 = (x+y)(x^2-xy+y^2)$.

Let $(x,y,z)$ be a solution for $x^3+y^3-z^6-3=0$. Then it is also a solution mod $7$. — Yanko, Jul 13 '19 at 19:14

score 3 · Accepted Answer · answered Jul 13 '19 at 19:19

3

Any solution for $x^3+y^3-z^6-3=0$ is also a solution mod $7$. Therefore there are no solutions to the original equation.

answered Jul 13 '19 at 19:19

Yanko

13,758

Thanks. I understand now. – Notsredt Jul 13 '19 at 19:25

score 2 · Answer 2 · answered Jul 13 '19 at 23:35

►If $z\equiv0\pmod7$ we have $x^3+y^3\equiv 3\pmod7$ but by Fermat's little theorem $w^6\equiv1\pmod7$ if $w\not \equiv 0\pmod7$ then $w^3\equiv\pm1\pmod7$ so either $\pm1\pm1\not\equiv3\pmod7$ or $0\pm1\not\equiv\pmod7$.

►If $z\not\equiv0\pmod7$ we have the equation $x^3+y^3\equiv 1+3=4\pmod7$ and similar reasoning show the impossibility.

►If $x^3+y^3=z^6+3$ have solution this solution must satisfy the congruence modulo $7$ so there is not solutions.

Oscar Lanzi · Answer 3 · 2019-07-13T23:18:03.087

1

Render any cube $\in\{-1,0,1\}\bmod 9$. Then $z$ can't be a multiple of $3$ because that makes $ x^6+3\equiv 3\bmod 9$. Try $z$ not a multiple of $3$, then $z^6=(z^2)^3$ where $z^2\equiv 1\bmod9$ forcing $z^6\equiv 1\bmod9$. Then $x^3+y^3 =x^6+3\equiv 4\bmod9$. Well, today is not the day Serena Williams ties that tennis record either.

edited Jul 13 '19 at 23:18

answered Jul 13 '19 at 19:51

Oscar Lanzi

39,403

Solving the diophantine equation $x^3+y^3 = z^6+3$

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