# Problem of the Week #359 - July 14, 2020

Staff member

#### Euge

##### MHB Global Moderator
Staff member
No one answered this week's problem. You can read my solution below.

Reducing the equation modulo $7$ yields $2a^3 \equiv 1 \pmod{7}$, or $a^3 \equiv 4 \pmod{7}$. Note that $\pm 1, \pm 2$, and $\pm 3$ all have cubes that are $\equiv \pm 1\pmod{7}$. Hence, the congruence $a^3 \equiv 3\pmod{7}$ has no solution. This implies the original Diophantine equation has no solution.