Is it true that all elements in $\mathbb{Z}/n\mathbb{Z}$ are representable as the sum of a square and a cube?

Example: ($n=7$)

$0 \equiv 0^2+0^3 \left( \text{mod } 7 \right)$

$1 \equiv 1^2+0^3 \left( \text{mod } 7 \right)$

$2 \equiv 1^2+1^3 \left( \text{mod } 7 \right)$

$3 \equiv 2^2+3^3 \left( \text{mod } 7 \right)$

$4 \equiv 5^2+0^3 \left( \text{mod } 7 \right)$

$5 \equiv 2^2+4^3 \left( \text{mod } 7 \right)$

$6 \equiv 0^2+6^3 \left( \text{mod } 7 \right)$

It is trivial to see that $0$, $1$, $2$, and $\left( n-1 \right)$ can be represented as the sum of a square and a cube. This is accomplished when considering the combinations like $\left( 0^2+0^3 \right)$, $\left( 0^2+1^3 \right)$, $\left( 1^2+0^3 \right)$, etc.

How can I prove this?

EDIT:

There is a slightly stronger version which states that all elements can be written as $a^2+b^3$, where $a \ne b$. I have verified this stronger version up to $n=1000$.