Sum of nonzero squares modulo p It is easy to prove that for prime $p$ every element of $\mathbb{Z} / p \mathbb{Z}$ can be written as the sum of two squares. An elementary proof is given here: Sum of two squares modulo p
How can we show that, provided further $p \geq 7$, any nonzero element of $\mathbb{Z}/ p \mathbb{Z}$ is the sum of two nonzero squares? I don't see how we could extend the counting argument used in the linked post to this case. Thanks

Let's prove that $1$ is the sum of two nonzero squares.
Think about the geometry of the plane. Any non-vertical line through
$(1,0)$ will meet the unit circle in another point. This argument
works over $k=\Bbb Z/p\Bbb Z$ too. The "non-vertical" lines through
$(1,0)$ are $y=m(x-1)$, and each meets the unit circle in the
further point $((m^2-1)/(m^2+1),2m/(m^2+1))$, as long as $m^2\ne-1$.
These coordinates are nonzero iff $m\notin\{0,1,-1\}$. There are
at most five "bad" values of $m$, so if $p>5$, then $1$ is the sum of
two non-zero squares in $k$.
By scaling, each non-zero square is the sum of
two non-zero squares in $k$. Of course, if $b$ is not a square, it is still the sum of two squares, and they are both nonzero.
