polynomials - How to solve Diophantine equations in $F_{p}$?

You want to know if the sum of $p-1$ squares can be equal to 0 mod $p$. I'll assume that you don't want to allow the trivial (all-zeroes) solution.

If $k$ is a quadratic residue mod $p$, not equal to $1$, then this is simple; take $x_1$ such that $x_1^2 = k$, take $x_2 = ldots = x_{p-k+1} = 1$, and take $x_{p-k+2} = ldots = x_{p-1} = 0$.

So the equation $x_1^2 + cdots + x_{p-1}^2 = 0$ has solutions mod $p$ as long as there exists a quadratic residue mod $p$ which is not equal to $1$. The number of quadratic residues mod $p$ is $phi(p)/2$, where $phi$ is Euler's totient function; if $phi(p)/2 ge 2$, or $phi(p) ge 4$, then there is at least one non-$1$ quadratic residue mod p. Now for a prime, $phi(p) = p-1$, so that means your equation has solutions when $p-1 ge 4$, i. e. when $p ge 5$. We can check by brute force that $x_1^2 = 0 mod 2$ and $x_1^2 + x_2^2 = 0 mod 3$ have only the trivial solutions. So the equation $x_1^2 + cdots + x_{p-1}^2 = 0 mod p$ has nontrivial solutions for all primes $p ge 5$.

(Basically, this is a more explicit version of the second paragraph of Bjorn Poonen's answer.)