Prove that there are infinitely many prime numbers $p$ such that $x_1^2+x_2^2+x_3^2 = p$ has no solutions.
So my attempt is the following. Let's look at residues modulo $8$. $x^2$ is either $4$ or $1$ mod $8$. So sum of $3$ squares can not give us residue for instance $7$. According to Dirichlet theorem $8n + 7$ contains infinitely many primes since $(8, 7) = 1$.
Is this a good solution?
The argument is a nice one. One can replace the appeal to Dirichlet's Theorem by an argument that uses facts about quadratic congruences that are proved in most introductory courses.
We will show there are infinitely many primes of the form $8k+7$. Let $q_1,q_2,\dots,q_n$ be primes of this form. We show there is a prime $p$ of this form that is different from all the $q_i$.
Let $N=(q_1q_2\cdots q_n)^2-2$. If $q$ is a prime divisor of $N$, then the congruence $x^2\equiv 2\pmod{q}$ has a solution $q_1q_2\cdots q_n$. Since $N$ is odd, by a standard result we must have $q\equiv \pm 1\pmod{8}$. But not all the prime divisors of $N$ can be of the form $8k+1$, else $N$ would be of that form. But it isn't.
Thus $N$ has a prime divisor $p$ of the form $8k+7$. It is clear that $p$ cannot be any of the $q_i$.