$x^2-x+1$ has a root $\!\bmod p\,$ for infinitely many primes $p$ Prove that the equation $$x^2 - x + 1 = p(x+y)$$ has integral solutions for infinitely many primes $p$.
First, we prove that there is a solution for at least one prime, $p$. Now, $x(x-1) + 1$ is always odd so there is no solution for $p=2$. We prove there is a solution for $p=3$. If $p=3$, $y = (x-2)^2/3-1$. We get integral solutions whenever we get $x = 3m +2$, where $m$ is any integer.$\\$
We provide a proof by contradiction which is   similar to Euclid's proof of there being infinitely many primes. \Let us assume that it is is true for only finitely many primes, and name the largest prime for which the equation is true $P$.\\We set $$x = 2\cdot3\cdot5\dots P$$ $x$ is the product of all primes upto $P.\\$
Then, the term $x(x-1) + 1$ is either prime or composite. If it is prime, then we set $p = x(x-1) + 1, y = 1 - x$ and get a solution. If it is composite, we write $x(x-1) + 1 = p\times q$, where $p$ is any prime factor of $x(x-1)+1$, and $q$ is an integer, $q = (x(x-1)+1)/p$. Now, $x(x-1) + 1$ is not divisible by any prime upto $P$ since it leaves a remainder of $1$ with all of them. So, $p > P$. We set $y=q-x$ for a solution.$\\$In either case, we get a solution for a prime $p > P$, which means there's no largest prime for which this equation has solutions. This contradicts the assumption that there are solutions for only finitely many primes.
I feel like I'm missing some step. Is this correct ?


Assumptions

$$0\notin\mathbb N\\n\in\mathbb N$$

$x^2-x+1$ is composite for infinitely many $x\in\mathbb N$

Proof: 
$$
x=3n+2\implies x^2-x+1=\color{red}{3}\times(3n^2+3n+1)
$$

There are infinitely many primes that can divide natural numbers of form $x^2-x+1$.

Proof: Suppose our claim is not true and in fact $q_1=\color{red}{3},q_2,\cdots,q_k$ are the only primes that can divide a natural number of form $x^2-x+1$. Borrowing Euclid's argument let $$n=\prod_{j=1}^kq_j\\ x=3n+2$$
Therefore, $x^2-x+1$ is not a prime and
$$
x^2-x+1\equiv3\not\equiv0\mod q_j\;\forall j\neq1\\
\implies q_j\not| x^2-x+1\;\forall j\neq1
$$
Therefore our assumption and the fundamental theorem of arithmatic indicate, for some $q\in\mathbb N$
$$\begin{align}
&x^2-x-1=\color{red}{3}^q \\
\implies& 3n^2+3n+1=3^{q-1}\\
\implies& 1\equiv 3^{q-1}\mod 3\\
\implies& q=1\\
\implies& x^2-x+1=3
\end{align}
$$
which contradicts
$$
x^2-x+1>(x-1)^2=(3n+1)^2\ge(3q_1+1)^2=100
$$

Solution to the current problem

Suppose $p_1,p_2,\cdots,p_m$ are the only primes for which integral solution exists. Using the above lemma we can find a prime $p$ not in this list such that $p| x^2-x+1$ for some $x\in\mathbb N$. Set 
$$
y={x^2-x+1\over p}-x
$$
which is definitely an integer (but not necessarily a natural number). Thus we have found a new prime $p$ for which an integral solution to the given equation exists. Therefore, our assumption that the given equation has integral solutions for finitely many choices of $p$ can't be true.   
