Find all primes x such that f(x) is also a prime number. There is a function
\begin{align*}
f(x)=x^3 + x^2 +11x +2 \\
\end{align*}
Find all prime $x$ such that $f(x)$ is also a prime number.
I found that this is satisfied with an x value of 3 then the function is equal to 71, so both are primes, but I am unsure how to find other values or prove that there are no other existing solutions. I tried to use modular arithmetic, but I did not go so far.
 A: Modular arithmetic is indeed the solution here. But first, we need to recall the following fact about prime numbers:

All primes $>3$ are $\pm1\pmod6$.

It's not too hard to see why. Numbers that are $0,2,4\pmod6$ are even while $3\pmod6$ is a multiple of $3$, leaving $1$ and $5$.
Now basically if $p$ is a prime that is $1\pmod6$, then $f(p)\equiv f(1)\equiv3\pmod6$, which cannot possibly be a prime (also $f(p)>3$). Similarly, if $p$ is $5\pmod6$ then $f(p)$ is also $3\pmod6$, which cannot be prime as well, so the only possible $x$ is $3$.
A: Mod 3:  $\mod 3$ we know by Fermat's little theorem $x^3 \equiv x\pmod 2$ and so $f(x) = x^3 + x^2 + 11x + 2 \equiv x + x^2 -x -1 \equiv x^2 - 1\equiv (x+1)(x-1)\pmod 3$.
What does this tell us?
It tells us that if $x$ is divisible by $3$ then $f(x)\equiv -1 \pmod 3$.  But if $x$ is not divisible by $3$ (i.e. if $x \equiv \pm 1 \pmod 3$) then $f(x)$ is divisible by $3$ (because $x\mp 1\equiv 0 \pmod 3$).
In other words $f(x)$ is divisible by $3$ if and only if $x$ is not divisible by three.
So for $x$ and $f(x)$ to both be prime we must have exactly one of them divisible by $3$.  So either $x = 3$ (the only prime divisible by $3$) and $f(x) = 71$ which just happens to be prime.  Or we have $f(x)=3$ and .... $x = ????$ well, if we assume $x\in \mathbb N$ then $x \ge 1$ and $f(x) \ge 1+1+11+2 > 3$ so that is not possible.
So $x = 3$ is the only case where both are prime.
