# 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.

• I think that, if you just worked a few values explicitly, the pattern would be obvious.
– lulu
Jul 5 at 20:49
• Okay. but how this could be proven without pattern recognition and stating that it is obvious. For example how to prove that there do not exist any prime values of x such that f(x) is prime also using modular arithmetic?
– user797753
Jul 5 at 20:51
• Of course, the proof goes by modular arithmetic. But you need to spot the pattern before it becomes clear what you want to prove. It's a really, really simple pattern.
– lulu
Jul 5 at 20:52
• Please change the tag to elementary-number-theory Jul 5 at 21:08
• $\!\!\bmod 3\!:\ f(\color{#c00}{\pm1})\equiv 0\,$ so $\,f(n)\,$ is a multiple of $3$ when $n\ \rm\color{#c00}{isn't}$ (i.e. when $\,n\equiv \color{#c00}{\pm 1},\,$ e.g. all primes $\,n\neq 3)\ \$ Jul 5 at 21:22

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 $$3$$.

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.