Find all integers x for which $ |x^3 + 6x^2 + 2x − 6| $ is prime.

Question

I think that this means that $|x^3 + 6x^2 + 2^x − 6| \equiv 0 \bmod p$. Ignoring the absolute value first, I get $ x(x^2+2)+6(x^2−1) \equiv 0 \mod p \Rightarrow x(x^2+2) \equiv -6(x^2−1) \bmod p$. But I'm not sure how to proceed.

Answer 1

Hint: Let $f(x)=x^3+6x^2+2x-6$. Then we have $$3\mid f(x)$$ for all $x\in \Bbb{Z}$.

Answer 2

We have that

$$x^3 + 6x^2 + 2x − 6\equiv x^3-x \pmod 3$$

and

$$x^3-x=(x-1)x(x+1)$$

Refer also to the related

• Show that if $a$ is an integer, then 3 divides $a^3 - a $

Answer 3

Following your thoughts of grouping the terms into:

$$x(x^2+2)+6(x^2−1)$$

consider the following:

• The second term, $6(x^2-1)$ is always divisible by $6$ and consequently is divisible by $3$.
• For the first term $x(x^2+2)$, it is also divisible by 3. Why? $x$ is either $0,-1,1 \pmod 3$. Hence, $x^2 \equiv 0,1 \pmod 3$. If $x^2 \equiv 0 \pmod 3$, then $3|x^2 \implies 3|x$. On the other hand, if $x^2 \equiv 1 \pmod 3$, then $x^2 + 2 \equiv 0 \pmod 3$.

Now, what can you conclude?