How to prove that $2x^3+6x+1=0$ has no integer solution?

Question

How can it be shown that $2x^3 + 6x + 1 = 0$ has no integer solutions?

I've used the intermediate value theorem to prove that there exists a real number $c$ between $f(-1)$ and $f(0)$ such that $f(c) = 0$, but I know that it's not enough since I need to prove that for all roots of $x$ (which I cannot assume to be only one). Based on Desmos Graphing Calculator, the function has only one roots. But how can I prove it?

Here is what the graph of the function looks like:

Answer 1

For an integer $x$, $2x^3+6x$ is always even, whereas $1$ is always odd. Thus, the sum can not be even. Especially it cant be $0$, which is an even number as well.

Answer 2

We have that for $x\in\mathbb Z$

• for $x\ge 0 \implies 2x^3+6x+1\ge 1$
• for $x< 0 \implies x(2x^2+6)+1\le -7$

Answer 3

Do you know Differential calculus?

We define

$f(x)=2x^3+6x+1$.

$f'(x)=6x^2+6>0$ $(x∈R)$

So $f(x)$ monotonically increases.

$f(0)>0$,$f(-1)<0$

Therefore $f(x)$ has one solution between $-1$ and $0$.