Ytukyg

$$1^2+1+1=3$$
$$2^2+2+1=7$$
$$8^2+8+1=73$$
$$10^2+10+1=111=3cdot37$$
There is no divisor which is square number.
Is it just coincidence? Or can be proved?

*I'm not english user, so my grammer might be wrong

What about $x=653$, where
$$x^2+x+1=427063=7cdot (13cdot 19)^2,?$$
How did I find this $x$? I first suppose that $x^2+x+1=ay^2$ for some positive integers $a$ and $y$. This means
$$(2x+1)^2-a(2y)^2=-3,.$$
Let $u:=2x+1$ and $v:=2y$. Then, we are to solve the Pell-type equation $$u^2-av^2=-3,,$$ where $u,vinmathbb{Z}_{>0}$. In particular, the case $x=2$ yields $a=7$ and $y=1$. Therefore, I attempt to solve
$$u^2-7v^2=-3,,tag{*}$$
where a minimal solution is $(u,v)=(5,2)$. Since $u^2-7v^2=1$ has the minimal solution $(u,v)=(8,3)$, we obtain an infinite family of solutions $(u,v)$ of (*):
$$u+vsqrt{7}=(5+2sqrt{7}),(8+3sqrt{7})^k,,tag{#}$$
where $k$ is a positive integer. We want $u$ to be odd, so $k=1$ does not work. With $k=2$, we get $(u,v)=(1307,494)$, so $$x=frac{1307-1}{2}=653text{ and }y=frac{494}{2}=13cdot 19$$
form a counterexample. (Using $k=-2$, we get lhf's counterexample $x=18$. Indeed, with even values of $k$ in $(#)$, we obtain infinitely many counterexamples.)

The square of any prime $pequiv1pmod3$ appears as a factor of $x^2+x+1$ for some choice of $x$.

This is seen as follows.

The multiplicative group $Bbb{Z}_{p^2}^*$ of coprime residue classes modulo $p^2$ is known to be cyclic of order $p(p-1)$. It follows that there is an element of order three in that group. Let the residue class of $x$ be one such. Because $p>3$ the order of $x$ modulo $p$ is also equal to three. Implying that $x-1$ is not a multiple of $p$. But $$x^3-1=(x-1)(x^2+x+1)equiv1-1=0pmod{p^2}$$
by construction,
so we can conclude that $$x^2+x+1equiv0pmod{p^2}.$$

A non-deterministic way of finding such an $x$ is to take a random integer $a$, and calculate the remainder $x$ of $a^{p(p-1)/3}$ modulo $p^2$. If the result is $xneq1$, then we have found the required element of order three.

For example, with $p=31$, $a=3$ we find that
$$
3^{31(31-1)/3}=3^{310}equiv521pmod{31^2}.
$$
And with $x=521$ we get
$$
521^2+521+1=271963=31^2cdot283
$$
as promised.

On the other hand no prime $pequiv-1pmod3$ will appear as a factor of $x^2+x+1$ for any integer $x>1$. This is because the factorization $(x^3-1)=(x-1)(x^2+x+1)equiv0pmod p$ implies that $x$ has order $1$ or $3$ in the group $Bbb{Z}_p^*$. In the former case $xequiv1pmod3$ and therefore
$x^2+x+1equiv1+1+1=3notequiv0pmod p$. In the latter case Lagrange's theorem from elementary group theory tells us that $3$ must be a factor of the order of the group $G=Bbb{Z}_p^*$. As $|G|=p-1$ we can conclude that $pequiv1pmod3$.

By more or less the same argument we can show that for all $pequiv1pmod3$ the number $x^2+x+1$ can be made divisible by any power $p^k$. This time the group has order $p^{k-1}(p-1)$. As an example consider $p=7,k=5$. The above method produces $$3^{7^4(7-1)/3}equiv1353pmod{7^5}.$$ And, predictably,
$$1353^2+1353+1=7^5cdot109.$$

No, for $x=18$ we get $x^2+x+1=343=7^3$.

Here are the first few counterexamples:

$$
begin{array}{rrl}
x & x^2+x+1 & text{factorization}\
18 & 343 & 7^3 \
22 & 507 & 3 cdot 13^2 \
30 & 931 & 7^2 cdot 19 \
67 & 4557 & 3 cdot 7^2 cdot 31 \
68 & 4693 & 13 cdot 19^2 \
79 & 6321 & 3 cdot 7^2 cdot 43 \
116 & 13573 & 7^2 cdot 277 \
128 & 16513 & 7^2 cdot 337 \
146 & 21463 & 13^2 cdot 127 \
165 & 27391 & 7^2 cdot 13 cdot 43 \
177 & 31507 & 7^2 cdot 643 \
191 & 36673 & 7 cdot 13^2 cdot 31 \
214 & 46011 & 3 cdot 7^2 cdot 313 \
end{array}
$$