Ring with three idempotents satisfying $\alpha+ \beta+ \gamma=0$. Prove that $\alpha=\beta= \gamma$. Problem

Let be $(R,+,\cdot)$ a ring.
Moreover, $(R,+,\cdot)$ has the property that if $$2022x=0$$ then $$x=0.$$
Let be $\alpha$, $\beta$, $\gamma$ three idempotent elements such that
$$\alpha+ \beta+ \gamma=0.$$
Prove that $\alpha=\beta= \gamma$.

My attempt:

We know that $$x^2=x, \forall x\in \{ \alpha, \beta, \gamma \}$$
I then proved via noetherian induction that:
$$x^n=x, \forall x\in \{ \alpha, \beta, \gamma \}, \forall n \text{ positive integer}$$
After that I observed that
$$\alpha ^n+ \beta^n+ \gamma^n=\alpha+ \beta+ \gamma=0$$
I got stuck there.
I didn't use the property that if $$2022x=0$$ then $$x=0$$
We have also got that $$\sum_{sym} \alpha \beta=0$$
I have not the faintest idea how to use it.
Any help would be appreciated.
How's it going?
I have the article only in a pdf file.
I could not upload the pdf file, so I attached the first page of the article.
Can it be shown that $R$ is an Algebra by using if $2022x=0$ then $x=0$?
What is an algebra? May $R$ be an algebra?  May we consider a homomorphism from the noncommutative polynomial ring $C[x_1, x_2, x_3]$ into $R$ defined by $\phi (x_1) = \alpha$ and the same for $x_2, x_3$?

 A: Here is a proof that works for non-commutative rings. As in Compacto's proof, you don't really need $2002r=0 \Rightarrow r=0$, you only need $2r=0 \Rightarrow r=0$ and $3r=0 \Rightarrow r=0$.
I'll write $a,b,c$ for $\alpha,\beta,\gamma$.
From $c=-(a+b)$ we deduce
$$c^2=(a+b)(a+b)=a^2+ab+ba+b^2=a+b+ab+ba\tag{1}$$,
and hence $-(a+b)=a+b+ab+ba$, or
$$
2(a+b)+ab+ba=0 \tag{2}
$$
Mutiplying by $a$ on the left in (2), we deduce $2(a^2+ab)+a^2b+aba=0$, or
$$
2a+3ab+aba=0 \tag{3}
$$
Mutiplying by $a$ on the right in (2), we deduce $2(a^2+ba)+aba+ba^2=0$, or
$$
2a+3ba+aba=0 \tag{4}
$$
Comparing (3) with (4), we see that $3(ba-ab)=0$, so $2022(ba-ab)=0$ and hence
$$
ba=ab \tag{5}
$$
Injecting this last identity into (2), we see that $2(a+b+ab)=0$, so $2022(a+b+ab)=0$ and hence
$$
ab=-(a+b) \tag{6}
$$
Mutiplying by $a$ on the left in (6), we deduce $a^2b=-(a^2+ab)$, or $ab=-(a+ab)$, or $2ab+a=0$. Combining this with (6), we obtain $a-2a-2b=0$, or
$$
a=-2b \tag {7}
$$
It follows that $a^2=4b^2$, or $a=4b$. From (7) we deduce $4b=a=-2b$, so $6b=0$, so $2022b=0$ and hence $b=0$. This finishes the proof.
A: This answer builds up on previous work by @shangq_tou and @winic92046.
First one observes that, for all $n\in \mathbb{N}$, we have that
$$\alpha^n + \beta^n + \gamma^n = 0$$
This can be used to show that all elementary symmetric polynomials:
$$a_1 =\alpha + \beta + \gamma,$$
$$a_2 = \alpha\beta+  \alpha \gamma + \beta\gamma,$$
$$a_3 =\alpha\beta\gamma$$
are equal to zero. The first one is zero by hypothesis. In order to prove that the other two are zero, we use the Newton's identities:
$$0 = \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta+ \gamma)^2 - 2(\alpha\beta+ \alpha\gamma + \beta \gamma) = - 2(\alpha\beta+ \alpha\gamma + \beta \gamma) $$
This implies that
$$2(\alpha\beta+ \alpha\gamma + \beta \gamma) = 0$$
Multiplying this equation by $1011$, we get that
$$2022(\alpha\beta+ \alpha\gamma + \beta \gamma) = 0$$
Which implies that $\alpha\beta+ \alpha\gamma + \beta \gamma = 0$.
Another one of Newton's identities says that:
$$0 = \alpha^3 + \beta^3 + \gamma^3 = (\alpha+  \beta+ \gamma)^3 - 3(\alpha+\beta+ \gamma)(\alpha\beta + \beta\gamma + \alpha\gamma) + 3\alpha\beta\gamma$$
And thus,
$$3\alpha\beta\gamma = 0$$
Multiplying this by $2022:3 = 674$ gives:
$$2022\alpha\beta\gamma = 0$$
And it follows that $\alpha\beta\gamma = 0$.
Now, consider the following polynomial:
$$p(x) = (x-\alpha)(x-\beta)(x-\gamma) = x^3 - a_1x^2 +a_2x -a_3 = x^3$$
(We have already shown that $a_1, a_2, a_3$ are all equal to zero).
Since obviously $\alpha, \beta, \gamma$ are roots of this polynomial, we get that
$$\alpha^3 = \beta^3 = \gamma^3 = 0$$
And, because all of them are idempotent, this implies that $\alpha = \beta = \gamma = 0$.
The hypothesis about $2022$ is useful to show that, for all $r\in R$, $2r = 0$ or $3r = 0$ implies $r=0$ (because $2$ and $3$ divide $2022$).
