First of all, let's not forget the trivial ring, the one with only one element (defined by $0=1$). Now let's assume that $0 \neq 1.$ We know that $-1=(-1)^3 \in \{0,1\}$ and $-1 \neq 0,$ so $-1=1,$ i.e. $2=0.$ We'll now prove the implication $x^3=0 \implies (x+1)^3=1.$ Assuming for contrary that there is $a$ such that $a^3=0$ and $(a+1)^3=0$, it would follow that $a^2+a+1=0.$ Multiplying the last relation by $a^2$ (to the left or to the right; it doesn't matter because of $a^2$ commutes with any element in that relation), we get $a^2=0.$ Hence $a+1=0,$ so $a=1,$ but this contradicts the fact that $a^3=0.$ Now that we know the implication is true, let's find all the elements which satisfy $x^3=0.$ If $x$ is such an element, than $(x+1)^3=1,$ so, after reducing, $x^2+x=0.$ Multiply this by $x$ to get $x^2=0$ and use this in the same relation to get $x=0.$ We proved that $0$ is the only solution for $x^3=0.$ Therefore, $x^3=1,\forall x \in A \setminus \{0\}.$ This implies that every nonzero element is invertible, hence $A$ is a division ring. Hence, the equation $x^3=1$ can have at most three distinct solutions, so $|A| \le 4.$ Clearly $\mathbb{F}_2$ is a solution, $\mathbb{F}_3$ is not a solution and $\mathbb{F}_4$ is a solution because $\mathbb{F}_4^{\times}$ is a multiplicative group of order $3,$ so $x^3=1,\forall x \in \mathbb{F}_4^{\times}.$ Edit (April 2024): I noticed there is a flaw in my solution, namely the part where I claim that the equation $x^3=1$ can have at most three distinct solutions. This is indeed true in a commutative division ring (aka field), but without commutativity we don't really have factorization of polynomials and so we can't deduce that. Correction: I inspected all the division rings with $|A| \le 4.$ It would then be enough to prove that $A$ cannot have more than $4$ elements. Let's say $A$ contains an element $a \not\in \{0,1\}.$ Then $a+1 \in A \setminus \{0,1,a\}.$ Now let's assume there is some $b \in A \setminus \{0,1,a,a+1\}.$ For any $x \not\in \{0,1\}$ we have $(x+1)^3=1,$ and $x^3=1,$ so $x^2+x+1=0.$ Since $a+b \not\in \{0,1\},$ we get $(a+b)^2+a+b+1=0,$ so $(a^2+a)+(b^2+b)+ab+ba+1=0,$ i.e. $ab+ba=1.$ $(*)$ This implies $a^2b+aba=aba+ba^2,$ so $a^2b=ba^2,$ or equivalently $(a+1)b=b(a+1),$ so $ab=ba,$ contradicting $(*).$

Romania National MO 2022 XII/2 wrote: Determine all rings $(A,+,\cdot)$ such that $x^3\in\{0,1\}$ for any $x\in A.$ Mihai Opincariu The answers are the trivial ring, $\mathbb{F}_2$ and $\mathbb{F}_4$. It is immediate to check that these work. We'll show that there are no other ring that works. Assume $|A| > 2$, since if $|A| \le 2$, $A$ is either $\{ 0, 1 \} \cong \mathbb{F}_2$, or the trivial ring. Claim 01. $A$ has no zero divisor. Proof. We first see that $\text{char}(A) = 2$ since $(-1)^3 \in \{ 0, 1 \}$. However, as $A$ is not trivial, we conclude $(-1)^3 = 1$ and hence $2 = 0$, proving $\text{char}(A) = 2$. Now, take any $x \in A$, note that as $1 \in A$, we also have $x + 1 \in A$ and hence $x^3 \in \{ 0, 1 \}$ and $x^3 + x^2 + x + 1 = (x + 1)^3 \in \{ 0, 1 \}$ as well, which gives us $x^2 + x \in \{0, 1 \}$ for any $x \in A$. Now, take any $a \in A$ such that $a^3 = 0$. As $a^2 \in A$, we conclude that $a^2 = (a^2)^2 + a^2 \in \{ 0, 1 \}$. Assuming $a^2 = 1$, then we have $0 = a^3 = a$, a contradiction as this forces $0 = 1$. Thus, we conclude that $a^2 = 0$. Therefore, $a = a^2 + a \in \{ 0, 1 \}$, which immediately implies $a = 0$. Therefore, $a^3 = 0 \implies a = 0$, i.e. $a \not= 0 \implies a^3 = 1$, and hence any nonzero element in $A$ is invertible. Now, take any $x, y \in A \setminus \{ 0, 1 \}$. We clearly have $x , x + 1 \in A \setminus \{ 0, 1 \}$ and therefore $x^3 = (x + 1)^3 = 1$ which gives us $x^2 + x = 1$. Similarly, $y^2 + y = 1$. Now, assuming $x + y \in A \setminus \{ 0, 1 \}$, we then have $xy + yx = (x^2 + x) + (y^2 + y) + xy + yx = (x + y)^2 + x + y = 1$. Therefore, we conclude that for any $x,y \in A \setminus \{ 0, 1 \}$, either: $x = y$ $x = y + 1$ $xy + yx = 1$ Assume there exists $x,y \in A \setminus \{ 0, 1 \}$ such that $xy + yx = 1$, then we have \[ x^{-1}yx^{-1} + xy = x ( xy + yx) x^2 = x^3 = 1 = yx + xy \implies x^{-1} yx^{-1} = yx \implies xy = x(yx)x^2 = x(x^{-1}yx^{-1})x^2 = yx \]But this gives you $0 = 1$, a contradiction. Therefore any distinct elements $x,y \in A \setminus \{ 0, 1 \}$ must satisfy $x = y + 1$. This forces $|A| = 4$, and $A = \{ 0, 1, x , x + 1 \} \cong \mathbb{F}_4$

We claim the answer is $\mathbb{F}_2$ and $\mathbb{F}_4$, which both work. Since $2^3\in\{0,1\}$, it follows that $8=0$ or $7=0$. If $7=0$ then $3^3=-1\not\in\{0,1\}$, a contradiction. Hence $8=0$. Since $3^3\in\{0,1\}$, it follows that $27=0$ or $26=0$. Clearly the former is not possible. Hence $26=0$ so $2=0$. We claim that $R$ is a division ring. Fix $a\neq 0,1$. We have $(a+1)^3=a^3+3a^2+3a+1=a^2+a+1\in\{0,1\}$ so $a^2=a$ or $a^2+a+1=0$. If the former then $a^3=a\not\in\{0,1\}$, a contradiction. Hence $a^3-1=(a-1)(a^2+a+1)=0$ so $a^3=1$. Thus $a$ is a unit, as desired. Note that $a^{-1}=a^2=a+1$ for $a\neq 0,1$. If there exists $a,b\neq 0,1$ such that $a+b\neq 0,1$, we have \[ ab=1+(ab)^{-1}=1+b^{-1}a^{-1}=1+(b+1)(a+1)=ba+a+b \]and \[ a+b+ab+ba=a^2+1+ab+ba+b^2+1=(a+b)^2=a+b+1 \]so $a+b=1$, a contradiction. Hence $|R|\in\{2,4\}$. If $|R|=2$ then $R\cong\mathbb{F}_2$. If $|R|=4$ then it is easy to check that $R\cong\mathbb{F}_4$. $\square$