Problem 1691

### Part (a)

1. Notation and Setup:
Let $x \in A$ be a non-invertible element. According to the problem, there exists an integer $m > 1$ such that:
$$x = x^2 + x^3 + \ldots + x^{2^m}$$
This can be rewritten as:
$$x(1 - x - x^2 - \ldots - x^{2^m - 1}) = 0$$

2. **Case 1: $2_A$ is invertible:**
Suppose $2_A$ is invertible. Then, we can multiply both sides of the equation by $2_A^{-1}$:
$$2x - 1 - x^{2^m} = 0$$
This implies:
$$2x - 1 = x^{2^m}$$
Since $2x - 1$ is non-invertible (otherwise $x$ would be invertible, contradicting our assumption), we can substitute $2x - 1$ into the original equation:
$$2 = x \cdot Q(x)$$
for some polynomial $Q(x)$ with integer coefficients. Since $2$ commutes with $x$, we have:
$$x \cdot Q(x) \cdot 2^{-1} = Q(x) \cdot 2^{-1} \cdot x = 1$$
This implies $x$ is invertible, which is a contradiction. Therefore, $2_A$ cannot be invertible.

3. **Case 2: $2_A$ is non-invertible and not equal to 0:**
Suppose $2_A$ is non-invertible and not equal to 0. Plugging $2$ into the equation, we get:
$$2(2^{2^m} - 3) = 0$$
Let $N$ be the least integer such that $N \cdot 1 = 0$. Then $N \mid 2(2^{2^m} - 3)$.

4. **Analyzing $N$:**
- If $N = 2$, then $2 = 0$, which is a contradiction.
- If $N$ is odd, then $2$ would be invertible, which is a contradiction.
- Let $N = 2k$, where $k > 1$ is odd. Then $k + k = 2k = 0$, and $k \cdot (k - 1) = 0$ because $k - 1$ is even. Hence, $k - 1$ is non-invertible, and there exists some $n \in \mathbb{N}^*$ such that:
$$(k - 1) = (k - 1)^2 + (k - 1)^3 + \ldots + (k - 1)^{2^n}$$
This implies:
$$k \cdot \ell = 2$$
for some $\ell \in \mathbb{Z}$. If $\ell$ is even, we get $2 = 0$, a contradiction. If $\ell$ is odd, it follows $k = 2$, a contradiction.

5. Conclusion:
The only remaining case is $2 = 0$, i.e., $1 + 1 = 0$, which means:
$$x + x = x \cdot (1 + 1) = 0 \quad \forall x \in A$$

### Part (b)

1. Given Condition:
For every non-invertible $x \in A$, we have:
$$x \cdot (x^{2^m} - 1) = 0$$
This implies:
$$x = x^{2^m + 1}$$

2. Simplification:
We can rewrite the above as:
$$x \cdot (x - 1) \cdot (x \cdot Q(x) + 1) = 0$$
where $Q(x)$ is a polynomial with integer coefficients. It is easy to see that $x^2 - x$ cannot be invertible.

3. Power Analysis:
Raising to the power $2^m$, we have:
$$(x^2 - x)^{2^m} = x^{2^{m+1}} - x^{2^m} = 0$$
(using $x = x^{2^m + 1}$).

4. Combining Results:
Since $x^2 - x$ is non-invertible, there exists $n$ such that:
$$(x^2 - x)^{2^n + 1} = x^2 - x$$
Combining with the previous result, we get:
$$x^2 - x = 0$$

$\blacksquare$