The equation is:

$$(x^{2} + 1)^{10} = 2x - x^{2} - 2$$

Solution:

First, check if $x = 2$ is a solution: $$\text{LHS} = (2^{2} + 1)^{10} = 5^{10},$$ $$\text{RHS} = 2(2) - 2^{2} - 2 = 4 - 4 - 2 = -2.$$ Clearly, $$5^{10} \neq -2.$$ So, $x = 2$ is not a solution.

Now consider the general RHS: $$2x - x^{2} - 2 = -\left( (x-1)^{2} + 1 \right).$$ Thus, $$\text{RHS} \leq -1 \quad \text{for all real } x.$$

Meanwhile, the LHS is: $$(x^{2} + 1)^{10} \geq 1 \quad \text{for all real } x.$$

Since the LHS is always positive (at least 1) and the RHS is always negative (at most $-1$), there can be no real solutions.

Answer: (b) No real solutions.