Let $$\lambda \in \mathbb{R}$$ and let the equation $$E$$ be $$|x|^2 - 2|x| + |\lambda - 3| = 0$$. Then the largest element in the set $$S = \{x + \lambda : x \text{ is an integer solution of } E\}$$ is ______

We need to find the largest element in $$S = \{x + \lambda : x \text{ is an integer solution of } E\}$$ where $$E: |x|^2 - 2|x| + |\lambda - 3| = 0$$.

We start by letting $$u = |x| \geq 0$$, which transforms the equation into:

$$u^2 - 2u + |\lambda - 3| = 0$$

This quadratic in $$u$$ yields $$u = \frac{2 \pm \sqrt{4 - 4|\lambda - 3|}}{2} = 1 \pm \sqrt{1 - |\lambda - 3|}$$.

For real solutions in $$u$$ we require $$|\lambda - 3| \leq 1$$, i.e., $$2 \leq \lambda \leq 4$$.

We also note that $$u \geq 0$$, and because $$\sqrt{1 - |\lambda - 3|} \leq 1$$, both expressions $$1 \pm \sqrt{1 - |\lambda - 3|}$$ are non-negative.

Since $$x$$ must be an integer, $$u = |x|$$ must be a non-negative integer, so we need $$u = 1 \pm \sqrt{1 - |\lambda - 3|}$$ to be an integer.

Introducing $$k = \sqrt{1 - |\lambda - 3|}$$ gives $$|\lambda - 3| = 1 - k^2$$ and hence $$u = 1 + k$$ or $$u = 1 - k$$.

For $$u$$ to be an integer, $$k$$ must also be an integer satisfying $$0 \leq k \leq 1$$, which leaves $$k = 0$$ or $$k = 1$$.

Case 1: If $$k = 0$$ then $$|\lambda - 3| = 1$$, yielding $$\lambda = 2$$ or $$\lambda = 4$$ and $$u = 1$$ so that $$x = \pm 1$$.

For $$\lambda = 4$$ this gives $$x + \lambda \in \{-1 + 4,\,1 + 4\} = \{3,\,5\}$$, and for $$\lambda = 2$$ it gives $$x + \lambda \in \{-1 + 2,\,1 + 2\} = \{1,\,3\}$$.

Case 2: If $$k = 1$$ then $$|\lambda - 3| = 0$$ so $$\lambda = 3$$ and $$u = 0$$ or $$u = 2$$, giving $$x \in \{0,\,\pm 2\}$$.

This leads to $$x + \lambda \in \{0 + 3,\,2 + 3,\,-2 + 3\} = \{3,\,5,\,1\}$$ for $$\lambda = 3$$.

Across all valid values of $$\lambda$$, the largest element in $$S$$ is $$5$$, which is the required answer.