Interpret the equation

The given equation is $|x|^2 - 2|x| + |\lambda-3| = 0.$

Since $|x|^2 = x^2$, let $y = |x|.$ Then $y \ge 0$, and the equation becomes $y^2 - 2y + |\lambda-3| = 0.$

Rewrite the quadratic

Complete the square: $y^2 - 2y + |\lambda-3| = (y-1)^2 + |\lambda-3| - 1.$

So the equation is $(|x|-1)^2 + |\lambda-3| - 1 = 0,$ or $(|x|-1)^2 = 1 - |\lambda-3|.$

Use the fact that $x$ is an integer solution

Since $x$ is an integer, $|x|$ is a non-negative integer, so $(|x|-1)^2$ can only be a non-negative integer.

Also, the right-hand side must be non-negative: $1 - |\lambda-3| \ge 0 \implies |\lambda-3| \le 1.$

Now for integer $x$, $(|x|-1)^2$ can be $0,1,4,\dots$ But since $(|x|-1)^2 = 1 - |\lambda-3| \le 1,$ the only possible values are $(|x|-1)^2 = 0 \quad \text{or} \quad 1.$

Case 1: $(|x|-1)^2 = 0$

Then $|x|=1 \implies x=\pm 1.$ Also, $1 - |\lambda-3| = 0 \implies |\lambda-3| = 1.$ Hence $\lambda = 2 \quad \text{or} \quad 4.$

Now compute $x+\lambda$:

• If $\lambda=2$, then $x+\lambda = -1+2=1$ or $1+2=3$.
• If $\lambda=4$, then $x+\lambda = -1+4=3$ or $1+4=5$.

Largest value from this case is $5$.

Case 2: $(|x|-1)^2 = 1$

Then $|x|-1 = \pm 1.$ So $|x|=0 \quad \text{or} \quad 2,$ giving $x=0, \pm 2.$

Also, $1 - |\lambda-3| = 1 \implies |\lambda-3| = 0 \implies \lambda=3.$

Now compute $x+\lambda$:

• For $x=0$, $x+\lambda = 3$.
• For $x=2$, $x+\lambda = 5$.
• For $x=-2$, $x+\lambda = 1$.

Largest value from this case is also $5$.

Conclusion

Collecting all possible values of $x+\lambda$, the maximum is $\boxed{5}.$

Comparison with stored answer

The derived answer is $5$, which matches the stored correct answer.