Square of Positive integers

Let $\begin{cases} N + 1537 & = x^2 \\ N - 474 & = y^2 \end{cases} \quad \Rightarrow x^2-y^2 = 2011 \quad \Rightarrow (x+y)(x-y) = 2011$

Since $2011$ is a prime,

$(x+y)(x-y) = (2011)(1)$

$\Rightarrow \begin{cases} x+y = 2011 \\ x-y = 1 \end{cases} \quad \Rightarrow 2x = 2012 \quad \Rightarrow x = 1006 \quad \Rightarrow y = 1005$

Therefore, $N + 1537 = 1006^2 \quad \Rightarrow N = 1012036 - 1537 = 1010499$

The last three digits of $N$ is $\boxed{499}$ .

The two squares differ by 1537 + 474 = 2011 a prime number. Hence the two integers are consecutive numbers (2011 +/- 1) /2. Smaller being 1005, its square contributes 25 to the answer along with given 474 to a sum total of 499.