$a$ and $b$ are two positive integers both less than $2010$; $a\ne b$. Find the number of ordered pairs $(a,b)$ such that $a^2+b^2$ is divisible by $5$.

Let's examine the squares of integers modulo 5: $$\begin{array}{rl}{1}^2& \equiv 1(\mathrm{mod}5)\\ 2^2& \equiv 4(\mathrm{mod}5)\\ 3^2& \equiv 4(\mathrm{mod}5)\\ 4^2& \equiv 1(\mathrm{mod}5)\\ 5^2& \equiv 0(\mathrm{mod}5)\end{array}$$ The pattern $(1,4,4,1,0)$ repeats for all integers.

Identify conditions for $a^2+b^2$ to be divisible by 5:

Case 1: Both numbers square to 0 (mod 5)

Case 2: One number squares to 1 and the other to 4 (mod 5)

Since the question asks for ordered pairs, we need to double our count for both cases.

There are 402 numbers that square to 0 (mod 5) within the range 1 to 2010 because there are 402 complete blocks of 5 numbers, and in each block, the last number squares to 0 (mod 5).

For each of 1 and 4 (mod 5), there are 804 numbers within the range 1 to 2010 that square to these values. This is because there are 402 complete blocks of 5 numbers, and in each block, there are 2 numbers that square to 1 (mod 5) and 2 numbers that square to 4 (mod 5).

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