Positive integers $$a$$, $$b$$ are such that both are relatively prime and less than or equal to 2008, $$a^2 + b^2$$ is a perfect square and that $$b$$ has the same digits as $$a$$ in the reverse order. The number of such ordered pairs $$(a, b)$$ is _________

Checking every $$a \le 2008$$ together with its digit reversal $$b$$, no coprime pair gives a perfect square; for example $$12^2 + 21^2 = 585$$, $$13^2 + 31^2 = 1130$$ and $$56^2 + 65^2 = 7361$$ all fall between consecutive squares. In particular, a coprime pair whose squares add to a square must contain a multiple of 4, which rules out most pairs at once. So the number of such pairs is 0.