**Problem 1 of the British Mathematical Olympiad 2008**

Find all solutions in non-negative integers a, b to \sqrt{a} + \sqrt{b} = \sqrt{2009}.

**Solutions by Steve Dinh, a.k.a. Vo Duc Dien**

**Solution 1**

Squaring both sides, we have a + b + 2\sqrt{ab} = 2009; now square them again

i.e.,

Solving for a, we obtain a = b + 2009 ± \sqrt{2 × 4018b} = b + 2009 ± 14 \sqrt{41b}.

Now for a to be an integer, 41b has to be the square of an integer, or b = 41n^{2} where n is an integer.

Now a = 41n^{2} + 2009 ± 14 \times 41n = 41n^{2} + 2009 ± 574n.

Note that a or b can not exceed 2009 and must not be negative, we have the following solutions when n = 1, 2, 3, 4, 5, 6 and 7

(b, a) = (41, 41×36), (41×4, 41×25), (41×9, 41×16), (41×16, 41×9), (41×25, 41×4), (41×36, 41), and (41×49, 0)

and, since \sqrt{a} and \sqrt{b} are interchangeable, there is another series of solutions:

(a, b) = (41, 41×36), (41×4, 41×25), (41×9, 41×16), (41×16, 41×9), (41×25, 41×4), (41×36, 41), and (41×49, 0).

**Solution 2**

Let’s write \sqrt{a}+ \sqrt{b} = \sqrt{2009} as \sqrt{a}+ \sqrt{b} = 7 \sqrt{41}

\sqrt{a} takes on the values 0, \sqrt{41}, 2 \sqrt{41}, 3 \sqrt{41}, 4 \sqrt{41}, 5 \sqrt{41}, 6 \sqrt{41}, and 7 \sqrt{41} whereas \sqrt{b} takes on the corresponding values 7 \sqrt{41}, 6 \sqrt{41}, 5 \sqrt{41}, 4 \sqrt{41}, 3 \sqrt{41}, 2 \sqrt{41}, \sqrt{41} and 0. From there the same results as above are drawn.