Irish Mathematical Olympiad

Let $f(a) = \left(\frac{a+1}{2}\right)$, then $f(1) = 1$ and $f(35) = 630$. Because $f(a) > 630$ for $a > 35$, any solution $(a, b)$ satisfies $1 \le a, b \le 35$ and $f(a)$ and $f(b)$ are divisors of $630 = 2 \cdot 3^2 \cdot 5 \cdot 7$. If $a$ is odd, $a$ is a factor of $f(a)$. If $a$ is even, $a+1$ is odd and a factor of $f(a)$. The odd positive divisors of 360 not exceeding 36 are 1, 3, 5, 7, 9, 15, 21 and 35. If $a$ is among these numbers and $f(a)$ is a factor of 630, $(a+1)/2$ is a factor of 630 as well. This is the case for the numbers 1, 2, 5, 9, 35. If $x = a + 1$ is among these numbers and $f(a)$ is a factor of 630, $a/2 = (x-1)/2$ will be a factor of 630. This is the case for $a$ being one of 2, 4, 6, 14, 20. The values $f(a)$ for these numbers are collected in the following table.

Because none of the quotients $630/10 = 63$, $630/15 = 42$, $630/21 = 30$ and $630/45 = 14$ occur in this table, no solution can have $a$ or $b$ equal to 4, 5, 6 or 9. The other values for $a$ lead to the following complete list of solutions:
$$(a, b) = (3, 14), (14, 3), (2, 20), (20, 2), (1, 35), (35, 1).$$