South African Mathematical Olympiad Third Round

(a) Suppose that
$$ax + \gcd(a, x) + \text{lcm}(a, x) = ay + \gcd(a, y) + \text{lcm}(a, y)$$
for certain positive integers $a$, $x$, $y$. It follows that
$$\gcd(a, ax + \gcd(a, x) + \text{lcm}(a, x)) = \gcd(a, ay + \gcd(a, y) + \text{lcm}(a, y)).$$
Since $a$ divides both $ax$ and $\text{lcm}(a, x)$, we have
$$\gcd(a, ax + \gcd(a, x) + \text{lcm}(a, x)) = \gcd(a, \gcd(a, x)) = \gcd(a, x)$$
and likewise
$$\gcd(a, ay + \gcd(a, y) + \text{lcm}(a, y)) = \gcd(a, \gcd(a, y)) = \gcd(a, y).$$
Therefore, we must have $\gcd(a, x) = \gcd(a, y) = d$ for some positive integer $d$. Since $\text{lcm}(a, x) = ax/\gcd(a, x)$ and $\text{lcm}(a, y) = ay/\gcd(a, y)$, this gives us
$$ax + d + \frac{ax}{d} = ay + d + \frac{ay}{d},$$
so
$$ax \left(1 + \frac{1}{d}\right) = ay \left(1 + \frac{1}{d}\right),$$
which implies $x = y$. This proves the first statement.

(b) Suppose that $\text{ax+ (a,x)+ (a,x) = 2014}$. Note that the left hand side is divisible by $\gcd(a,x)$, so $\gcd(a,x)$ has to be a divisor of 2014, i.e., one of $1, 2, 19, 38, 53, 106, 1007, 2014$. On the other hand,
$$\text{( (a, x) + 1)( (a, x) + 1) = ax + (a, x) + (a, x) + 1 = 2015,}$$
so $\gcd(a, x) + 1$ has to divide 2015. Since 2, 3, 20, 39, 54, 107, 1008 are all not divisors of 2015, this leaves us with $\gcd(a, x) = 2014$. But then $\text{ax + (a, x) = 0}$, which is clearly impossible since the left hand side is positive.