Canadian Mathematical Olympiad

Solution:
Let $f(x) = x^2 + x$. The equation is $4 f(a) = f(b)$, i.e.,

$$4(a^2 + a) = b^2 + b.$$

Expanding:
$$4a^2 + 4a = b^2 + b$$
Bring all terms to one side:
$$4a^2 + 4a - b^2 - b = 0$$
$$4a^2 + 4a = b^2 + b$$

Let us try to solve for $b$ in terms of $a$:

$$b^2 + b - 4a^2 - 4a = 0$$
This is a quadratic in $b$:
$$b^2 + b - (4a^2 + 4a) = 0$$
The discriminant must be a perfect square for integer solutions:

The quadratic formula gives:
$$b = \frac{-1 \pm \sqrt{1 + 16a^2 + 16a}}{2}$$
So $1 + 16a^2 + 16a$ must be a perfect square.
Let $1 + 16a^2 + 16a = k^2$ for some integer $k$.

So:
$$k^2 - 16a^2 - 16a = 1$$
$$k^2 - 16(a^2 + a) = 1$$

Let $a$ be a positive integer. Then $a^2 + a$ is always even, so $16(a^2 + a)$ is divisible by $32$.

Let us try small values of $a$:

If $a = 1$: $1^2 + 1 = 2$, $16 \times 2 = 32$, $k^2 = 1 + 32 = 33$ (not a perfect square).

If $a = 2$: $2^2 + 2 = 6$, $16 \times 6 = 96$, $k^2 = 1 + 96 = 97$ (not a perfect square).

If $a = 3$: $3^2 + 3 = 12$, $16 \times 12 = 192$, $k^2 = 1 + 192 = 193$ (not a perfect square).

If $a = 4$: $4^2 + 4 = 20$, $16 \times 20 = 320$, $k^2 = 1 + 320 = 321$ (not a perfect square).

If $a = 5$: $5^2 + 5 = 30$, $16 \times 30 = 480$, $k^2 = 1 + 480 = 481$ (not a perfect square).

In general, $k^2 = 16a^2 + 16a + 1 = (4a + 2)^2 - 3$.

So $k^2 + 3 = (4a + 2)^2$.

Thus, $(4a + 2)^2 - k^2 = 3$.

So $(4a + 2 - k)(4a + 2 + k) = 3$.

The only integer factorizations of $3$ are $(1, 3)$ and $(-1, -3)$.

Case 1: $4a + 2 - k = 1$, $4a + 2 + k = 3$

Add: $2(4a + 2) = 4$, so $4a + 2 = 2$, so $a = 0$ (not positive integer).

Case 2: $4a + 2 - k = -1$, $4a + 2 + k = -3$

Add: $2(4a + 2) = -4$, so $4a + 2 = -2$, so $a = -1$ (not positive integer).

Case 3: $4a + 2 - k = 3$, $4a + 2 + k = 1$

Add: $2(4a + 2) = 4$, so $4a + 2 = 2$, $a = 0$ (not positive integer).

Case 4: $4a + 2 - k = -3$, $4a + 2 + k = -1$

Add: $2(4a + 2) = -4$, $4a + 2 = -2$, $a = -1$ (not positive integer).

Therefore, there are no positive integer solutions for $a$.

Therefore, the equation $4 f(a) = f(b)$ has no solutions in positive integers $a$ and $b$.