1977 Canadian MO Problems/Problem 1

Problem

If $f(x)=x^2+x,$ prove that the equation $4f(a)=f(b)$ has no solutions in positive integers $a$ and $b.$

Solution

Directly plugging $a$ and $b$ into the function, $4a^2+4a=b^2+b.$ We now have a quadratic in $a.$

Applying the quadratic formula, $a=\frac{-1\pm \sqrt{b^2+b+1}}{2}.$

In order for both $a$ and $b$ to be integers, the discriminant must be a perfect square. However, since $b^2< b^2+b+1 <(b+1)^2,$ the quantity $b^2+b+1$ cannot be a perfect square when $b$ is an integer. Hence, when $b$ is a positive integer, $a$ cannot be.

Solution 2

Suppose there exist positive integers $a$ and $b$ such that $4f(a) = f(b)$ .

Thus, $4a^2 + 4a = b^2 + b$ , or $(2a+1)^2 = b^2 + b + 1$ . Then in order for the original equation to be true, $b^{2} + b + 1$ would have to be a perfect square. Completing the square of $b^{2} + b + 1$ results in $(b+1/2)^{2} + 3/4$ . Thus, $b^{2} + b + 1$ is not a perfect square, and thus there is no $b$ that satisfies $4f(a) = f(b)$ .

Solution 3

Suppose there exist positive integers $a$ and $b$ such that $4f(a) = f(b)$ .

Then, $4a^2 + 4a = b^2 + b$ , or $(4a^2 - b^2) + (4a - b) = 0$ . This LHS can be factored to give $(4a - b)(4a + b + 1) = 0$ . Now, there are 2 cases.

Case 1: $4a - b$ = 0. Then $4a = b$ . Plugging this into $4a^2 + 4a = b^2 + b$ give $4a^2 + 4a = 16a^2 + 4a$ , so $4a^2 = 16a^2$ , meaning $a = 0$ , so $a$ isn't positive.

Case 2: $4a + b + 1 = 0$ , so $4a = -(b + 1)$ . Since $a$ must be positive, $4a$ must be be positive, meaning $b + 1 < 0$ , so $b + 1 < -1$ , meaning b must be negative.