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

The given equation is $4f(a) = f(b)$ or, $4a^2 + 4a = b^2 + b$ or, $4a(a+1) = b(b+1)$ .

For positive integers $a$ and $b$ , $4a(a+1)$ is an even number because it is divisible by $2$ whereas $b(b+1)$ is an odd number (the product of two consecutive numbers is an odd number). Therefore, $a$ and $b$ both cannot be positive integers.

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Alternate Solutions?

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1977 Canadian MO (Problems)
Preceded by First question	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 •	Followed by Problem 2

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