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

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

Thus, $4a^2 + 4a = b^2 + b$ , or $(2a+1)^2 = b^2 + b + 1$ . But $b^2 < b^2 + b + 1 < (b+1)^2$ , so $b^2 + b + 1$ cannot be a perfect square, contradiction. The conclusion follows. 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

Page

Toolbox

Search

1977 Canadian MO Problems/Problem 1

Contents

Problem

Solution

Solution

See also