# Difference between revisions of "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.

## Alternate Solution

Write out the expanded form of $4f(a)=f(b)$:

$4a^2+4a=b^2+b$

Now simply factor it to get: $4a(a+1)=b(b+1)$

Subtract $b(b+1)$ from both sides: $4a(a+1)-b(b+1)=0$

For the left side to equal $0$, $4a$ or $a+1$ must be $0$ AND $b$ or $b+1$ must be $0$.

Set each one equal to $0$ to find the possible solutions:

$4a=0$

$a+1=0$

$b=0$

$b+1=0$

Thus, $a$ must be $0$ or $-1$. The same applies to $b$. None of these solutions are greater than $0$.

$\therefore$ There are no positive solutions for $a$ or $b$, Q.E.D.