1977 Canadian MO Problems/Problem 1

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


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 $4f(a) = f(b)$.

Alternate Solutions?

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

See also

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