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Difference between revisions of "1977 Canadian MO Problems/Problem 1"

m (unfortunately, the equation does not imply that either term is zero, just that their difference is zero)
 
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In order for both <math>a</math> and <math>b</math> to be integers, the [[discriminant]] must be a [[perfect square]].  However, since <math>b^2< b^2+b+1 <(b+1)^2,</math> the quantity <math>b^2+b+1</math> cannot be a perfect square when <math>b</math> is an integer. Hence, when <math>b</math> is a positive integer, <math>a</math> cannot be.
 
In order for both <math>a</math> and <math>b</math> to be integers, the [[discriminant]] must be a [[perfect square]].  However, since <math>b^2< b^2+b+1 <(b+1)^2,</math> the quantity <math>b^2+b+1</math> cannot be a perfect square when <math>b</math> is an integer. Hence, when <math>b</math> is a positive integer, <math>a</math> cannot be.
  
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==Solution 2==
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Suppose there exist positive integers <math>a</math> and <math>b</math> such that <math>4f(a) = f(b)</math>.
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 +
Thus, <math>4a^2 + 4a = b^2 + b</math>, or <math>(2a+1)^2 = b^2 + b + 1</math>. Then in order for the original equation to be true, <math>b^{2} + b + 1</math> would have to be a perfect square. Completing the square of <math>b^{2} + b + 1</math> results in <math>(b+1/2)^{2} + 3/4</math>. Thus, <math>b^{2} + b + 1</math>  is not a perfect square, and thus there is no <math>b</math> that satisfies <math>4f(a) = f(b)</math>.
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==Alternate Solutions?==
 
{{alternate solutions}}
 
{{alternate solutions}}
  

Latest revision as of 09:04, 28 September 2023

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


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
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