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

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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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== Alternate Solution ==
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Write out the expanded form of <math>4f(a)=f(b)</math>:
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<math>4a^2+4a=b^2+b</math>
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Now simply factor it to get: <math>4a(a+1)=b(b+1)</math>
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Subtract <math>b(b+1)</math> from both sides: <math>4a(a+1)-b(b+1)=0</math>
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For the left side to equal <math>0</math>, <math>4a</math> or <math>a+1</math> must be <math>0</math> AND <math>b</math> or <math>b+1</math> must be <math>0</math>.
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Set each one equal to <math>0</math> to find the possible solutions:
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<math>4a=0</math>
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<math>a+1=0</math>
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<math>b=0</math>
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<math>b+1=0</math>
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Thus, <math>a</math> must be <math>0</math> or <math>-1</math>. The same applies to <math>b</math>. None of these solutions are greater than <math>0</math>.
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<math>\therefore</math> There are no positive solutions for <math>a</math> or <math>b</math>, Q.E.D.
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 17:46, 6 September 2008

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.

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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