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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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==Solution==
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==Solution 2==
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Suppose there exist positive integral <math>a</math> and <math>b</math> such that <math>4f(a) = f(b)</math>.
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Suppose there exist positive integers <math>a</math> and <math>b</math> such that <math>4f(a) = f(b)</math>.
  
−
Thus, <math>4a^2 + 4a = b^2 + b</math>, or <math>(2a+1)^2 = b^2 + b + 1</math>. But <math>b^2 < b^2 + b + 1 < (b+1)^2</math>, so <math>b^2 + b + 1</math> cannot be a perfect square, contradiction. The conclusion follows.
+
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>.
 +
 
 +
==Solution 3==
 +
Suppose there exist positive integers <math>a</math> and <math>b</math> such that <math>4f(a) = f(b)</math>.
 +
 
 +
Then, <math>4a^2 + 4a = b^2 + b</math>, or <math>(4a^2 - b^2) + (4a - b) = 0</math>. This LHS can be factored to give <math>(4a - b)(4a + b + 1) = 0</math>. Now, there are 2 cases.
 +
 
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Case 1: <math>4a - b</math> = 0. Then <math>4a = b</math>. Plugging this into <math>4a^2 + 4a = b^2 + b</math> give <math>4a^2 + 4a = 16a^2 + 4a</math>, so <math>4a^2 = 16a^2</math>, meaning <math>a = 0</math>, so <math>a</math> isn't positive.
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Case 2: <math>4a + b + 1 = 0</math>, so <math>4a = -(b + 1)</math>. Since <math>a</math> must be positive, <math>4a</math> must be be positive, meaning <math>b + 1 < 0</math>, so <math>b + 1 < -1</math>, meaning b must be negative.
 +
 
 +
Therefore, there is no <math>b</math> that satisfies <math>4f(a) = f(b)</math>.
 +
 
 +
~adi2011
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==Alternate Solutions?==
 
{{alternate solutions}}
 
{{alternate solutions}}
  

Latest revision as of 02:00, 18 April 2025

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

Solution 3

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

Then, $4a^2 + 4a = b^2 + b$, or $(4a^2 - b^2) + (4a - b) = 0$. This LHS can be factored to give $(4a - b)(4a + b + 1) = 0$. Now, there are 2 cases.

Case 1: $4a - b$ = 0. Then $4a = b$. Plugging this into $4a^2 + 4a = b^2 + b$ give $4a^2 + 4a = 16a^2 + 4a$, so $4a^2 = 16a^2$, meaning $a = 0$, so $a$ isn't positive.

Case 2: $4a + b + 1 = 0$, so $4a = -(b + 1)$. Since $a$ must be positive, $4a$ must be be positive, meaning $b + 1 < 0$, so $b + 1 < -1$, meaning b must be negative.

Therefore, there is no $b$ that satisfies $4f(a) = f(b)$.

~adi2011


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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