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

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.

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 == Problem ==
-If <math>\displaystyle f(x)=x^2+x,</math> prove that the equation <math>\displaystyle 4f(a)=f(b)</math> has no solutions in positive integers <math>\displaystyle a</math> and <math>\displaystyle b.</math>
+If <math>f(x)=x^2+x,</math> prove that the equation <math>4f(a)=f(b)</math> has no solutions in positive integers <math>a</math> and <math>b.</math>
 == Solution ==
-Directly plugging <math>\displaystyle a</math> and <math>\displaystyle b</math> into the function, <math>\displaystyle 4a^2+4a=b^2+b.</math> We now have a quadratic in <math>\displaystyle a.</math>
+Directly plugging <math>a</math> and <math>b</math> into the function, <math>4a^2+4a=b^2+b.</math> We now have a quadratic in <math>a.</math>
-Applying the quadratic formula, <math> \displaystyle a=\frac{-1\pm \sqrt{b^2+b+1}}{2}. </math>
+Applying the quadratic formula, <math>a=\frac{-1\pm \sqrt{b^2+b+1}}{2}. </math>
-Because, <math>\displaystyle b^2< b^2+b+1 <(b+1)^2,</math> the quantity <math>\displaystyle b^2+b+1</math> cannot be a perfect square when <math>\displaystyle b</math> is an integer. Hence, when <math>\displaystyle b</math> is a positive integer, <math>\displaystyle 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.
+==Solution 2==
+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>. 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>.
+==Alternate Solutions?==
+{{alternate solutions}}
 == See also ==
+{{Old CanadaMO box|before=First question|num-a=2|year=1977}}
+[[Category:Intermediate Algebra Problems]]

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

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