Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1977 Canadian MO Problems/Problem 1: Difference between revisions

← Older editNewer edit →
Added another solution that uses simpler algebra.
m unfortunately, the equation does not imply that either term is zero, just that their difference is zero
Line 1: Line 1:
== Problem ==
== Problem ==
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>
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 ==
== Solution ==
Line 10: Line 8:


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.
== Alternate Solution ==
Write out the expanded form of <math>4f(a)=f(b)</math>:
<math>4a^2+4a=b^2+b</math>
Now simply factor it to get: <math>4a(a+1)=b(b+1)</math>
Subtract <math>b(b+1)</math> from both sides: <math>4a(a+1)-b(b+1)=0</math>
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>.
Set each one equal to <math>0</math> to find the possible solutions:
<math>4a=0</math>
<math>a+1=0</math>
<math>b=0</math>
<math>b+1=0</math>
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>.
<math>\therefore</math> There are no positive solutions for <math>a</math> or <math>b</math>, Q.E.D.


{{alternate solutions}}
{{alternate solutions}}


== See also ==
{{Old CanadaMO box|before=First question|num-a=2|year=1977}}
{{Old CanadaMO box|before=First question|num-a=2|year=1977}}


[[Category:Olympiad Algebra Problems]]
[[Category:Intermediate Algebra Problems]]

Revision as of 09:26, 7 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 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
Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=1977_Canadian_MO_Problems/Problem_1&oldid=27840"