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Difference between revisions of "2005 Canadian MO Problems/Problem 2"

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* Prove that there does not exist any integer <math>n</math> for which we can find a Pythagorean triple <math>(a,b,c)</math> satisfying <math>(c/a + c/b)^2 = n</math>.
 
* Prove that there does not exist any integer <math>n</math> for which we can find a Pythagorean triple <math>(a,b,c)</math> satisfying <math>(c/a + c/b)^2 = n</math>.
 
==Solution==
 
==Solution==
−
 
+
* We have  
−
We have  
 
  
 
:<math>\left(\frac ca + \frac cb\right)^2 = \frac{c^2}{a^2} + 2\frac{c^2}{ab} + \frac{c^2}{b^2} = \frac{a^2 + b^2}{a^2} + 2\frac{a^2 + b^2}{ab} + \frac{a^2+b^2}{b^2} = 2 + \left(\frac{a^2}{b^2} + \frac{b^2}{a^2}\right) + 2\left(\frac ab + \frac ba\right)</math>
 
:<math>\left(\frac ca + \frac cb\right)^2 = \frac{c^2}{a^2} + 2\frac{c^2}{ab} + \frac{c^2}{b^2} = \frac{a^2 + b^2}{a^2} + 2\frac{a^2 + b^2}{ab} + \frac{a^2+b^2}{b^2} = 2 + \left(\frac{a^2}{b^2} + \frac{b^2}{a^2}\right) + 2\left(\frac ab + \frac ba\right)</math>
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:<math>x + \frac 1x > 2,</math>
 
:<math>x + \frac 1x > 2,</math>
  
−
where <math>x</math> is a [[positive]] [[real number]] not equal to one.  If <math>a = b</math>, then <math>c \not\in \mathbb{Z}</math>. Thus <math>\displaystyle a \neq b</math> and <math>\frac ab \neq 1\implies \frac{a^2}{b^2}\neq 1</math>. Therefore,  
+
where <math>x</math> is a [[positive]] [[real number]] not equal to one.  If <math>a = b</math>, then <math>c \not\in \mathbb{Z}</math>. Thus <math>a \neq b</math> and <math>\frac ab \neq 1\implies \frac{a^2}{b^2}\neq 1</math>. Therefore,  
  
 
:<math>\left(\frac ca + \frac cb\right)^2 > 2 + 2 + 2(2) = 8.</math>
 
:<math>\left(\frac ca + \frac cb\right)^2 > 2 + 2 + 2(2) = 8.</math>
 +
 +
* Now since <math>a</math>, <math>b</math>, and <math>c</math> are positive integers, <math>c/a + c/b</math> is a rational number <math>p/q</math>, where <math>p</math> and <math>q</math> are positive integers. Now if <math>p^2/q^2=n</math>, where <math>n</math> is an integer, then <math>p/q</math> must also be an integer. Thus <math>c(a+b)/ab</math> must be an integer.
 +
 +
Now every pythagorean triple can be written in the form <math>(2mn, m^2-n^2, m^2+n^2)</math>, with <math>m</math> and <math>n</math> positive integers. Thus one of <math>a</math> or <math>b</math> must be even. If <math>a</math> and <math>b</math> are both even, then <math>c</math> is even too. Factors of 4 can be cancelled from the numerator and the denominator(since every time one of <math>a</math>, <math>b</math>, <math>c</math>, and <math>a+b</math> increase by a factor of 2, they all increase by a factor of 2) repeatedly until one of <math>a</math>, <math>b</math>, or <math>c</math> is odd, and we can continue from there. Thus the <math>m^2-n^2</math> term is odd, and thus <math>c</math> is odd. Now <math>c</math> and <math>a+b</math> are odd, and <math>ab</math> is even. Thus <math>c(a+b)/ab</math> is not an integer. Now we have reached a contradiction, and thus there does not exist any integer <math>n</math> for which we can find a Pythagorean triple <math>(a,b,c)</math> satisfying <math>(c/a + c/b)^2 = n</math>.
  
 
==See also==
 
==See also==
−
*[[2005 Canadian MO Problems/Problem 3|Next problem]]
 
−
*[[2005 Canadian MO Problems/Problem 1|Previous problem]]
 
 
*[[2005 Canadian MO]]
 
*[[2005 Canadian MO]]
  
 +
{{CanadaMO box|year=2005|num-b=1|num-a=3}}
  
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Number Theory Problems]]
 
[[Category:Olympiad Number Theory Problems]]

Latest revision as of 13:50, 20 September 2008

Problem

Let $(a,b,c)$ be a Pythagorean triple, i.e., a triplet of positive integers with ${a}^2+{b}^2={c}^2$.

  • Prove that $(c/a + c/b)^2 > 8$.
  • Prove that there does not exist any integer $n$ for which we can find a Pythagorean triple $(a,b,c)$ satisfying $(c/a + c/b)^2 = n$.

Solution

  • We have
$\left(\frac ca + \frac cb\right)^2 = \frac{c^2}{a^2} + 2\frac{c^2}{ab} + \frac{c^2}{b^2} = \frac{a^2 + b^2}{a^2} + 2\frac{a^2 + b^2}{ab} + \frac{a^2+b^2}{b^2} = 2 + \left(\frac{a^2}{b^2} + \frac{b^2}{a^2}\right) + 2\left(\frac ab + \frac ba\right)$

By AM-GM, we have

$x + \frac 1x > 2,$

where $x$ is a positive real number not equal to one. If $a = b$, then $c \not\in \mathbb{Z}$. Thus $a \neq b$ and $\frac ab \neq 1\implies \frac{a^2}{b^2}\neq 1$. Therefore,

$\left(\frac ca + \frac cb\right)^2 > 2 + 2 + 2(2) = 8.$
  • Now since $a$, $b$, and $c$ are positive integers, $c/a + c/b$ is a rational number $p/q$, where $p$ and $q$ are positive integers. Now if $p^2/q^2=n$, where $n$ is an integer, then $p/q$ must also be an integer. Thus $c(a+b)/ab$ must be an integer.

Now every pythagorean triple can be written in the form $(2mn, m^2-n^2, m^2+n^2)$, with $m$ and $n$ positive integers. Thus one of $a$ or $b$ must be even. If $a$ and $b$ are both even, then $c$ is even too. Factors of 4 can be cancelled from the numerator and the denominator(since every time one of $a$, $b$, $c$, and $a+b$ increase by a factor of 2, they all increase by a factor of 2) repeatedly until one of $a$, $b$, or $c$ is odd, and we can continue from there. Thus the $m^2-n^2$ term is odd, and thus $c$ is odd. Now $c$ and $a+b$ are odd, and $ab$ is even. Thus $c(a+b)/ab$ is not an integer. Now we have reached a contradiction, and thus there does not exist any integer $n$ for which we can find a Pythagorean triple $(a,b,c)$ satisfying $(c/a + c/b)^2 = n$.

See also

2005 Canadian MO (Problems)
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 Followed by
Problem 3
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