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2005 Canadian MO Problems/Problem 2: Difference between revisions

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[[Category:Olympiad Algebra Problems]]
[[Category:Olympiad Number Theory Problems]]
[[Category:Olympiad Number Theory Problems]]

Revision as of 12:58, 4 September 2006

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

First part:

$\left(\frac ca + \frac cb\right)^2 = \frac{c^2}{a^2} + 2\frac{c^2}{ab} + \frac{c^2}{b^2} = \frac{a^2 + b^2}{a^2} + 2\frac{a^2 + b^2}{ab} + \frac{a^2+b^2}{b^2} = 2 + \left(\frac{a^2}{b^2} + \frac{b^2}{a^2}\right) + 2\left(\frac ab + \frac ba\right)$. By AM-GM we have $x + \frac 1x > 2$ if $x$ is a positive real number other than 1. If $a = b$ then $c \not\in \mathbb{Z}$ so $\displaystyle a \neq b$ and $\frac ab \neq 1$ and $\frac{a^2}{b^2}\neq 1$ and thus $\left(\frac ca + \frac cb\right)^2 > 2 + 2 + 2(2) = 8$.


See also

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