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 | ||
− | + | :<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> | |
+ | |||
+ | By [[AM-GM]], we have | ||
+ | |||
+ | :<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>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> | ||
+ | |||
+ | * 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]] | *[[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 be a Pythagorean triple, i.e., a triplet of positive integers with
.
- Prove that
.
- Prove that there does not exist any integer
for which we can find a Pythagorean triple
satisfying
.
Solution
- We have
By AM-GM, we have
where is a positive real number not equal to one. If
, then
. Thus
and
. Therefore,
- Now since
,
, and
are positive integers,
is a rational number
, where
and
are positive integers. Now if
, where
is an integer, then
must also be an integer. Thus
must be an integer.
Now every pythagorean triple can be written in the form , with
and
positive integers. Thus one of
or
must be even. If
and
are both even, then
is even too. Factors of 4 can be cancelled from the numerator and the denominator(since every time one of
,
,
, and
increase by a factor of 2, they all increase by a factor of 2) repeatedly until one of
,
, or
is odd, and we can continue from there. Thus the
term is odd, and thus
is odd. Now
and
are odd, and
is even. Thus
is not an integer. Now we have reached a contradiction, and thus there does not exist any integer
for which we can find a Pythagorean triple
satisfying
.
See also
2005 Canadian MO (Problems) | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 | Followed by Problem 3 |