APMO 1989-2

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Masum
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APMO 1989-2

Unread post by Masum » Sat Mar 05, 2011 1:51 pm

Prove that $5n^2=36a^2+18b^2+6c^2$ has no integer solutions except $a = b = c = n = 0.$
One one thing is neutral in the universe, that is $0$.

sourav das
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Re: APMO 1989-2

Unread post by sourav das » Thu Dec 22, 2011 4:18 pm

As $6|R.H.S$, $6|n^2$. Set $n=6n'$ Now, $30n'^2=6a^2+3b^2+c^2$ So, $3|c$ Set $c=3c'$ Now,
$10n'^2=2a^2+b^2+3c'^2$.........(i)
Now, $2(5n'^2-a^2)=b^2+3c'^2$
If $b,c$ both are odd then, $b^2+3c^2\equiv 1+3\equiv 4(mod$ $8)$
So,$5n'^2-a^2\equiv \pm 2 (mod$ $8) $
$a^2\equiv 5n'^2\pm 2(mod$ $8)$ But as $n'^2 \equiv 0,1,4(mod$ $8)$, $a^2 \equiv 5n'^2\pm 2 \equiv (\pm 2,-1 or 3,\mp 2)(mod$ $8)$ a contradiction.
So both must be even. Set $b=2b'$ and $c'=2c''$
Now,$5n'^2-a^2=2(b'^2+3c''^2)$
Again assume $n',a$ both are odd. then again, $5n'^2 - a^2 \equiv5-1\equiv 4(mod$ $8)$
So $b'^2+3c''^2\equiv \pm 2(mod$ $8)$ But as $c''^2 \equiv 0,1,4(mod$ $8)$ ; $b'^2\equiv \pm 2-3c''^2\equiv \pm 2,-1or3,\mp 2(mod$ $8)$ a contradiction.
So $n',a$ both must be even. Set $n'= 2n''$ and $a=2a'$. And so,
$10n''^2=2a'^2+b'^2+3c''^2$.......(ii)
But (i) and (ii) are in same form. So using infinite descent, we'll find that equality holds only when $a=b=c=n=0$

And we are done
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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Masum
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Re: APMO 1989-2

Unread post by Masum » Thu Dec 22, 2011 7:17 pm

sourav das wrote:As $6|R.H.S$, $6|n^2$. Set $n=6n'$
Careful, you could conclude this because $6$ is square-free.
One one thing is neutral in the universe, that is $0$.

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bristy1588
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Re: APMO 1989-2

Unread post by bristy1588 » Thu Dec 22, 2011 9:47 pm

What is the infinite descent technique, where can i find out about it?
Bristy Sikder

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bristy1588
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Re: APMO 1989-2

Unread post by bristy1588 » Thu Dec 22, 2011 9:49 pm

If anyone knows any good source, pls do tell me.
Bristy Sikder

sourav das
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Re: APMO 1989-2

Unread post by sourav das » Thu Dec 22, 2011 10:44 pm

@ Masum bhai: Thank you :) . Next time I'll try to be more careful in dealing with these little but very serious parts. (I wish i could do so in IMO :cry: )

@Brishty : http://www.math.ust.hk/excalibur/v10_n4.pdf
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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Masum
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Re: APMO 1989-2

Unread post by Masum » Fri Dec 23, 2011 12:11 am

sourav das wrote:@ Masum bhai: Thank you :) . Next time I'll try to be more careful in dealing with these little but very serious parts. (I wish i could do so in IMO :cry: )

@Brishty : http://www.math.ust.hk/excalibur/v10_n4.pdf
In fact, I gave this post because we(obviously including me) make very silly but severe mistakes while we are thinking. For example, if $a|bc$ with $\gcd(b,c)=1$ then $a|b$ or $a|c$. But this is false, and so many such errors. Always be aware of them, specially in number theory because making silly mistakes in number theory rather than geometry or algebra is the easiest I think. Otherwise you may lose something very great. :)
@Brishty, initially you need to know that: Infinite Descent or Descent Inifini or Method of infinite descent is a contradiction, mainly based on the principle that: "There are no solution less than the smallest or largest solution". Specially it is useful for divisibility or diophantine equations. We generally assume a solution to an equation to be the smallest one, and then show a contradiction reaching the point that we get another solution smaller than the one we assumed.
In this particular example, we can assume $(n,a,b,c)$ is a solution such that it is the smallest. Then when we say that, say $a$ is even, and $a=2a_1$ then we have $a_1<a$. So by the process when we get an equation of the same form as the previous one, we have a smaller solution (in this case $(n'',a',b',c'')$) we can conclude in the way above. Now read any article or wiki, hope you will understand how to implement it. Make sure you understand it.
One one thing is neutral in the universe, that is $0$.

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