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APMO 1989-2

Posted: Sat Mar 05, 2011 1:51 pm
by Masum
Prove that $5n^2=36a^2+18b^2+6c^2$ has no integer solutions except $a = b = c = n = 0.$

Re: APMO 1989-2

Posted: Thu Dec 22, 2011 4:18 pm
by sourav das
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

Re: APMO 1989-2

Posted: Thu Dec 22, 2011 7:17 pm
by Masum
sourav das wrote:As $6|R.H.S$, $6|n^2$. Set $n=6n'$
Careful, you could conclude this because $6$ is square-free.

Re: APMO 1989-2

Posted: Thu Dec 22, 2011 9:47 pm
by bristy1588
What is the infinite descent technique, where can i find out about it?

Re: APMO 1989-2

Posted: Thu Dec 22, 2011 9:49 pm
by bristy1588
If anyone knows any good source, pls do tell me.

Re: APMO 1989-2

Posted: Thu Dec 22, 2011 10:44 pm
by sourav das
@ 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

Re: APMO 1989-2

Posted: Fri Dec 23, 2011 12:11 am
by Masum
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.