## APMO 1989-2

Discussion on Asian Pacific Mathematical Olympiad (APMO)
Masum
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Joined: Tue Dec 07, 2010 1:12 pm

### APMO 1989-2

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

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......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm

### Re: APMO 1989-2

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 $0$.

bristy1588
Posts: 92
Joined: Sun Jun 19, 2011 10:31 am

### Re: APMO 1989-2

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

bristy1588
Posts: 92
Joined: Sun Jun 19, 2011 10:31 am

### Re: APMO 1989-2

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

sourav das
Posts: 461
Joined: Wed Dec 15, 2010 10:05 am
Location: Dhaka
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### Re: APMO 1989-2

@ 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 )

@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......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm
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.
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 $0$.