Cool but may be tough(relatively) (BOMC-2)

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sourav das
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Cool but may be tough(relatively) (BOMC-2)

Unread post by sourav das » Thu Mar 29, 2012 9:49 pm

(Russian Math Olympiad 1999, Grade 11, #5) Four natural numbers have the property that
the square of the sum of any two of the numbers is divisible by the product of the other two.
Show that at least three of the four numbers are equal.
You spin my head right round right round,
When you go down, when you go down down......
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*Mahi*
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Re: Cool but may be tough(relatively) (BOMC-2)

Unread post by *Mahi* » Thu Apr 05, 2012 12:56 am

Sorry for being late. I did not notice it earlier.
Steps:
1. Let us assume $\gcd (a,b,c,d)=1$. Then it can be shown that $\text{cyclic } \gcd(a,b)=1$
2. Then we have to show that $\text{cyclic} ab \mid 2(c+d)$
3.From that, inequality gives us $\text{cyclic }(a-1)(c+d) \leq 2b\;$, which can be manipulated to show that at least three of $a,b,c,d$ is $1$.
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SANZEED
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Re: Cool but may be tough(relatively) (BOMC-2)

Unread post by SANZEED » Thu Apr 05, 2012 12:59 am

I haven't solved it fully yet,but a useful hint:
If an integer $p$ divides $a$,then find the property of $p$ for this problem.
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*Mahi*
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Re: Cool but may be tough(relatively) (BOMC-2)

Unread post by *Mahi* » Thu Apr 05, 2012 1:18 am

Share your full thought and progress... no need for "hints of progress".
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zadid xcalibured
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Re: Cool but may be tough(relatively) (BOMC-2)

Unread post by zadid xcalibured » Thu Mar 07, 2013 6:16 pm

SANZEED wrote:I haven't solved it fully yet,but a useful hint:
If an integer $p$ divides $a$,then find the property of $p$ for this problem.
সত্যি এই ধরনের কমেন্ট দেখলে মেজাজ গরম হয়। :evil:

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