posting apmo problems

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Masum
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posting apmo problems

Unread post by Masum » Fri May 20, 2011 10:01 am

I think now the ban time of Apmo is over. We can post the problems. But I am not sure.
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Re: posting apmo problems

Unread post by Moon » Fri May 20, 2011 2:32 pm

Feel free to post them! :D
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Masum
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Re: posting apmo problems

Unread post by Masum » Sat May 21, 2011 5:48 pm

I know that the following problem (though it seems to be too trivial for APMO) appeared in APMO 2011.
Problem :
Prove that for no positive integers $a,b,c$ all of $a^2+b+c,b^2+c+a,c^2+a+b$ are squares.
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Tahmid Hasan
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Re: posting apmo problems

Unread post by Tahmid Hasan » Sat May 21, 2011 7:28 pm

let $$a^2+b+c=(a+x)^2$$
$b^2+c+a=(b+y)2$
$c^2+a+b=(c+z)^2$
where $x,y,z$ are positive integers
adding these equations gives us
$a^2+b^2+c^2+2a+2b+2c=a^2+b^2+c^2+2ax+2by+2cz+x^2+y^2+z^2$
so $2a+2b+2c=2ax+2by+2cz+x^2+y^2+z^2$
but $ax+by+cz>a+b+c$
hence$2ax+2by+2cz+x^2+y^2+z^2>2a+2b+2c$
which is a contradiction.
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Masum
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Re: posting apmo problems

Unread post by Masum » Sun May 22, 2011 2:18 pm

Here are solutions other than the above.
First Solution :
For symmetry, assume $a\ge b\ge c>0$. Then \[a^2<a^2+b+c\le a^2+2a<(a+1)^2\]

Second Solution :
Obviously, we have that \[a^2+b+c\ge (a+1)^2\]
Then, \[b+c\ge 2a+1\]
Similarly, \[c+a\ge 2b+1\]
\[a+b\ge 2c+1\]
Adding them, \[0\ge3\]
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