posting apmo problems
I think now the ban time of Apmo is over. We can post the problems. But I am not sure.
One one thing is neutral in the universe, that is $0$.
Re: posting apmo problems
Feel free to post them!
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Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
Re: posting apmo problems
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.
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.
One one thing is neutral in the universe, that is $0$.
- Tahmid Hasan
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Re: posting apmo problems
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.
$b^2+c+a=(b+y)2$
$c^2+a+b=(c+z)^2$
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.
বড় ভালবাসি তোমায়,মা
Re: posting apmo problems
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, \[0\ge3\]
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, \[0\ge3\]
One one thing is neutral in the universe, that is $0$.