1998 Romanian TST
Find all positive integers $(x, n)$ such that $x^n +2^n +1$ is a divisor of $x^{n+1}+ 2^{n+1}+1$.
Last edited by Phlembac Adib Hasan on Fri Jun 21, 2013 9:32 pm, edited 1 time in total.
Reason: Fixed some issues in Latex
Reason: Fixed some issues in Latex
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Phlembac Adib Hasan
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Re: 1998 Romanian IMO Team Selection Test
Putting $x=1,2,3$ we find that there is no solution. (Use inequalities)
So assume $x\ge 4$.
$x^n+2^n+1|x^{n+1}+2^{n+1}+1\Longleftrightarrow x^n+2^n+1|x(x^n+2^n+1)-(x^{n+1}+2^{n+1}+1)$
$\Longrightarrow x^n+2\leq (x-3)2^n+x$
By putting $x=4,5$ we find that this inequality is false if $n>1$.
And by induction, \[(x+1)^n+2=x^n+n\cdot x^{n-1}+...+1+2> x^n+2^n+1+2>(x-3)2^n+2^n+x+1\]$=(x-2)2^n+x+1$
So we must have $n=1$
Now the first condition becomes $x+3|x^2+5\Longleftrightarrow x+3|(x^2+5)-(x^2-9)=14$
As $x\ge 4$, we have $x+3=7,14\Longleftrightarrow x=4,11$. It's easy to see both of them are correct. So all the solutions are $(x,n)=(4,1),(11,1)$
So assume $x\ge 4$.
$x^n+2^n+1|x^{n+1}+2^{n+1}+1\Longleftrightarrow x^n+2^n+1|x(x^n+2^n+1)-(x^{n+1}+2^{n+1}+1)$
$\Longrightarrow x^n+2\leq (x-3)2^n+x$
By putting $x=4,5$ we find that this inequality is false if $n>1$.
And by induction, \[(x+1)^n+2=x^n+n\cdot x^{n-1}+...+1+2> x^n+2^n+1+2>(x-3)2^n+2^n+x+1\]$=(x-2)2^n+x+1$
So we must have $n=1$
Now the first condition becomes $x+3|x^2+5\Longleftrightarrow x+3|(x^2+5)-(x^2-9)=14$
As $x\ge 4$, we have $x+3=7,14\Longleftrightarrow x=4,11$. It's easy to see both of them are correct. So all the solutions are $(x,n)=(4,1),(11,1)$