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Find all positive integers $(x,n)$ such that $x^{n} + 2^{n} + 1$ is a divisor of $x^{n+1} +2^{n+1} +1$

I encountered this question in one of my monthly assignments. Unfortunately, I don't know how to proceed about this question at all. Please help.
Thanks in advance!

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Detailed solutions are not appropriate for homework. –  Robert Israel Jun 11 '14 at 6:57

3 Answers 3

up vote 3 down vote accepted

Check that $n=1$ gives two solutions $x=4$ and $x=11$. From now on $n>1$.

For each single case $x=1$, $x=2$ check that there is no solution.

Now we will consider $$ x(x^n+2^n+1)-(x^{n+1}+2^{n+1}+1)=2^n(x-2)+x-1 $$ instead of $x^{n+1}+2^{n+1}+1$.

Check that $x=3$ gives no solution. From now on $x>3$ and $n>1$, hence $$ x^{n-1}(x-2)\geq 2^n(x-2), $$
$$ x^{n-1}\cdot 2\geq x-1, $$ $$ 2^n+1>0. $$ Summing last three lines we get $$ x^n+2^n+1>2^n(x-2)+x-1 $$ and the left hand side is not a divisor of the right hand side, there is no solution for $n>1$, $x>3$.

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"Now we will consider $$x(x^n+2^n+1)-(x^{n+1}+2^{n+1}+1)=2^n(x-2)+x-1 $$ instead of $x^{n+1}+2^{n+1}+1$" Why is this true? –  Samurai Jun 14 '14 at 11:04
Because A is a divisor of B if and only if A is a divisor of AC-B. –  IBazhov Jun 14 '14 at 12:06


  • Consider $f(x,n)= \dfrac{x^{n+1} +2^{n+1} +1}{x^{n} + 2^{n} + 1}$ and the behaviour of $f(x,n)-x$
  • Put bounds on $f(x,n) - x $ for $x \ge 2, n \ge 2$
  • Put bounds on $f(x,1) - x $ i.e. for $n = 1$
  • Find limit of $f(1,n) - 1 $ as $n\to \infty$ i.e. for $x=1$
    • Consider cases where the absolute value of $\displaystyle f(1,n) - 1 - \lim_{n\to \infty} (f(1,n) - 1)$ is greater than or equal to $1$
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