Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
Detailed solutions are not appropriate for homework. – Robert Israel Jun 11 '14 at 6:57
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 $n>1$, $x>3$.

share|cite|improve this answer
"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? – Henry 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 $1$
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.