# Problem Of The Week #431 August 24th, 2020

Status
Not open for further replies.

#### anemone

##### MHB POTW Director
Staff member
Here is this week's POTW:

-----

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

-----

#### anemone

##### MHB POTW Director
Staff member
No one answered last week's POTW. You can find the suggested solution below:
For $x=1$, $2(1^n+2^n+1)>1^{n+1}+2^{n+1}+1>1^n+2^n+1$.

For $x=2$, $2(2^n+2^n+1)>2^{n+1}+2^{n+1}+1>2^n+2^n+1$.

For $x=3$, $3(3^n+2^n+1)>3^{n+1}+2^{n+1}+1>2(3^n+2^n+1)$.

So there are no solutions with $x=1,\,2,\,3$.

For $x\ge 4$, if $n\ge 2$, then we get $x(x^n+2^n+1)>x^{n+1}+2^{n+1}+1$.

Now,

$x^{n+1}+2^{n+1}+1=(x-1)(x^n+2^n+1)+x^n-(2^n+1)x+3\cdot 2^n+2>(x-1)(x^n+2^n+1)$

because for $n=2$, $x^n-(2^{n}+1)x+2^{n+1}=x^2-5x+8>0$ and for $n\ge3$, $x^n-(2^{n}+1)x+2^{n+1}\ge x(4^{n-1}-2^n-1)>0$

Hence, only $n=1$ and $x\ge 4$ are possible.

Now, $x^n+2^n+1=x+3$ is a divisor of $x^{n+1}+2^ {n+1}+1=x^2+5=(x-3)(x+3)+14$ if and only if $x+3$ is a divisor of 14.

Since $x+3\ge 7$, $x=4$ or 11. So the solutions are $(x,\,y)=(4,\,1)$ and $(11,\,1)$.

• 