# Determine all positive integers

Determine all positive integers $$(x,n)$$ so that $$x^n+2^n+1$$ is divisor of $$x^{n+1}+2^{n+1}+1$$

Note by Idham Muqoddas
4 years, 11 months ago

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Interesting problem but very messy .

I will show that only solutions are $$(11,1),(4,1)$$ . First if $$n=1$$ then $$x+3 \mid x^2+5 \implies x+3 \mid x^2−9+14 \implies x+3 \mid 14 \implies x=11,4$$ Thus $$(11,1)$$ and $$(4,1)$$ are two solutions. .

Now suppose $$x \ge 4$$
We will prove that $(x-1)(x^n+2^n+1) < x^{n+1}+2^{n+1}+1 < x ( x^n+2^n+1)$ Which can be verified by seeing that , $x^n=\frac{x^n}{2}+ \frac{x^n}{2} \ge \frac{4^n}{2}+\frac{x^2}{2}$ And , $(x-1)(x^n+2^n+1) = x^{n+1}+x(2^n+1) - (x^n+2^n+1) < \\ x^{n+1}+\frac{x^2+(2^n+1)^2}{2} - (x^n+2^n+1) = \\ x^{n+1}+\frac{4^n+2 \cdot 2^n+1+x^2}{2} - (x^n+2^n+1) < \\ x^{n+1}+ 2^{n+1}+x^n+2^n+2 - (x^n+2^n+1) < x^{n+1}+2^{n+1}+1$ And $x(x^n+2^n+1) > x^{n+1}+2^{n+1}+1$ Is quite obvious. So there are no solutions , for the case when $$x\ge 4$$ Now suppose $$x \in 1,2,3$$ then ,

$\text{If} \ \ x=1 \implies 1^n+2^n+1<2+2^n+1<2+2^{n+1}+1<2(1+2^n+1)$ $\text{If} \ \ x=2 \implies 2^n+2^n+1<2^{n+1}+2^{n+1}+1<2(2^n+2^n+1)$ and $\text{If} \ \ x=3 \implies 2(3^n+2^n+1)<3^{n+1}+2^{n+1}+1<3(3^n+2^n+1)$ So for $$x=1,2,3$$ also there are no solutions.

So Only possible solutions are $$(11,1) , (4,1)$$ . $$\Box$$.

PS : Comment , if there is any mistake or if any-part is not explained well

- 4 years, 11 months ago

how do you get this? $x\left(x^n+2^n+1\right)>x^{n+1}+2^{n+1}+1$

- 4 years, 11 months ago

Clearly, $$x \cdot x^n = x^{n+1}$$, and $$x \cdot 2^n > 2 \cdot 2^n = 2^{n+1}$$, and $$x \cdot 1 = x > 1$$.

Add these to get $$x(x^n+2^n+1) > x^{n+1}+2^{n+1}+1$$.

- 4 years, 11 months ago