How Far Can You Go?

Note that $p^n+(p+2)^n=(p+1)(p^{n-1}+(p+2)^{n-1})+((p+2)^{n-1}-p^{n-1})$

Thus $\dfrac{p^n+(p+2)^n}{p^{n-1}+(p+2)^{n-1}}=p+1+\dfrac{(p+2)^{n-1}-p^{n-1}}{p^{n-1}+(p+2)^{n-1}}$

Since $p^{n-1}+(p+2)^{n-1}\mid p^n+(p+2)^n$ we must have that $\dfrac{(p+2)^{n-1}-p^{n-1}}{p^{n-1}+(p+2)^{n-1}}$ is an integer.

However, unless $n=1$ at which $\dfrac{(p+2)^{n-1}-p^{n-1}}{p^{n-1}+(p+2)^{n-1}}=0$ , the numerator, which is a positive integer, is always less than the denominator, which is also a positive integer.

Thus our only solution is $\boxed{1}$ .

Here's my solution.

Claim : Only $n=1$ works.

We prove our claim in $3$ steps. First we define $f:\mathbb{N}\to\mathbb{R}$ by $f(n)=\dfrac{p^n+(p+2)^n}{p^{n-1}+(p+2)^{n-1}}$ for a fixed $p\in\mathbb{N}$ .

Step 1 : We show that $f(n)$ is increasing in $[1,\infty)$ , that is, $f(n+1)>f(n)$ for all $n\in\mathbb{N}$ . Letting $p^{n-1}=x$ and $(p+2)^{n-1}=y$ we have $f(n+1)>f(n)\implies \dfrac{p^2 x+(p+2)^2 y}{px+(p+2)y}>\dfrac{px+(p+2)y}{x+y}.$ Cross multiplying and cancelling gives $p^2+(p+2)^2>2p(p+2)\implies 4>0$ which is trivially true.

Step 2 : We bound the range of $f$ by computing the minimum and the maximum. Since $f$ is increasing, the minimum occurs at the minimum value of $n$ and the maximum occurs at the maximum value of $n$ in the domain.

So we have $\min{f}=f(1)=\dfrac{p+(p+2)}{2}=p+1$ . To compute the maximum, we let $n\to \infty$ :

$\lim_{n\to\infty} \dfrac{p^n+(p+2)^n}{p^{n-1}+(p+2)^{n-1}}=\lim_{n\to\infty} \dfrac{p\left(\frac{p}{p+2}\right)^{n-1}+(p+2)}{\left(\frac{p}{p+2}\right)^{n-1}+1}=p+2.$ So the range of $f$ is $[p+1,p+2)$ .

Step 3 : Step 2 implies that $f$ can give only two integer values, $p+1$ and $p+2$ . Time to deal with them one by one.

If $f(n)=p+2$ , then $p^n+(p+2)^n=(p+2)\left(p^{n-1}+(p+2)^{n-1}\right)\implies p^{n-1}=0$ which is absurd.

If $f(n)=p+1$ , then $p^n+(p+2)^n=(p+1)\left(p^{n-1}+(p+2)^{n-1}\right)$ which after expanding and manipulating becomes $p^{n-1}=(p+2)^{n-1}\implies n-1=0\implies n=1.$

Therefore $n=\boxed{1}$ is the only solution over all such $p$ .

EDIT : A way more easier solution: Let $x=p^n+(p+2)^n$ and $y=p^{n-1}+(p+2)^{n-1}$ . Now notice that: $x=(p+2)y-2p^{n-1}$ . If $y\mid x$ then $y\mid 2p^{n-1}$ . Then $y\le 2p^{n-1}\implies p^{n-1}\geq (p+2)^{n-1}$ which is true only for $n=1$ .

$\quad \ p^{n-1} + \left(p+2\right)^{n-1} \mid p^n + \left(p+2\right)^n \\ \Leftrightarrow p^{n-1} + \left(p+2\right)^{n-1} \mid p^n + \left(p+2\right)^n - \left(p+2\right)\left(p^{n-1} + \left(p+2\right)^{n-1}\right) \\ \Leftrightarrow p^{n-1} + \left(p+2\right)^{n-1} \mid 2p^{n-1} \\ \Rightarrow \left(p+2\right)^{n-1} \leq p^{n-1} \\ \Rightarrow n = 1$

Hence, the answer is $\boxed{1}$ .

Let $a|b$ denote that $a$ divides $b$ . Now $[p^{n-1}+(p+2)^{n-1}]|[p^{n}+(p+2)^{n}-(p+2)p^{n-1}+(p+2)(p+2)^{n-1}]=-2p^{n-1}$ . So, $[p^{n-1}+(p+2)^{n-1}]|[2p^{n-1}-p^{n-1}-(p+2)^{n-1}]=p^{n-1}-(p+2)^{n-1}$ . But $p^{n-1}+(p+2)^{n-1}> p^{n-1}-(p+2)^{n-1}$ . Thus $n-1=0$ which gives us the only solution.