Positive integers expressed in the form : $\frac{a^b+c}{a+c}$ Note : This problem has no specific source .
Is it true that :

Every positive integer $n$ greater than $1$ can be expressed in the form :
$n=\frac{a^b+c}{a+c}$ , where $a,b>1$ , and $c \in \mathbb{Z} \backslash \{0\}$

I am able to prove following :
$\forall a , \exists ~b,c$ such that : $a+c \mid a^b+c$ , where $b>1$
Proof :
$n=\frac{a^b+c}{a+c}=\frac{a^b-a}{a+c}+1$ , therefore it is sufficient to prove :
$a+c \mid a(a^{b-1}-1)$
If $a+c \mid a$ then $b$ can be any positive integer .
If $a+c \not \mid a$ then we have to prove : $a+c \mid a^{b-1}-1$
or , in other words :
$a^{b-1} \equiv 1 \pmod {a+c}$
Now , for every $a$ there is a number $c$ such that : $\gcd(a,a+c)=1$
According to Euler Theorem :
$a^{\varphi(a+c)} \equiv 1 \pmod {a+c}$
Hence , we can write :
$b-1=\varphi(a+c) \Rightarrow b=1+\varphi(a+c)$
Since $\varphi(a+c)\geq 1$ it follows $b>1$
Q.E.D.
But , this is only necessary condition .

Let $a = n$, $b = 3$, $c = n^2$. This does it.
