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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 $b>1$


But , this is only necessary condition .

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up vote 5 down vote accepted

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

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Hint $\ $ If $\rm\:b,c\:$ are independent of $\rm\:n\:$ and $\rm\:a\:$ is a polynomial in $\rm\:n\:$ of degree $\rm\:d\:$ then comparing degrees in $\rm\:a^b+c = n\:(a+c)\:$ $\Rightarrow$ $\rm\:db = d+1,\:$ so $\rm\:d(b-1) = 1,\:$ hence $\rm\:d=1,\:b=2.\:$ Plugging in undetermined coefficients $\rm\:a\: =\: a_1\: n + a_0 \:$ yields $\rm\:a_1 = 1,\:a_0 = -1,\:$ hence

$$\rm\ \frac{a^b+c}{a+c}\ =\ \frac{(n-1)^2 - 1}{(n-1)-1}\ =\ \frac{n^2 - 2n}{n-2}\ =\ n$$

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Do you know Fermat's Little Theorem? It is a useful tool here...

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