Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 .

share|improve this question

3 Answers 3

up vote 5 down vote accepted

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

share|improve this answer

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

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

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.