# JOMO 6, Long 1

Find, with proof, all values of $x\in\mathbb{N}$ such that $\frac{x^x+1}{x+1}$ is a natural number.

Note by Yan Yau Cheng
7 years ago

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For uneven values of $y$, the expression $\dfrac{x^y+1}{x+1}$ is always a natural number. Having now $x=y$, we have that our desired expression $\dfrac{x^x+1}{x+1}$ is a natural number for $\boxed{x = 2k + 1, k \in \mathbb{N}.}$

- 7 years ago

Could you explain your first assumption?

- 7 years ago

Finn, please note that $\dfrac{a^{2k+1} + b^{2k+1} }{a+b} = a^{2k} + a^{2k-1} b + a^{2k-2} b^2 + \cdots + a^2 b^{2k-2} + a b^{2k-1} +b^{2k}$ (you can quickly check this by trying some small cases).

Since both $a,b$ are integers, we have then that $\dfrac{a^c + b^c }{a+b}$ is an integer for odd = uneven values of $c$. Now let $a=c$ and $b=1$.

Note: math fixed by mod.

- 7 years ago

Guilherme, I would prefer a proof rather than

"(you can quickly check this by trying some small cases)"

please. This is the assumption I am referring to.

- 7 years ago

The proof is right there: just clear denominators of

$\dfrac{a^{2k+1} + b^{2k+1} }{a+b} = a^{2k} + a^{2k-1} b + a^{2k-2} b^2 + \cdots + a^2 b^{2k-2} + a b^{2k-1} +b^{2k}$

and you have proved it.

- 7 years ago

Oh shoot. I was just skimming, it didn't seem like much. Also dude do you play League?

- 7 years ago

There is one flaw in your proof: you did not prove that the number $\dfrac{x^x+1}{x+1}$ is not an integer or all even $x$.

Thankfully, this is pretty straightforward to do.

- 7 years ago

It can be expressed as a sum of a GP for all values of x belong to natural number.

- 6 years, 6 months ago

sorry, it would be so if there would be -1 instead of +1

- 6 years, 6 months ago

For a positive integer x,n (where n=x+1), L=(1+x^x)/(1+x)= [1+(n-1)^(n-1)]/[n]. By remainder theorem, (n-1)^(n-1)=n(m)+(-1)^(n-1), for some integer m. L=[n(m)+((-1)^(n-1)+1)]/n. Since L is integer, (-1)^(n-1)+1 should be 0. That means, n-1 should be a odd number. Therefore, x=n-1 should be a odd number. ~

- 7 years ago