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# 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
2 years, 8 months ago

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sorry, it would be so if there would be -1 instead of +1 · 2 years, 2 months ago

It can be expressed as a sum of a GP for all values of x belong to natural number. · 2 years, 2 months ago

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 $$\boxed{x = 2k + 1, k \in \mathbb{N}.}$$ · 2 years, 8 months 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. · 2 years, 8 months ago

Could you explain your first assumption? · 2 years, 8 months 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. · 2 years, 8 months 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. · 2 years, 8 months 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. · 2 years, 8 months ago

Oh shoot. I was just skimming, it didn't seem like much. Also dude do you play League? · 2 years, 8 months ago