JOMO 6, Long 1

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

Note by Yan Yau Cheng
5 years, 3 months ago

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

bhavya jain - 4 years, 9 months ago

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

bhavya jain - 4 years, 9 months ago

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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. ~

汶汶 樂 - 5 years, 2 months ago

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

Guilherme Dela Corte - 5 years, 3 months ago

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

Thankfully, this is pretty straightforward to do.

Daniel Liu - 5 years, 2 months ago

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Could you explain your first assumption?

Finn Hulse - 5 years, 3 months ago

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Finn, please note that a2k+1+b2k+1a+b=a2k+a2k1b+a2k2b2++a2b2k2+ab2k1+b2k\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,ba,b are integers, we have then that ac+bca+b\dfrac{a^c + b^c }{a+b} is an integer for odd = uneven values of cc. Now let a=ca=c and b=1 b=1.

Note: math fixed by mod.

Guilherme Dela Corte - 5 years, 2 months ago

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@Guilherme Dela Corte 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.

Finn Hulse - 5 years, 2 months ago

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@Finn Hulse The proof is right there: just clear denominators of

a2k+1+b2k+1a+b=a2k+a2k1b+a2k2b2++a2b2k2+ab2k1+b2k\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.

Daniel Liu - 5 years, 2 months ago

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

Finn Hulse - 5 years, 2 months ago

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