sorry,
it would be so if there would be -1 instead of +1
–
Bhavya Jain
·
2 years, 2 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
·
2 years, 2 months 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 \(\boxed{x = 2k + 1, k \in \mathbb{N}.}\)
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Guilherme Dela Corte
·
2 years, 8 months ago

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@Guilherme Dela Corte
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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.
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Daniel Liu
·
2 years, 8 months ago

@Finn Hulse
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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 \( b=1\).

and you have proved it.
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Daniel Liu
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2 years, 8 months ago

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@Daniel Liu
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Oh shoot. I was just skimming, it didn't seem like much. Also dude do you play League?
–
Finn Hulse
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2 years, 8 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. ~
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汶汶 樂
·
2 years, 8 months ago

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TopNewestsorry, it would be so if there would be -1 instead of +1 – Bhavya Jain · 2 years, 2 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 · 2 years, 2 months 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 \(\boxed{x = 2k + 1, k \in \mathbb{N}.}\) – Guilherme Dela Corte · 2 years, 8 months ago

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Thankfully, this is pretty straightforward to do. – Daniel Liu · 2 years, 8 months ago

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– Finn Hulse · 2 years, 8 months ago

Could you explain your first assumption?Log in to reply

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.– Guilherme Dela Corte · 2 years, 8 months agoLog in to reply

please. This is the assumption I am referring to. – Finn Hulse · 2 years, 8 months ago

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and you have proved it. – Daniel Liu · 2 years, 8 months ago

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– Finn Hulse · 2 years, 8 months ago

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

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 · 2 years, 8 months ago

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