Sum of integers divide sum of integer powers

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Moon
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Sum of integers divide sum of integer powers

Unread post by Moon » Tue Dec 14, 2010 12:29 am

Show that if $k$ is odd,
\[ 1+2+\cdots+n \]
divides
\[1^k +2^k +\cdots+n^k\]
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Masum
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Re: Sum of integers divide sum of integer powers

Unread post by Masum » Wed Dec 15, 2010 11:41 am

$1+2+....+n=\frac {n(n+1)} 2$
It is well known that $a+b|a^k+b^k$ for $k$ odd.
Using this $n+1|n^k+1^k,(n-1)^k+2^k,......$
$n|n^k,(n-1)^k+1^k,...$ and $gcd(n,n+1)=1$,so if both of them divides this then there product divide $2(1^k+....+n^k)$
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Dipan
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Re: Sum of integers divide sum of integer powers

Unread post by Dipan » Sun Dec 19, 2010 9:43 am

I am posting a very easy solution....it can't be accepted by all....

we know that if k is odd then ...

a^k + b^k/a + b

so why we can't say that

a^k + b^k + ............. +n^k/a + b +.............+n [a,b belong to N and k is an odd]

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Masum
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Re: Sum of integers divide sum of integer powers

Unread post by Masum » Sun Dec 19, 2010 4:56 pm

I don't think it is true.How did you get this?
One one thing is neutral in the universe, that is $0$.

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