## 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

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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Joined: Tue Dec 07, 2010 1:12 pm

### Re: Sum of integers divide sum of integer powers

$1+2+....+n=\frac {n(n+1)} 2$
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 $2(1^k+....+n^k)$
One one thing is neutral in the universe, that is $0$.

Dipan
Posts: 158
Joined: Wed Dec 08, 2010 5:36 pm

### Re: Sum of integers divide sum of integer powers

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]

Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm
One one thing is neutral in the universe, that is $0$.