Show that if $k$ is odd,
\[ 1+2+\cdots+n \]
divides
\[1^k +2^k +\cdots+n^k\]
Sum of integers divide sum of integer powers
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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 $2(1^k+....+n^k)$
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 $0$.
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]
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]
Re: Sum of integers divide sum of integer powers
I don't think it is true.How did you get this?
One one thing is neutral in the universe, that is $0$.