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I'm working on a proof for my number theory course and I am a bit confused on how to prove a certain case of the question..

  • If $n$ is a odd positive integer or if $n$ is divisible by $4$ then $$1^3 + 2^3 + 3^3 + ... + (n-1)^3 \equiv 0 \pmod n$$is this statement true if $n$ is even but not divisible by $ 4$?

So for the case "$n$ is a positive integer divisible by $4$" I set $n = 4k$ and try to plug $n$ into $\frac{n^{2}(n-1)^{2}}{4}$

But I can't seem find a way to simplify it and otherwise prove its congruent to $0 \pmod n$.

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    If $n=4k$ then $n^2=n\times n=4k\times n$. Thus $\frac {n^2}4=k\times n$. – 2017-02-17
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    I'm so sorry, I get to kn(n-1)^2 but how does that show that it is congruent to 0 mod n? – 2017-02-17
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    $k\times n\times (n-1)^2$ is obviously a multiple of $n$. – 2017-02-17

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If $n=4k$ then $n^2 = 4kn$; hence $n^2 (n-1)^2 / 4 = n \cdot k \cdot (n-1)^2$ which is divisible by n.

If n is odd, $(n-1)^2$ is 0 mod 4, so $n^2(n-1)^2/4 = n (n(n-1)^2/4) which is 0 mod n.