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Equation $(36)$ at Mathworld's Prim Sums page reads: $$ \sum_{k=1}^{p-1}\left\lfloor \frac{k^3}{p}\right\rfloor=\frac{(p-2)(p-1)(p+1)}4 $$ I'm curious how this can be proven, but I have no idea...

draks ...
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2 Answers2

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The idea is surely to compare the two sums $$ A=\sum_{k=1}^{p-1}\frac{k^3}p\qquad\text{and}\qquad B=\sum_{k=1}^{p-1}\left\lfloor\frac{k^3}p\right\rfloor. $$ We see that the difference $A-B$ equals the sum of least positive remainders of the cubes $k^3$ modulo $p$ divided by $p$.

I claim that the sum of those remainders is $p(p-1)/2$. If $p\not\equiv1\pmod 3$, then this follows from the fact that $k\mapsto k^3$ is a permutation of the elements of the group $\mathbb{Z}_p^*$. OTOH, if $p\equiv1\pmod3$, then $k\mapsto k^3$ is a 3-1 mapping attaining all cubic residues modulo $p$ as a value thrice. Because $-1$ is a cubic residue, the $(p-1)/3$ cubic residues come in pairs:$\{a,p-a\}$. The claim follows in this case as well.

Thus $$ A-B=\frac{p-1}2. $$ It is known that $$ A=\frac14p(p-1)^2. $$ The claim follows from the calculation $$ B=A-\frac{p-1}2=\frac14[p(p-1)^2-2(p-1)]=\frac14(p-1)(p^2-p-2). $$

Jyrki Lahtonen
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  • Just to make sure. I'm using on the fact that $k\mapsto k^3$ is a homomorphism from the cyclic group $\Bbb{Z}_p^*$ of order $p-1$ to itself. The two cases vary according to whether there are elements of order $3$, or equivalently whether $3\mid(p-1)$ or not. – Jyrki Lahtonen Sep 08 '13 at 05:27
  • Jyrki: could you look at my answer? Seems it cannot get simpler than $x^3+(p-x)^3\equiv 0\mod p$. – Pedro Sep 10 '13 at 19:28
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For each $1\leq k\leq p-1$ we can write $$k^3=q_kp+r_k$$and $$q_k=\left\lfloor\frac{k^3}{p}\right\rfloor\;\;,\;\;0\leqslant r_k<p-1$$

Now, we obtain that $$\sum_{k=1}^{p-1}k^3=p\sum_{k=1}^{p-1}\left\lfloor\frac{k^3}{p}\right\rfloor+\sum_{k=1}^{p-1}r_k$$

It is well known that $$\sum_{k=1}^{p-1}k^3=\left[\frac{p(p-1)}{2}\right]^2$$

so we get that $$\frac{{p{{\left( {p - 1} \right)}^2}}}{4} - \frac{1}{p}\sum\limits_{k = 1}^{p - 1} {{r_k}} = \sum\limits_{k = 1}^{p - 1} {\left\lfloor {\frac{{{k^3}}}{p}} \right\rfloor } $$ Thus, it remains to determine $$\sum_{k=1}^{p-1}r_k$$ which in light of your formula must be $$\frac{{p\left( {p - 1} \right)}}{2} = \sum\limits_{k = 1}^{p - 1} {{r_k}} $$

Now, we make the simple observation that, mod $p$, we have $$x^3+(p-x)^3=0\mod p$$

Since residues are $<p$ but here they sum to a multiple of $p$, this multiple must be $p$, whence we have $(p-1)/2$ pairs that sum to $p$, whence their sum must be $$\frac{p(p-1)}{2}$$ and we're done.

Pedro
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  • +1 Looks good to me now! I was tacitly using the facts that A) cubic residues form a subgroup of $\mathbb{Z}_p^*$, B) $-1$ is in that subgroup, C) when the subgroup is proper, all the elements occur as cubic residues exactly thrice. Anyway the idea is the same that the cubic residues form pairs that add up to $p$. – Jyrki Lahtonen Sep 10 '13 at 19:36
  • @JyrkiLahtonen I see. I was at loss since I didn't know such details! =) – Pedro Sep 10 '13 at 19:45