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## 12S A Generalization of an Identity

12. PEN I10 Show that for all primes $p$,

$\sum_{k=1}^{p-1}{\left\lfloor\frac{k^{3}}{p}\right\rfloor}=\frac{(p+1)(p-1)(p-2)}{4}$.

Here is the official solution file: pen12.pdf
You can also discuss the problems here!

### 6 Responses

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