Problem of the Week #216 - Jul 19, 2016

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Euge

MHB Global Moderator
Staff member
Here is this week's POTW:

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Let $p$ be a prime greater than $3$. Compute the sum of the quadratic residues in $\Bbb Z/p\Bbb Z$.

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Euge

MHB Global Moderator
Staff member
Opalg submitted a correct solution to the problem, although he admits it comes from another. Here is the solution.

Since $a^2 = (p-a)^2$ in $\mathbb{Z}/p\mathbb{Z}$, the quadratic residues are the squares of the elements in the "first half" of $\mathbb{Z}/p\mathbb{Z}$, namely the elements $1^2, 2^2,\ldots, k^2$, where $k = \frac12(p-1).$ Their sum is therefore $$\sum_{r=1}^k r^2 = \tfrac16k(k+1)(2k+1) = \tfrac1{24}(p-1)(p+1)p.$$ Since $p>3$, and the only prime factors of $24$ are $2$ and $3$, it follows that this sum is a multiple of $p$ and is therefore the zero element of $\mathbb{Z}/p\mathbb{Z}.$

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