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For any prime $p \equiv 1 \pmod{5}$ do there exist 5 integers $\{a_0, \dots, a_4\}$, each of absolute value less than $p$, satisfying $\sum_{i=0}^{4}{a_i}=p$, $(\sum_{i=0}^{4}{a_ig^i})(\sum_{i=0}^{4}{a_ig^{-i}})=p^2$ AND the $a_i$ are coset representatives of roots of a polynomial $x^5-\alpha$ in $\mathbb{Z/pZ}[x]$? Above $g$ is the generator of the cyclic group of order 5 written multiplicatively. Such integers are easily found by computer for primes less than 10,000, is there a general proof?

A similar statement about primes $p \equiv 1\pmod{4}$ is easily proven using Fermat's theorem on sums of squares.

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Use \pmod{x} to produce $\pmod{x}$. –  Arturo Magidin Dec 9 '11 at 22:09
The statement needs some work. First, $g$ is a generator, not the generator, since there are several. Due to the symmetries, it doesn't matter which. Second, are we to assume that all those $g^i$ and $g^{-i}$ have been reduced modulo $p$ to the interval from $1$ to $p-1$? If so, that should be stated in the problem. Third, the description of the $a_i$ is unclear. I think you mean that they are the roots, taken as integers between $1$ and $p-1$, of some equation $x^5\equiv\alpha\pmod p$ for some $\alpha$. –  Gerry Myerson Dec 10 '11 at 4:54
Agree with @GerryMyerson in that the problem description needs to be improved. It is impossible to pick all the numbers in the range $1\ldots p-1$. For then we would surely have $\sum a_ig^i>\sum a_i$ as well as $\sum a_ig^{-i}>\sum a_i$, because the non-trivial powers of $g$ are all $>1$. Fomin, please clarify which (if any) variables are constrained to what range, and which (if any) can be freely selected within their respective cosets modulo $p$. Alternatively, show us the argument in $p\equiv 1\pmod 4$ case, so that we have a chance to decipher your intent. –  Jyrki Lahtonen Dec 11 '11 at 6:01
To elaborate: If we look at this from the finite field side alone, then it is clear that the sets $\{a_0,a_1,\ldots,a_4\}$, $\{a_ig^i\mid i=0,1,\ldots,4\}$ and $\{a_ig^{-i}\mid i=0,1,\ldots,4\}$ all coincide, because they consist of the zeros of the polynomial $x^5-\alpha$. IOW they all cover the same five cosets modulo $p$. The question is all about, whether we can select the representatives in such a way that they sum up to $(\pm) p$ for all three sets. The choices for the representatives of $a_i$:s are limited by the absolute value, but is $g^{-i}$ the same as $g^{5-i}$. Normally yes, but? –  Jyrki Lahtonen Dec 13 '11 at 6:22
Oops! My preceding comment has a point, but it may easily happen (unless we are careful), that the powers of $g$ introduce repetitions among the five roots. That may be the whole point. Thinking... –  Jyrki Lahtonen Dec 13 '11 at 10:44

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