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Let $a$ be an element of some ring or field, possibly finite.

Is there closed form for $\sum_{i=1}^n{a^{i^2}}$?

sage and wolframalpha couldn't solve it.

If $a$ is primitive n-th root of unity this is Gauss sum and the result is $\sqrt{n}$ or $i\sqrt{n}$ for odd prime $n$.

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    $\begingroup$ You mean, if $a$ is a primitive $n$-th root of unity. $\endgroup$ Jan 6, 2021 at 11:40
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    $\begingroup$ Can you perhaps clarify in what class of expressions you consider a "closed form""? $\endgroup$ Jan 6, 2021 at 12:02
  • $\begingroup$ @GerryMyerson Thanks, indeed. I edited. $\endgroup$
    – joro
    Jan 6, 2021 at 12:23

2 Answers 2

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Some partial information.

If $g$ is a primitive root modulo prime $p$, then for $p\equiv 3 \pmod{4}$ $$\sum\limits_{x=1}^{p-1}g^{x^2}\equiv 0\pmod{p}.$$

If $p\equiv 1 \pmod{4}$ and $$S_1=\sum\limits_{x=1}^{p-1}g^{x^2},\quad S_2=\sum\limits_{x=1}^{p-1}g^{-x^2}$$ then $$S_1S_2+2\equiv 0\pmod{p}.$$

Sorry, I forget the source. Probably one can find related things in the papers of Arne Winterhof. For example he studied A Polynomial Representation of the Diffie-Hellman Mapping.

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  • $\begingroup$ Thanks. The general case is hopeless, so only partial results are possible. $\endgroup$
    – joro
    Jan 6, 2021 at 13:42
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I do not know of a closed form for this finite series. For an infinite series like this, we can use a Jacobi theta function $$ \vartheta_3\left( z,q \right) =1+2\,\sum _{n=1}^{\infty }{q}^{ {n}^{2}}\cos \left( 2\,nz \right) $$ with $z=0$.

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  • $\begingroup$ Thanks, probably other thetas might give similar results. $\endgroup$
    – joro
    Jan 6, 2021 at 12:25

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