Let $\xi_{19} \in G_{19}$ be a primitive root of unity. Find $\Re\sum_{k=1}^9\xi_{19}^{k^2}$.

Question

Question

Let $\xi_{19} \in G_{19}$ be a primitive root of unity. Find $\Re\sum_{k=1}^9\xi_{19}^{k^2}$.

Attempt

I'm having doubts about how I'm solving this exercise, this is what I did:

$$ \Re\left(\sum_{k=1}^9\xi_{19}^{k^2}\right)=\sum_{k=1}^9\Re\left(\xi_{19}^{k^2}\right)\\ =\Re\xi^1+\Re\xi^4+\Re\xi^9+\Re\xi^{16}+\Re\xi^{25}+\Re\xi^{36}+\Re\xi^{49}+\Re\xi^{64}+\Re\xi^{81}\\ =\Re\xi^1+\Re\xi^4+\Re\xi^9+\Re\xi^{-3}+\Re\xi^{6}+\Re\xi^{-2}+\Re\xi^{-8}+\Re\xi^{7}+\Re\xi^{5}\\ =\Re\xi^1+\Re\xi^4+\Re\xi^9+\Re (\overline\xi)^{3}+\Re\xi^{6}+\Re(\overline\xi)^{2}+\Re(\overline \xi)^{8}+\Re\xi^{7}+\Re\xi^{5} $$

Also, using $\Re(z)=\Re(\overline z)$, the sum is then simplified to:

$$\Re \sum_{k=1}^{9}\xi^i=\Re \frac {\xi^{10}-\xi}{\xi-1} .$$

Is what I did thus far correct? Can this be simplified further?

I assume that $\xi$ is meant to be a complex number. The key here is that not all integers in the range $0\le i<19$ can be gotten as remainders squares $k^2$ - only so called quadratic residues (modulo 19) come out. Test this first with smaller primes such as $5$ or $7$ first to see what's going on. — Jyrki Lahtonen, Nov 29 '15 at 09:34
This is a special case of a Gauss sums. They are interesting objects. See my earlier answer about the case $p=7$. — Jyrki Lahtonen, Nov 29 '15 at 09:39
I expanded my explanation of my attempt. Could you guys tell me if I made any mistakes? — YoTengoUnLCD, Nov 29 '15 at 17:28
Ok. You only look at the real parts and the summation only goes up to 9. Sorry about "seeing what I'm used to seeing" :-) . As $-1$ is not a quadratic residue you do get all the exponents $1,2,3,4,5,6,7,8,9$ but with varying signs. A minor error is that because the geometric sum begins from $\xi$, the sum is $$\frac{\xi^{10}-\xi}{\xi-1}.$$ — Jyrki Lahtonen, Nov 29 '15 at 17:39
Thanks! I Fixed it. I still wonder, however, why doesn't this help in actually finding the value $-\frac 1 2$ which I should get... — YoTengoUnLCD, Nov 29 '15 at 17:56

Myself · Answer 1 · 2015-11-29T09:31:20.143

I'm not exactly sure how exactly you got rid of the squares in the exponents. That step definitely lacks justification and is probably incorrect.

Here's what you can do:

Let $A = \sum_{k=1}^{9}\xi_{19}^{k^2}$ and $B = \sum_{k=1}^{9}\xi_{19}^{-k^2}$. Note that the exponents in $A$ and $B$ are precisely the sets of all squares and non-squares mod $19$. (Excluding $0$.) Then we can verify that $$ AB = 9+4\sum_{k=0}^{18}\xi_{19}^{k} = 5,$$ by verifying that the pairwise sums of all squares and non-squares in $\mathbb F_{19}$ will give every $x\neq 0$ precisely $4$ times and $0$ precisely $9$ times.(*)

On the other hand $1+A+B = \sum_{k=0}^{18}\xi_{19}^{k} = 0$. So we may solve the system for $A$ and $B$. We find $$ A,B = \frac{-1\pm\sqrt{-19}}{2}, $$ and thus $\Re A=\Re B=-1/2$.

(*) I have checked this fact using the programming language Python, but it is probably not too hard to check it by hand.

L1 = sorted([(i^2)%19 for i in range(1,10)]]) L2 = sorted([(-i^2)%19 for i in range(1,10)]) sorted([(i+j)%19 for i in L1 for j in L2])

@JyrkiLahtonen A very good remark! Let me add that in this case, the theory of Gauss sums immediately tells us that $1+2A=\sum_{i=1}^{19}\xi^{i^2}=i\sqrt{19}$, from which one can immediately deduce that $\Re A=-1/2$. — Myself, Nov 29 '15 at 09:40

Let $\xi_{19} \in G_{19}$ be a primitive root of unity. Find $\Re\sum_{k=1}^9\xi_{19}^{k^2}$.

1 Answers1