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Is the following true (and if yes, where the best proof is written?)?

For any $c>0$ for large enough positive integers $N$ we have $\sum_{k=0}^{N-1} \cos(k^2t)\geqslant -cN$ for all real $t$?

Hm, if true, it should be hard: it allows to get signs of certain Gauss type sums, for example.

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  • $\begingroup$ It is hard to believe that the Gaus-sums have nonnegative real part... Maple shows that $\sum_{k=0}^{q-1}\exp(2 \pi i \frac{p}{q} k^{2})$ has negative real part for $q=5$ and $p=2$. $\endgroup$
    – Paata Ivanishvili
    Commented Jul 21, 2016 at 17:42

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The answer is negative.

As you said if you take $\frac{t}{2\pi}$ to be a rational number, say $\frac{p}{q}$ then it converges to the Gaus-sums: $$ \lim_{N \to \infty} \frac{1}{N} \sum_{k=0}^{N-1} \cos(k^{2} t) = \Re \frac{1}{q} \sum_{k=0}^{q-1} \exp\left( 2 \pi i \cdot \frac{p}{q} k^{2}\right) $$ Now if you take $t=\frac{4 \pi}{5}$ (i.e., $p=2, q=5$) I think it should no be difficult to show that the latter sum is a negative number $< - 0.4$ (even without taking the real part).

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