How can I prove that $3(\sum_{i=1}^{n}{\lfloor \frac{i^2}{n} \rfloor}) - n^2 \geq 2\ $ for all $n\geq 2?$

Asked Sep 19 '24 at 06:50

Active Sep 19 '24 at 16:33

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The OEIS sequence A175908 represents a sequence derived from this formula:

Using the Legendre $\operatorname{L}$-Function, I was able to prove that
- for primes $n$ of the form $4k + 1$, the value becomes $2$,
- and for primes of the form $4k + 3$, the value becomes a positive integer greater than or equal $\mbox{to}\ 2$.
However, I am unsure how to calculate this value for general composite numbers $n$.
For prime numbers, I could express this formula using the Legendre Symbol and complete the proof, but how should I approach it for composite numbers ?.

edited Sep 19 '24 at 16:33

Felix Marin

asked Sep 19 '24 at 06:50

추민서

You can't because it is false, take eg $n=1$ – Olivier Roche Sep 19 '24 at 09:06
it is correct as long as $n\ge 2$. – van der Wolf Sep 19 '24 at 09:42
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You should attach your attempts and sufficient context for this question. – Sahaj Sep 19 '24 at 11:38
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@OlivierRoche If n=1, left is 2. So it is also true when $n=1$ – 추민서 Sep 19 '24 at 12:50
Equivalently $\sum_{i=1}^{n}(i^2 \bmod n)\leq \frac{n(n-1)}{2}$, simply using $a\bmod b=a-b\cdot \lfloor \frac{a}{b} \rfloor$. Then search reveals some additional resources: Conjecture $ \sum _{i=1}^{n-1}\lfloor \frac{i^2}{n}\rfloor \ge \frac{\left(n-1\right)\left(n-2\right)}3$ where $n\in\mathbb N^*$, also asked on MO – Sil Sep 19 '24 at 16:22
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David Speyer's proof at MO works for $n$ odd and square-free, and hints that same approach should work for all $n$ with some modifications (though it seems on one worked that out explicitly). – Sil Sep 19 '24 at 16:29
I tried with an elementary approach, but it didn't work. I think heavier machinery is required here, probably like the ones in the link. – Adam Rubinson Sep 19 '24 at 16:36

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