Finite sum of reciprocals of squares of integers modulo primes of the form 6k + 1

Posted: January 17, 2024 in Basic Number Theory
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Let’s begin this post with an extension of the concept of divisibility of integers by a prime number to divisibility of positive rational numbers by a prime number.

Definition. Let p be a prime number, and r=\frac{a}{b}, where a,b are positive integers and p \nmid b. Then we say that p \mid r or, equivalently, r \equiv 0 \mod p if p \mid a.

The following problem was posted on the Art of Problem Solving website a couple of years ago and remained unsolved until I saw it a couple of months ago! haha …

Problem. Let k be a positive integer such that p=6k+1 is a prime number. Show that

\displaystyle \sum_{i=1}^{2k}\frac{1}{i^2}\equiv 0 \mod p \iff \sum_{i=1}^{k}\frac{1}{i^2} \equiv 0 \mod p.

Solution. I wrote the problem here as it was posted on the AoPS website but I think a much better problem is to prove that

\displaystyle 5\sum_{i=1}^{2k}\frac{1}{i^2}-\sum_{i=1}^k\frac{1}{i^2} \equiv 0 \mod p, \ \ \ \ \ \ \ \ \ \ (*)

which obviously solves the problem posted on AoPS too.

Proof of (*). For a positive integer i < p, we’ll write i^{-1} for the inverse of i modulo p. Also, for simplicity, we’ll write \equiv instead of \equiv \mod p. It is clear that \frac{1}{i} \equiv i^{-1}. Now

\displaystyle \sum_{i=1}^{p-1}\frac{1}{i^2} \equiv \sum_{i=1}^{p-1}(i^{-1})^2=\sum_{i=1}^{p-1}i^2=\frac{p(p-1)(2p-1)}{6}=k(2p-1)p \equiv 0

and so

\displaystyle 2\sum_{i=1}^{3k}\frac{1}{i^2}=2\sum_{i=1}^{\frac{p-1}{2}}\frac{1}{i^2} \equiv \sum_{i=1}^{p-1}\frac{1}{i^2} \equiv 0,

which gives

\displaystyle \sum_{i=1}^{3k}\frac{1}{i^2} \equiv 0. \ \ \ \ \ \ \ \ \ \ \ (1)

On the other hand,

\displaystyle \sum_{i=1}^{3k}\frac{1}{i^2}=\sum_{i=1}^{2k}\frac{1}{i^2}+\sum_{i=2k+1}^{3k}\frac{1}{i^2}=\sum_{i=1}^{2k}\frac{1}{i^2}+\sum_{i=1}^{k}\frac{1}{(3k+1-i)^2} \equiv \sum_{i=1}^{2k}\frac{1}{i^2} + \sum_{i=1}^{k}\frac{1}{(2^{-1}(2i-1))^2}

\displaystyle \equiv \sum_{i=1}^{2k}\frac{1}{i^2} + 4\sum_{i=1}^{k}\frac{1}{(2i-1)^2} =\sum_{i=1}^{2k}\frac{1}{i^2} + 4\left(\sum_{i=1}^{2k}\frac{1}{i^2}-\sum_{i=1}^k\frac{1}{(2i)^2}\right)

\displaystyle =5\sum_{i=1}^{2k}\frac{1}{i^2}-\sum_{i=1}^k\frac{1}{i^2}. \ \ \ \ \ \ \ \ \ \ \ (2)

The equivalence (*) now follows from (1),(2). \ \Box

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