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2 Add conjecture and evidence that $(l/p)=(l/3)$ is unique

[Added later] In fact this seems to be the only construction, in thefollowing sense:

Conjecture. For every prime $l \neq 3$ there exists $N$ with the following property: for all primes $p>N$ such that $(l/p) \neq (l/3)$ there is some $n \in \lbrace 3, 4, \ldots, N \rbrace$ such that each of $n$, $n-1$, and $n-2$ is a quadratic residue of $p$.

For example, if $l \in \lbrace 2, 5, 7, 11, 13, 17 \rbrace$ thenwe can take $N=121$. For $19 \leq l \leq 43$ we can use $N = 325$,and $N = 376$ works for $l=47$ and several larger $l$.

This can be checked as follows. For a positive integer $n$ let$s(n)$ be the unique squarefree number such that $n/s(n)$ is a square;e.g. for $n=24,25,26,27,28$ we have $s(n)=6,1,26,3,7$ respectively.Then $(n/p) = (s(n)/p)$ for all $p>n$.Given a small set $S$ of primes containing $l$ and a bound $N$,let $\cal N\phantom.$ be the set of all $n \in \lbrace 3, 4, \ldots, N \rbrace$such that each of $s(n)$, $s(n-1)$, and $s(n-2)$ is a product of primes in $S$.Now try all $2^{|S|}$ ways to assign $\pm 1$ to each $(l'/p)$ with $l' \in S$,and see which ones make at least one of $s(n),s(n-1),s(n-2)$a quadratic nonresidue for each $n \in \cal N$.

For $S = \lbrace 2, 3, 5, 7, 11, 13, 17 \rbrace$ and $N = 121$, we compute$${\cal N} = \lbrace 3, 4, 5, \ldots, 17, 18, 22, 26, 27, 28, 34, 35, 36, 50,51, 52, 56, 65, 66, 100, 121 \rbrace,$$and find that the only choices that work are the two that make$(l/p) = (l/3)$ for each $l \in S - \lbrace 3 \rbrace$.

Then if we put $l=19$ into $S$ and increase $N$ to $325$ we find that${\cal N} \ni 325$, with$323 = 17 \cdot 19$, $324 = 18^2$, and $325 = 13 \cdot 5^2$. Sothe only way to avoid $(323/p) = (324/p) = (325/p) = 1$ is to make$(19/p) = +1$. We then incorporate $l=23$ by considering $n=92$,and $l=29$ using $n=290$, "etc." Computation suggests that there arelots of choices to make this work once we get past $l=19$,but I don't know how feasible it might be to prove this.

[The exhaustive computation over $2^{|S|}$ choices of $(l'/p)$is what led me to the pattern $(l/p) = (l/3)$ in the first place.Once only two choices remained for $S = \lbrace 2, 3, 5, 7, 11, 13, 17 \rbrace$I thought that a few more primes might whittle it down to zero anddisprove the claim, but I kept seeing only two choices that differedonly in the value of $(3/p)$, and the pattern in the other $(l/p)$ valuessoon became clear.]

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By Dirichlet's theorem, there exists $p>N$ such that each prime $l\leq N$, with the exception of $l=3$, satisfies $(l/p) = (l/3)$. I claim that this $p$ is a counterexample. Indeed by multiplicativity $(m/p) = (m/3)$ for each $m \leq N$ that is not a multiple of 3. In particular $(m/p) = -1$ if $m \equiv -1 \bmod 3$. Each triple ${ n, n-1, n-2 }$ with $n \leq N$ contains one such $m$, and therefore cannot comprise three quadratic residues of $p$, QED.

What's the context? Seems rather tricky for homework; hope it's not a problem from an ongoing contest...