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For arbitrarily large $k$, can you pick a set of $k$ primes $p_i$ satisfying $$p_i \equiv 1 \pmod 8 \text{ for all } i $$ and $$\left(\frac{p_i}{p_j} \right) = 1 \quad \text{ for all } i \neq j$$

My guess is that you can, because if you have picked $n$ such primes $p_1 < p_2 < \dotsm < p_n$, then among the infinitely many primes $p$ with $p > p_n$ and $p \equiv 1 \pmod 8$, the Legendre symbol conditions $\left(\frac{p}{p_i} \right)$ should be quite random (I think), and so eventually you will find a prime where all the $\left(\frac{p}{p_i} \right) = 1$ and thus managed to increase the size of your set of primes by one.

Other than this intuition I have no idea how to approach this question.

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You don't need to depend on randomness. You can choose $p_{n+1}$ to be $1$ modulo $8p_1p_2\cdots p_n$ so $p_{n+1}$ will be a square modulo each of the previously selected $p_i$. And they will be squares modulo it, since they're all $1$ modulo $4$.

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  • $\begingroup$ When you put it like that it seems so obvious hahaha thank you! $\endgroup$
    – eatfood
    Sep 1 '20 at 7:44
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Hint: If you have picked $p_1<p_2<\cdots<p_n$, any prime $p$ that is $1$ modulo $p_1\cdots p_n$ will be a quadratic residue modulo $p_i$ for each $i$.

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    $\begingroup$ Thank you for putting the time to make an answer! This definitely answers my question but both you and Gerry posted similar answers at the same time so I just flipped a coin to decide. $\endgroup$
    – eatfood
    Sep 1 '20 at 7:43
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    $\begingroup$ @eatfood Makes sense. Glad I was able to help :) $\endgroup$ Sep 1 '20 at 7:47

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