# Can you pick a set of $k$ primes $p_i$ with all $p_i \equiv 1 \pmod 8$ and $(\frac{p_i}{p_j})=1$ for all $i \neq j$?

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.

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 $$4$$.
Hint: If you have picked $$p_1, any prime $$p$$ that is $$1$$ modulo $$p_1\cdots p_n$$ will be a quadratic residue modulo $$p_i$$ for each $$i$$.