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It is better to ask one question per post. Here is the answer to Q1. Assume that $(x,y)$ is a positive integer solution of $(1)$. Then $(x,ay)$ is a solution of $X^2-aY^2=1$. Let us restrict to $a=u^2 …

Let us denote $$\sigma_\ell:=\sum_{i=1}^\ell q_i\qquad\text{and}\qquad\tau_\ell:=\sum_{i=\ell+1}^s\frac{1}{q_i}.$$ Then $$\prod_{\ell=1}^{s-1}\left(\frac{q_\ell}{q_{\ell+1}}\cdot\frac{\sigma_{\ell+1}} …

Such a lower bound is not possible, even when $k=1$. Indeed, there exists a sequence $(n_1,n_2,\dotsc)$ containing every natural number such that for infinitely many positive integers $N$ we have $n_i …

Let us denote the left hand side of $(1)$ by $\psi(x)$. It is known that $|\psi(x)-x|$ is not bounded by a constant times $x^{1/2}$. In fact Littlewood (1914) proved that $$\psi(x)-x=\Omega_{\pm}(x^{1 …

Your conclusion is valid, and for this we only need to assume that $\varphi(3n)$ is divisible by $3$. Indeed, this weaker condition is equivalent to the existence of a prime divisor $p\mid n$ that is …

As hinted in Alex Kruckman's comment, it is easier to work with $$N'(a) = \left|\{x\in \{1,\dots,a-1\} : \sqrt{x(a-x)}\in \mathbb{Z}\}\right|,$$ which is the same as $N(a)$ up to a factor of $2$. It i …