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A user on MSE, @martin , asked http://math.stackexchange.com/questions/1611411/pell-equations-upper-bound about an upper bound for $x$ in $x^2 - p y^2 = 1,$ when $p$ is prime. I checked, it appears reasonable to guess that $$ x < p^{\sqrt p} $$ when $p > 2.$ I had the computer solve by Lagrange's method, no continued fractions, no decimal accuracy required, no memory required, but the method is still elementary. I had the machine print out whenever $\log_p(\log_p(x))$ increased. It was necessary to take $p > 2$ because $x=3$ gives an overly large logarithm. Meanwhile, if all we do is print whenever $x$ itself increases, there are several composite numbers below $100$ that get included, after that they give way to primes $p \equiv 1 \pmod 4.$ I put in a fair amount of effort but was unable to draw any firm conclusions.

A user on MSE, @martin , asked https://math.stackexchange.com/questions/1611411/pell-equations-upper-bound about an upper bound for $x$ in $x^2 - p y^2 = 1,$ when $p$ is prime. I checked, it appears reasonable to guess that $$ x < p^{\sqrt p} $$ when $p > 2.$ I had the computer solve by Lagrange's method, no continued fractions, no decimal accuracy required, no memory required, but the method is still elementary. I had the machine print out whenever $\log_p(\log_p(x))$ increased. It was necessary to take $p > 2$ because $x=3$ gives an overly large logarithm. Meanwhile, if all we do is print whenever $x$ itself increases, there are several composite numbers below $100$ that get included, after that they give way to primes $p \equiv 1 \pmod 4.$ I put in a fair amount of effort but was unable to draw any firm conclusions.

Someone asked on MSE about an upper bound for $x$ in $x^2 - p y^2 = 1,$ when $p$ is prime. I checked, it appears reasonable to guess that $$ x < p^{\sqrt p} $$ when $p > 2.$ I had the computer solve by Lagrange's method, no continued fractions, no decimal accuracy required, no memory required, but the method is still elementary. I had the machine print out whenever $\log_p(\log_p(x))$ increased. It was necessary to take $p > 2$ because $x=3$ gives an overly large logarithm. Meanwhile, if all we do is print whenever $x$ itself increases, there are several composite numbers below $100$ that get included, after that they give way to primes $p \equiv 1 \pmod 4.$

A user on MSE, @martin , asked http://math.stackexchange.com/questions/1611411/pell-equations-upper-bound about an upper bound for $x$ in $x^2 - p y^2 = 1,$ when $p$ is prime. I checked, it appears reasonable to guess that $$ x < p^{\sqrt p} $$ when $p > 2.$ I had the computer solve by Lagrange's method, no continued fractions, no decimal accuracy required, no memory required, but the method is still elementary. I had the machine print out whenever $\log_p(\log_p(x))$ increased. It was necessary to take $p > 2$ because $x=3$ gives an overly large logarithm. Meanwhile, if all we do is print whenever $x$ itself increases, there are several composite numbers below $100$ that get included, after that they give way to primes $p \equiv 1 \pmod 4.$ I put in a fair amount of effort but was unable to draw any firm conclusions.

Why not, here is how it begins if we print every time $x$ increases and make no requirement about loglog, taking $x^2 - n y^2$ with $2 \leq n \leq 500$

Why not, here is how it begins if we print every time $x$ increases and make no requirement about loglog, allowing $n$ composite in $x^2 - n y^2$ with $2 \leq n \leq 500$