Prime Factors Bounds in a Recurrence

[Matt Rolo]

Let $$g(n)$$ be the numerators of the elements of the recursion $$i(n)=i(n-1)+\frac{1}{i(n-1)}$$ when they are expressed in simplest form, with $$i(0)=1$$. Let $$p$$ be the smallest prime factor of $$g(m)$$. Show that $$p>4m-4$$.

Homework Equations

Euler's Theorem?

The Attempt at a Solution

OEIS yielded the following recursion for $$g(n)$$:
$$0 = g(n)^2(g(n+1) - g(n)^2) - (g(n+2) - g(n+1)^2)$$, which I found to be equivalent to $$g(n)=g^2(n-1)+(\prod_{i=0}^{n-2} g(i))^2$$. This seems suggestive of Euler's contradiction proof of infinitely many primes, but I'm not sure how to proceed.

 

[Matt Rolo]

A few observations: It’s quite clear that all $g$ for $n>2$ are congruent to 1 modulo 4. In addition I’ve found that all $g$ are relatively prime. It’s also not quite true that the $p$ increase as $m$ increases; for instance, 941 has $p$ of 941, while 969581 has $p$ of 521. Further the first few elements are part of Pythagorean triples:
$$5^2=3^2+4^2$$
$$29^2=20^2+21^2$$

$$941^2=580^2+741^2$$
$$969581^2=492100^2+835419^2=545780^2+801381^2$$
Each of which is primitive. This is of course necessary by Euclid’s formula, since each $g(n)$ is the sum of two squares, but the fact might be useful.
Still, I can’t put together a proof for the original problem

Here’s the first few $g(n)$:
1,2,5,29,941, 969581.
OEIS link:
https://oeis.org/search?q=1,2,5,29,941&sort=&language=english&go=Search