I am just reading at MSE some links to tryouts done to prove Legendre's conjecture, all of them containing (as kindly explained by other MSE users) dubious points, and also reviewed the paper by Chen in 1975, "On the distribution of almost primes in an interval" that seems to be so far (please correct me if this is not true) the closest accepted fact regarding the conjecture, which is that, at least, there is an almost prime $pq$ where $p,q \in \Bbb P$ in a $[n^2,(n+1)^2]$ interval (it would also be true for $[n^2,(n+1)^2-1]$).
I would like to know if the following approach has been tried: trying to prove first that at least there is a Fermat's base $2$ prime (applying Fermat's little theorem for $a=2$, this is $2^{n^2+t}\equiv{2} \pmod{n^2+t}$ for some $t \lt 2n+1$) in the interval $[n^2,(n+1)^2]$. "Fermat's base $2$ prime" means that there is at least a number inside the interval that pass the primality test of Fermat's little theorem for $a=2$. That would include true primes or Fermat pseudoprimes in the interval.
These are some thoughts about the topic, and the questions are at the end:
(a) If we assume that the conjecture is true for the interval $[n^2,(n+1)^2-1]$ then the following would hold:
$$\prod_{t \lt 2n+1} (2^{n^2+t}-(\lfloor\frac{2^{n^2+t}}{n^2+t}\rfloor(n^2+t))-2)=0$$
The expression means that the product of the moduli should be $0$ if one "Fermat base $2$ prime" exists in the interval (be aware that the above expression shows $2^{n^2+t}-2 \equiv{0} \pmod{n^2+t}$ for some $t \lt 2n+1$).
The expression would be true, assuming $\lfloor\frac{2^{n^2+t}}{n^2+t}\rfloor \gt \frac{2^{n^2+t}}{n^2+t-1}$, only if (let us assume a big $n$ initially but for a small $n$ I think it should be fine as well):
$$\prod_{t \lt 2n+1} (2^{n^2+t}-((\frac{2^{n^2+t}}{n^2+t-1})(n^2+t))-2) \lt 0$$
Reordering:
$$\prod_{t \lt 2n+1} (2^{n^2+t}-((2^{n^2+t})(\frac{n^2+t}{n^2+t-1}))-2) \lt 0$$
The term $(\frac{n^2+t}{n^2+t-1})$ is monotone decreasing and its limit is $1$ so basically the whole term $2^{n^2+t}-(2^{n^2+t})(\frac{n^2+t}{n^2+t-1})$ is very close to $1$ as $n \to \infty$, so the quantity less $2$ will always be negative. As the quantity of the elements in $t \lt 2n+1$ is an odd number (from $0$ to $2n$ there are $2n+1$ elements), the whole product is negative.
(b) In the other hand, if we assume that the conjecture is false for the interval $[n^2,(n+1)^2-1]$ then the following would hold:
$$\prod_{t \lt 2n+1} (2^{n^2+t}-(\lfloor\frac{2^{n^2+t}}{n^2+t}\rfloor(n^2+t))-2) \gt 0$$
But in this case, assuming $\lfloor\frac{2^{n^2+t}}{n^2+t}\rfloor \gt \frac{2^{n^2+t}}{n^2+t-1}$ would it be possible for the following expression being positive for the interval?:
$$\prod_{t \lt 2n+1} (2^{n^2+t}-((2^{n^2+t})(\frac{n^2+t}{n^2+t-1}))-2)$$
I am not sure if this kind of approach might show at least empirically something useful or not, but the idea of playing with the moduli of the whole interval seemed interesting, at least to think about it.
I would like to ask the following questions:
Can Legendre's conjecture be studied from the perspective of a Fermat's little theorem problem (a weaker version of the conjecture)? e.g. to obtain some facts, at least empirically, about the existence of Fermat's base $2$ primes in quadratic intervals?
Is it possible to conclude something about the existence of at least Fermat's "base $2$" (primes or pseudoprimes) in $[n^2,(n+1)^2]$ intervals, for a big $n$? Are there some papers regarding this topic? thank you!