Today I found a strange phenomenon that I want to ask about.
If $$f(n)=4729494n+1,$$ is square, where $n$ is positive integers. Then I found $n=4729492$, because $$f(4729492)=4729493^2$$ In fact, note $$(x-1)(x+1)+1=x^2$$
Now I conjecture: $$n_{\min}=4729492?$$
No, it's $n=17372$, because $4729494\cdot 17372 +1 = 286637^2$.
The question can be restated as solving $x^2 \equiv 1 \bmod 4729494$.
Since $4729494 =2 \cdot 3 \cdot 7 \cdot 11 \cdot 29 \cdot 353$, you first solve $x^2 \equiv 1 \bmod p$ for each of these primes. The solution is of course $x \equiv \pm1 \bmod p$ and they can be combined using the Chinese remainder theorem. It's a lot of work to do by hand because there are $2^5$ solutions mod $4729494$, one for each combination of signs $\pm1$. It's easier to write a small program and search for the smallest solution of the original problem, which is what I did.