Rom

your answer to (1) is correct

You all want to post chatty stuff... Fine.

Put the off-topic flag on it so those of us who aren't interested can ignore it.

Thank you.

what is a re-randomization?

\(\text{You start with a configuration of 3 charges separated by a distance }s\\ \text{By symmetry each pair of charges generates the same energy and thus }\\ \text{each configuration generates }\dfrac{15}{3} = 5J\\ \text{Now we move the charge to the midpoint between two other charges}\\ \text{Our new configuration is one pair separated by distance }s\\ \text{and 2 other pairs separated by distance }\dfrac s 2\)

\(\text{The pair separated by }s \text{ generates }5J \text{ as before}\\ \text{The energy stored is inversely proportional to the distance between them}\\ \text{so halving the distance doubles the energy}\\ \text{thus we have two pairs that generate }2 \cdot 5J = 10J \text{ each}\\ \text{Totaling these up we have } E = (5+10+10) = 25J\)

Do you intend a finite width setup as pictured?

In this case the pegs at either end only deflect back towards the center.

I get 6

\(\dfrac 1 e = e^{-1}\\ \ln\left(e^{-1}\right) = -1\\ \ln\left(\dfrac 1 e\right) = -1\)

\(\text{the roots are }\\ x = \dfrac{m \pm \sqrt{m^2-4n}}{2}\)

\(\text{as these are integers }m^2 - 4n \text{ must be a perfect square}\\ m<20 \Rightarrow m^2 \leq 19^2 = 361\\ 4n

It appears there are indeed 90 valid values of n

done

Just because this board has the off-topic flag doesn't mean that it's facebook or reddit.

This site is here to allow folks to get "help" with their math problems. 

There are a zillion other sites out there for idle chatter.

Use one of them.