Let S3 = {p1⇔(¬p2), p2⇔(¬p3), p3⇔(¬p1)}. Using that a triangle does not admit a 2-coloring, show that the set S3 is not satisfiable.
Let n≥2. Show that the set S(n) defined by S(n) = {p1⇔(¬p2), p2⇔(¬p3), …, pn−1⇔(¬pn), pn⇔(¬p1)} is satisfiable if and only if n is even.
I think my solution for 1 is in the right direction but it does not look like a very strong/clear argument, however I'm not sure how to improve it in line with the question's intent.
I am very shaky with 2. I think it could be proven nicely by relating it to even/odd 2-regular cycles similar to the way 1 was proven but I lost at how to formulate this thinking in words.
Attempt at solution, 1:
Let ABC be a triangle:
A is color 1 iff B is color 2.
B is color 2 iff C is color 1.
C is color 1 iff A is color 2.
This is a contradiction stemming from the fact that ABC is not 2-colorable.
Let p1 be the node A colored 1. Let p2 be the node B colored 1. Let p3 be the node C colored 1. And let there be 2 colors. From above:
p1⇔(¬p2)
p2⇔(¬p3)
p3⇔(¬p1)
This set is equivalent to the previous set. The previous set is not satisfiable because a triangle is not 2 colorable. So the set S3 is not satisfiable.
Attempt at solution, 2:
First direction:
Let S(n) = {p1⇔(¬p2), p2⇔(¬p3), …, pn−1⇔(¬pn), pn⇔(¬p1)} be satisfiable.
Based on the fact that a cycle is 2 colorable if and only if the cycle has an even number of nodes, we know that n is even.
Other direction:
If n=2 then S(2) = {p1⇔(¬p2), p2⇔(¬p1)} is clearly satisfiable. If n=3 then S(3) is not satisfiable, as shown in 1.
Let k be an integer greater than 1. Assume the hypothesis holds for all n≤k.
Consider n+1.
If n+1 is even then the cycle is 2 colorable and S(n+1) is satisfiable.
if n+1 is odd then the cycle is not 2 colorable and S(n+1) is not satisfiable.
I think you have the right idea for (1), but I would restructure the proof. In particular, I would assume that S is satisfiable, and use that to produce a two-coloring of the triangle, which would be a contradiction. I'll get you started.
Assume that $S_3$ is satisfiable, so that there is a truth-assignment to the propositional variables $p_1,p_2,p_3$ such that every member of $S_3$ is true under this truth assignment. Fix such an assignment. Let $T$ be a triangle and label its vertices $v_1,v_2,v_3$. Define a coloring of $T$ by setting $v_i$ to be red if $p_i$ is true under the true under the truth assignment and is blue otherwise. I claim that this is a 2-coloring of $T$. Now go on to prove this claim.
This proof structure can be adapted to help with your (2).