# Pure Strategy Nash Equilibria Repeated Game

I am trying to solve the above problem. I know that the two Nash Equilibria are $$(D,L)$$ and $$(C,R)$$, so the first four pure strategy Nash equilibria would just involve playing either of these strategies in both periods. I understand that, for all other possible pure strategy Nash equilibria, the second/last game must still end with one of these Nash equilibria as well but could start in the first game with something that is not a Nash equilibria.

However, I am somewhat confused by what the other two Nash equilibria are.

One of them is:

If $$(U,L)$$ is played in the first period, then $$(D,L)$$ is played in the second period. Otherwise, $$(C,R)$$ is played in the second period.

The other one is:

If $$(C,M)$$ is played in the first period, then $$(C,R)$$ is played in the second period. Otherwise, $$(D,L)$$ is played in the second period.

I have noted the discount factor of $$\frac{3}{4}$$, so understand that if, for instance if $$(U,L)$$ is played in the first period and $$(D,L)$$ in the second, then the payoff for the row player would be $$5+\frac{3}{4}\left(6\right)\:=\:9.5$$, but I am not sure how to apply this here, although I know I have to.

Any help would be highly appreciated!

"If $$(U,L)$$ is played in the first period, then $$(D,L)$$ is played in the second period. Otherwise, $$(C,R)$$ is played in the second period."
The idea here is that Row player "sacrifices" something in the first period (he gets $$5$$ instead of $$6$$) so that his favorite NE will be played in the second period. The $$(C,R)$$ equilibrium is the way the column player "punishes" him for deviations.
So the payoff of the row player in this scenario is indeed $$5+\tfrac{3}{4}\cdot 6=9.5$$. If he deviates, however, and plays $$D$$ in the first round, he gets $$6+\tfrac{3}{4}\cdot 4=9$$. You can repeat this calculation for Player 2 and verify that this is indeed an equilibrium.