Differential case,
as the input is differential, voltage at emitters is zero (so no AC current through the emitter resistor), Ic1 = -Ic2 (assuming hfe>>1)
Seen from the inputs, you have two transistors in series, so the overall Gm is half the value of each transistor. You can also say that each transistor experiences half the input voltage.
So Ic1 = 0.5*gm*(v1-v2), vo1 = -RC*0.5*gm*vdiff vdiff = v1-v2. Assuming hfe>>>1, ic2 = -ic1 (as there is only DC through Re),
So v02 = Rc*0.5*gm*vdiff, v01-v02 = vout = -Rc*gm*vdiff, Ad = -Rc*Gm.
the common mode case, here both inputs are connected together and you can see this as a single-transistor case with an emitter resistor and a collector resistor of 0.5*Rc.
Gm of the equivalent transistor is now twice as high as they are in parallel. The input voltage divides across both Re and 1/(2*gm).
Ic1+Ic2 = Ic = vin/( 1/(2*gm) + Re ) So vout = -0.5*Rc*vin/( 1/(2*gm) + Re ), hence the gain is Acm = -0.5*Rc/( 1/(2*gm) + Re )
Multiply with gm/gm (that is 1) gives:Acm = -0.5*Rc*gm/(0.5 + gm*Re) = -Rc*gm/(1+2*gm*Re).
When 2*gm*Re >>1, then Acm = -Rc/(2*Re) = -0.5*Rc/Re.
As you can see, this looks like the equation for a single BJT stage with emitter degeneration.
I hope if there is a typo, somebody will respond.