A question about understanding a BJT differential amplifier?
If the circuit is symmetric (both components and excitations), then the current from the current source will be divided equally between the left and the right branches. Therefore, even though Vin+ and Vin- both increase, the current through Q1 will stay approximately constant at 0.5mA. This requires that the emitter voltage increases to keep the base-emitter voltage approximately constant.

If the excitations are instead differential, then the current through one transistor will increase and the current through the other will decrease, so that the sum is always equal to 1mA. This results in a large change in Vout.

EDIT:

Suppose that Vin+ increases and Vin- decreases. The current through the BJT on the left will therefore increase, and the current through the BJT on the left will decrease. Due to symmetry, it turns out that the emitter voltage will be constant with differential excitations. Then we can find $$\Delta I_C = g_m \Delta V_{b}.$$
On the left, this change in collector current will be positive, and on the left negative. The gain of the amplifier is then (roughly):
$$ A_v = g_m R_1,$$
since the output voltage will change by the product of the resistor and the change in collector current.