The problem is the amplification factor, and the voltage of the source, and/or the power supply.
The gain of the differential amplifier is: \[\frac{V_{out}}{V_{2}-V_{1}} = (1+\frac{2*R_{2}}{R_{1}})*\frac{R_{6}}{R_{4}}\], with the resistor values of your circuit you get: \[\frac{V_{out}}{V_{2}-V_{1}} = 4\], the output you will get is: \[V_{out} = 4*({V_{2}-V_{1}})\].
keith1200rs is trying to say is that if you put a source of 10V, you will get an output signal of 40V, and this is greater than your power supply (+5V and -5V, it is 10V). That's the reason of the square wave, the amplifier is saturated, because they can't go outside the limits impossed by the power supply.
To solve this, you can try lowering the amplification factor of the differential amplifier (changing the values of the resistors), or lowering your input signal, or using a greater power supply. ¡What solution is the best?, it depends of your particular situation.
Regards