Branch 1 |a+b|, |b+c|, |c+d| and |d+a| are integers, so there are two possibilities:

3 is 0, 1 is 2.

Two are 0 and two are 1.

So |a+d| can only take 0, 1 and 2.

If it is 2, then

|a+b|=|b+c|=|c+d|=0

It is not difficult to get a =-d, so | a+d | = 0, which contradicts the assumption that | a+d | = 2.

So |a+d| can only take 0 and 1.

A=0，b=0，c=- 1，d= 1 | A+D | = 1

| a+d | = 0 when a =- 1, b = 0, c = 0 and d = 1.