[tex]\begin{cases} \rm {a}^{2} + abcd = 2a \\ \rm b {}^{2} + bcda = 2 b\\ \rm c {}^{2} + cdab = 2c\\ \rm d {}^{2} + dabc = 2d \end{cases}[/tex]

a,b,c,d are roots to quadratic equation

[tex] \rm x^2-2x+t = 0[/tex]

, where t=abcd

If a,b,c,d all equal, then

[tex] \rm {a + a}^{2} = 2[/tex]

[tex] \rm {a}^{2} + a - 2 = 0[/tex]

[tex] \rm (a - 1)( {a}^{2} + a + 2) = 0[/tex]

[tex] \therefore \rm a = 1[/tex]

otherwise WOWG, If a [tex]\ne[/tex] b, then a+b=2, ab=t.

However, abcd=t, so cd=1

1) c [tex]\ne d [/tex]

2) c=d

Then cd=t, so t=1,

Our equation is [tex] \rm {x}^{2} - 2x + 1 = 0 [/tex]

[tex] \rm(x - {1)}^{2} = 0[/tex]

x=1

a=b=c=d=1

[tex] \because \rm cd = 1 \\ \therefore \rm c = d = 1 \: or \: c = d = - 1[/tex]

[tex]\begin{cases} \rm a + b = 2 \\ \rm ab = 1 \end{cases} \\ \rm a = b = 1[/tex]

[tex]\begin{cases} \rm a + b = 2 \\ \rm ab = - 3 \end{cases} \\ \rm a = 3 ,b = 1[/tex]

[tex] \rm(a,b,c,d) = (1,1,1,1) \: or \: (3, - 1, - 1, - 1)[/tex]