Subtracting the first two equations from each other gives x2−x3=x32−x22+6x4(x3−x2), which we can factor as 0=(x3−x2)(x3+x2+1+6x4). We see that x2=x3 or x2+x3+1+6x4=0. Similarly, it also holds that x2=x3 or x2+x3+1+6x1=0. Therefore, if x2=x3, then in both cases the second equality holds; subtracting these from each other gives x1=x4. We conclude that x2=x3 or x1=x4. Similarly, for any permutation (i,j,k,l) of (1,2,3,4), it holds that xi=xj or xk=xl.
We now prove that at least three of the xi are equal. If they are all equal, this is obviously true. Otherwise, there are two that are unequal, say without loss of generality x1=x2. Then x3=x4. If now x1=x3, then there are three elements equal. Otherwise, x1=x3, so x2=x4 and again there are three elements equal. The quadruple (x1,x2,x3,x4) is thus, up to order, of the form (x,x,x,y), where it may be that x=y.
Substituting this into the given equations gives x+y=8x2 and 2x=x2+y2+6xy. Adding these two equations: 3x+y=9x2+y2+6xy. The right-hand side can be factored as (3x+y)2. With s=3x+y we thus have s=s2, from which it follows that s=0 or s=1. It holds that s=3x+y=2x+(x+y)=2x+8x2. Therefore, 8x2+2x=0 or 8x2+2x=1.
In the first case, x=0 or x=−41. We find y=0−3x=0 respectively y=0−3x=43. In the second case, we can factor the equation as (4x−1)(2x+1)=0, so x=41 or x=−21. We find y=1−3x=41 respectively y=1−3x=25.
All together, we have found the following quadruples: (0,0,0,0),(−41,−41,−41,43) and the permutations thereof, (41,41,41,41) and (−21,−21,−21,25) and the permutations thereof. Checking shows that all these quadruples satisfy the equations.