More details in addition to the last post.

1) subtracting the 1st equation from the 2nd one gives:

(b-a)*(b+a-2c)＝2011

2) 2011 is prime, so there are only two possibilities:

b-a=1, b+a-2c=2011; or

b-a=2011, b+a-2c=1

3) subtracting the 1st equation from the 3rd equation gives:

(c-a)(c+a-2b)=x-1

4) plugging c-a=0.5*[(b-a)-(b+a-2c)], c+a-2b=(c-a)-2(b-a) into 3) gives you x

2011 is prime