If a is not 0, we can divide through by it and get a- (b+c)= 0 so a= b+c. Putting that into the first equation we get so .
Putting a= b+ c into the last eqaution gives or and, now, . That means we must have and so .
If c= -b, we have, from a-(b+c)= a= 0 again.
If c= b, a-(b+c)= 0 becomes a- 2b= 0 so a= 2b. Given any integer, k, take a= 2k, b= c= k. The equation becomes .
If n> 0, that is impossible. The only solution is a=0, b, c any integers.
If n= 0, that gives k= 0. That is a= b= c= 0, which is included in "a= 0, b, c any integers above".
If n< 0, so that which is integer only if n is -3 times a square: [tex]n= -3m^2[tex], in which case .
If or n< 0 is not of the form for some integer m, then a=0, b, c any numbers is the only solution.
If , then a= 2m, b= c= m.