Cyclically 91

Since 91 is not a perfect square, all variables are non-zero. Not all of them can be positive, since from $b\ge a$ and $c\ge a$ we would get in that case $a^2-bc \leq 0 < 91$ . Not all of them can be negative either, because if $(a,b,c)$ is a solution, then so is $(-c,-b,-a)$ . It follows that $c > 0$ and $a < 0$ , and we can restrict our search to positive $b$ 's.

Now observe that $91(b - c) = b(b^2-ac)-c(c^2-ab)=b^3-c^3,$ so $91 = \frac{b^3-c^3}{b-c} = b^2 + bc + c^2=\left(c+\frac{b}{2}\right)^2+\frac{3b^2}{4} \geq \left(\frac{3b}{2} \right)^2 + \frac{3b^2}{4} = 3b^2 .$ This means that $91 \ge 3 b^2$ , so $b \in\{1,2,3,4,5\}$ . The only values that result in integral $c$ are $b=1,c=9$ and $b=5,c=6$ . This leads to the solutions $a=-10,b=1,c=9$ and $a=-11,b=1,c=6$ , and the other two that can be obtained from negating all variables. The sum of all $c$ s is then $9+6+10+11=36$ .

(a-b)(a+b+c) = a^2 - b^2 + ac - bc = 0

(b-c)(a+b+c) = 0

(c-a)(a+b+c) = 0

Note that if a+b+c is not equal to 0, then a = b = c; but 91 = a^2 - bc = 0 is a contradiction.

Thus we have a+b+c = 0

(b+c)^2 - bc = 91

b^2 + c^2 + bc = 91

Clearly, a <= b <= 0 <= c or a<= 0 <= b <= c. If a <= 0 <= b <= c. If b = 0, ac = - 91; an impossibility since a = -c.

Thus either a <= b < 0 <= c or a<= 0 < b <= c.

Every solution (a,b,c) corresponds to another valid solution (-c,-b,-a). Thus it suffices to consider the cases where b and c >0

Going back to b^2 + c^2 + bc = 91, simple exhausting of cases gives us:

(-10,1,9) (-11,5,6)

as the only solutions with b and c >0. We thus have (-9,-1,10) and (-6,-5,11) are the other solutions. Summing gives 36

Subtract any 2 of the equations, eg, 1st and 2nd. (a-b) (a+b+c)=0 If possible, let a=b and further, =k. Hence, subtracting other 2nd and 3rd, (b-c) (a+b+c)=0, gives c= -2k <since a=b=c is not possible as it gives 0=91 upon substituting in given equation> Further, putting c= -2k and a= k in 1st equation, we get 3 {k^2} = 91, which is not possible as k is an integer. Thus our only possibility is a+b+c=0 and none of the two being equal. Now, ab= c^2 -91, and a+b= -c. Thus, x^2 + cx + c^2-91 = 0 has roots a and b. Hence its discriminant should be an integer. 364- 3 {c^2} is a perfect square. We proceed by trial and error with values of c starting from c=1 until this becomes negative. (Note that if c= x satisfies, then c= -x shall also satisfy) We evaluate both the roots of the equation, and in each case, choose the greatest among c and the roots, as a possible value of c. To further shorten this trial and error procedure, suppose we have already obtained the roots x and y for some value c= z. Now we can avoid going for c= x and c= y as we know owing to the symmetry between a, b, and c, they too must satisfy the equations. The obtained triplets are: (-10, -1, 9), (-9, 1, 10), (-6, 5, 11) (-11, -5, 6)

Subtracting the second equation from the first one, we get $(a-b)(a+b+c)=0.$ Similarly, subtracting the third equation from the second and the first from the third, we get $(b-c)(a+b+c)=0$ and $(c-a)(a+b+c)=0.$ If $a+b+c\neq 0,$ then $a=b=c,$ which implies that $a^2-bc=a^2-a^2=0.$ Because $a^2-bc=91,$ this is impossible, so $a+b+c=0.$ Thinking of $a$ as a parameter, we have the following system of equations for $b$ and $c$ : $\begin{cases} b+c=-a\\ bc=a^2-91 \end{cases}$ Note that the original system is equivalent to this one, because when $a+b+c=0$ all three equations become equivalent.

From the Vieta's formula, $b$ and $c$ are roots of the following quadratic equation in $x$ : $x^2+ax+(a^2-91)=0.$ For the roots to be integers, the discriminant $a^2-4(a^2-91)=364-3a^2$ must be a complete square.

Suppose $364-3a^2=m^2,$ $m\geq 0.$ Then, because $b\leq c,$ $b=\frac{-a-m}{2},\ c=\frac{-a+m}{2}.$ Because $a\leq b \leq c$ and $a+b+c=0,$ $a\leq 0.$ Because $3a^2=364-m^2\leq 364,$ $|a|\leq 11.$

Case 1. $a=-11.$ Then $m=1,$ and we get $(a,b,c)=(-11,5,6).$

Case 2. $a=-10.$ Then $m=8,$ and we get $(a,b,c)=(-10,1,9).$

Case 3. $a=-9.$ Then $m=11,$ and we get $(a,b,c)=(-9,-1,10).$

Case 4. $a=-8.$ Then $364-3a^2$ is not a complete square, no solutions.

Case 5. $a=-7.$ Then $364-3a^2$ is not a complete square, no solutions.

Case 6. $a=-6.$ Then $m=16,$ and we get $(a,b,c)=(-6,-5,11).$

Case 7. $a=-5.$ Then $m=17,$ and we get $b=-6,$ which is less than $a$ , so no solutions.

Case 8. $a=-4.$ Then $364-3a^2$ is not a complete square, no solutions.

Case 9. $a=-3.$ Then $364-3a^2$ is not a complete square, no solutions.

Case 10. $a=-2.$ Then $364-3a^2$ is not a complete square, no solutions.

Case 11. $a=-1.$ Then $m=19,$ and we get $b=-9,$ which is less than $a$ , so no solutions.

Case 12. $a=0.$ Then $364-3a^2$ is not a complete square, no solutions.

So, overall we got four solutions: $(-11,5,6),\ (-10,1,9),\ (-9,-1,10),\ (-6,-5,11).$ Therefore, the answer is $6+9+10+11=36.$

a²-b²-bc+ac=(a+b)(a-b)+c(a-b)=(a-b)(a+b+c)=0,

we get, a=b or a+b+c=0;

Similarly, b²-c²-ac+ab=(b+c)(b-c)+a(b-c)=(b-c)(a+b+c)=0,

b=c, or a+b+c=0;

if a=b=c, we obtain 0=91 (not possible), so a≠b≠c and a+b+c=0,

c=-(a+b);

since a≤b≤c,

therefore, a<0, a+b<0, c>0.

and we have c²-ab=(a+b)²-ab=91,

solving the indefinite equation,

all possible different values of a,b,c that satisfies the equations are

a=-11, b=5, c=6

a=-10, b=1, c=9

a=-9, b=-1, c=10

a=-6, b=-5, c=11

the sum of integers c = 6+9+10+11=36

6+9+10+11=36