3 variable diophantine system

Solution: We see that since $a^3 = b+ c = a + b + b$ (from equation 1) that $a^3 = a + 2b$ Also, we plug in $a+b = c$ into equation 2 to get $a^4 = a^2 + 2ab$ . This is equivalent to our previous equation, so in our system of equations one equation can be discarded as it is built off the others. So we discard $a^4 + b^2 = c^2$ Now with our remaining solutions and our positive integers restriction we can plug in a=2,3,4,5 to get (a,b,c)=(2,3,5)(3,12,15)(4,30,34)(5,60,65). If a=6 then we get (6,105, 111) which is greater than 1--. So then our answer is $2+3+5+3+12+15+4+30+34+5+60+65=238$ .

$\begin{cases} a + b = c & ...(1) \\ a^4 + b^2 = c^2 & ...(2) \\ a^3 = b + c & ...(3) \end{cases}$

\(\begin{array} {} (1)+(2): & a^3 + a = 2c & \Rightarrow c = \frac{1}{2}(a^3+a) \\ (2)-(1): & a^3 - a = 2b & \Rightarrow b = \frac{1}{2}(a^3-a) \end{array} \)

Since $c \le 100 \quad \Rightarrow \frac{1}{2}(a^3+a) \le 100 \quad \Rightarrow a^3+a \le 200 \quad \Rightarrow a \le 5$ since $\sqrt[3]{200} \approx 5.8$ . We note that $b = \frac{1}{2}(a^3-a)$ . If $a = 1$ , $b = 0$ which is unacceptable. Therefore, $a \in \{ 2,3,4,5\}$ . And we have:

\(\begin{array} {} a & + & b & + & c & & \\ 2& + & 3& + & 5& = & 10 \\ 3& + & 12& + & 15& = & 30 \\ 4& + & 30& + & 34& = & 68 \\ 5& + & 60& + & 65& = & 130 \end{array} \)

The sum of all possible values of $a+b+c$ is $10+30+68+130 = \boxed{238}$

from #1: a =c - b

from #3: a^3 = c + b

b = 1/2 *a(a^2 - 1)

c = 1/2 * a(a^2 + 1)

a+b+c = a +a^3

for a = 2, b = 3, c = 5

for a = 3, b = 12, c = 15

for a = 4, b = 30, c = 34

for a = 5, b = 60, c = 65

for a = 6, b = 105, c = 111, this is rejected.