We are given two equations involving positive integers $a$, $b$, and $c$:

1. $a^2 + b - c = 100$
2. $a + b^2 - c = 124$

Our goal is to find the values of $a$, $b$, and $c$.

Step 1: Express $c$ from both equations

From equation (1): $c = a^2 + b - 100$ From equation (2): $c = a + b^2 - 124$

Now we can set these two expressions for $c$ equal to each other: $a^2 + b - 100 = a + b^2 - 124$

Step 2: Simplify the equation

Rearrange terms to move everything to one side: $a^2 - a + b - b^2 = -124 + 100$ $a^2 - a + b - b^2 = -24$ This is the equation we need to solve.

Step 3: Try values for $a$ and $b$

Let's try different values of $a$ and $b$, as the equation is quadratic and involves integers.

Try $a = 5$:

$5^2 - 5 + b - b^2 = -24$ $25 - 5 + b - b^2 = -24$ $20 + b - b^2 = -24$ $b - b^2 = -44$ $b^2 - b - 44 = 0$ Solve this quadratic equation using the quadratic formula: $b = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-44)}}{2(1)}$ $b = \frac{1 \pm \sqrt{1 + 176}}{2}$ $b = \frac{1 \pm \sqrt{177}}{2}$

Since $\sqrt{177}$ is not an integer, $a = 5$ does not work.

Try $a = 6$:

$6^2 - 6 + b - b^2 = -24$ $36 - 6 + b - b^2 = -24$ $30 + b - b^2 = -24$ $b - b^2 = -54$ $b^2 - b - 54 = 0$ Solve this quadratic equation using the quadratic formula: $b = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-54)}}{2(1)}$ $b = \frac{1 \pm \sqrt{1 + 216}}{2}$ $b = \frac{1 \pm \sqrt{217}}{2}$

Again, $\sqrt{217}$ is not an integer, so $a = 6$ does not work.

Try $a = 7$:

$7^2 - 7 + b - b^2 = -24$ $49 - 7 + b - b^2 = -24$ $42 + b - b^2 = -24$ $b - b^2 = -66$ $b^2 - b - 66 = 0$ Solve this quadratic equation using the quadratic formula: $b = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-66)}}{2(1)}$ $b = \frac{1 \pm \sqrt{1 + 264}}{2}$ $b = \frac{1 \pm \sqrt{265}}{2}$

Since $\sqrt{265}$ is not an integer either, $a = 7$ does not work.

Try $a = 8$:

$8^2 - 8 + b - b^2 = -24$ $64 - 8 + b - b^2 = -24$ $56 + b - b^2 = -24$ $b - b^2 = -80$ $b^2 - b - 80 = 0$ Solve this quadratic equation using the quadratic formula: $b = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-80)}}{2(1)}$ $b = \frac{1 \pm \sqrt{1 + 320}}{2}$ $b = \frac{1 \pm \sqrt{321}}{2}$

Since $\sqrt{321}$ is also not an integer, $a = 8$ does not work.

At this point, trying different values