The first challenge is to spot the neat trick.

1 The first equation

X^2 = A^2 + 4B + 1

Consider (A + 1)^2 will equal A^2 + 2A + 1

Hence if we solve for X being 1 + A then 4B +1 = 2A + 1 Hence A = 2B (This is the neat trick)

2 We now need to discount X being A + n where n is greater than 1.

If n = 2 then A^2 + 4A + 4

the 2B +1 = 4A + 4 hence 2B = 4A + 3 this can be discounted as B would be greater than A.

3 The second equation using the same logic as above B = 2C

4 Hence A = 2B = 4C

5 Then start with the third formula find the value of Z^2 which has an integer square root.

6 Then do the same with the other equations.

7 The answer is C = 24, B=48 and A=96 giving Z=31, Y=49 and X-97

8 NOTE there is almost but not quite another solution C, B, A = 4, 8 ,16 give Z, Y, X = 8, 8, 17.