500 Followers Special

From $x^{\frac{1}{7}} + y^{\frac{1}{7}} + z^{\frac{1}{7}} = 12$ , we know that $x$ , $y$ and $z$ must be integers of perfect $7$ -power.

We also know that $y = x - 61741$ , $z = x -75938$ and $z < y < x < 12^7$ . We can list out $n^7$ for $n = 1,2,3...12$ and find for n such $(n^7 - 61741)$ and $(n^7 -75938)$ are also perfect $7$ -powers. It is readily done with a spreadsheet and mine is as follows:

$\begin {matrix} n & n^7 & n^7 -61741 & n^7-75938 \\ 1 & 1 & & \\ 2 & 128 & & \\ 3 & \boxed{2187} & & \\ 4 & \boxed{16384} & & \\ 5 &78125 & \boxed {16384} & \boxed {2187} \\ 6 & 279936 & 218195 & 203998 \\ 7 & 823543 & 761802 & 747605 \\ 8 & 2097152 & 2035411 & 2021214 \\ 9 & 4782969 & 4721228 & 4707031 \\ 10 & 10000000 & 9938259 & 9924062 \\ 11 &19487171 & 19425430 & 19411233 \\ 12 & 35831808 & 35770067 & 35755870 \end {matrix}$

It can be seen that $x = 5^7$ , $y = 4^7$ and $z = 3^7$ . Therefore,

$x+y+x = 5^7 + 4^7 + 3^7 = 78125+16384+2187 = \boxed{96696 }$

When we distribute 12 as sum of three integers, then from all the possibilities we can remove those sets of integers which have : 3 or 2 same integers. 1 as an integer. All even integers. Thus we get 3+4+5. As per conditions given in question , when crosschecked we will get x^1/7=5, y^1/7=4, z^1/7=3.

look the answer upside down