Relation between positive integers greater than 2

Question

$x_1, x_2,\ldots, x_{96}$ are positive integers greater than 2, which satisfy the relation: $$ \frac{1}{x_1^4}+\frac{1}{x_2^4}+\cdots+\frac{1}{x_{96}^4}=\frac{1}{6}$$ I have two questions: 1. At least 8 numbers are equal? 2. Which numbers satisfy this relation?

Answer 1

If for each $n\ge 3$ there are at most $7$ indices $i$ with $x_i=n$, then we can bound the left hand sum by $$\frac1{x_1^4}+\ldots \frac1{x_{96}^4}<7\cdot\sum_{n=3}^{14} \frac1{n^4}\approx 0.138<\frac16$$ This is because we may assume wlog. that $x_1\le x_2\le\ldots\le x_{96}$. Then by the assumption of no more than seven equal numbers we get $x_{k+7}\ge x_k+1$ and hence, starting from $x_1\ge 3$ the estimate.