Must the number k be the same for all triples?

For the first part just take $a=-\frac{16}{7}, b=-\frac{9}{7}, c=\frac{5}{7} $ ( we will work only modulo $p$) As for the second part note by FLT that p divides $f_k(a,b,c)$ if and only if p divides $\frac{a}{(b-c)^{k-1}}+\frac{b}{(c-a)^{k-1}}+\frac{c}{(a-b)^{k-1}}$ Claim: for $k=3$ p divides $f_k(a,b,c)$ provided that p divides $f_2(a,b,c)$\ Solution: well this is just algebra We have p divides $(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})*(\frac{1}{b-c}+\frac{1}{c-a}+\frac{1}{a-b})=(\sum_{cyc} \frac{a}{(b-c)^{2}})+\sum_{cyc} \frac{a(c-b)}{(b-c)(c-a)(a-b)}=\sum_{cyc} \frac{a}{(b-c)^{2}}$ and the claim is done Claim(that finishes the problem): p never divides $f_4(a,b,c)$ so the answer will be 4 Solution:Suppose by contradiction that p divides $\sum_{cyc} \frac{a}{(b-c)^{3}}$ Look at the following product (which is divisible by p) $(\sum_{cyc} \frac{a}{(b-c)^{2}})*(\sum_{cyc} \frac{1}{b-c}))=(\sum_{cyc} \frac{a}{(b-c)^{3}})+ \sum_{cyc} ((\frac{a}{(b-c)^{2}})*(\frac{1}{c-a}+\frac{1}{a-b}))$ and after some calculation you will get that $p$ divides $a+b+c$ Look again at $f_2(a,b,c)$ after you make a common denominator and using that $p$ divides $a+b+c$ you will get that $p$ divides $9abc$ so p divides one of a,b,c which is a contradiction (sorry for not posting all calculation but they are easy and annoying) and we are done