Let A={1^(p^2 −p) , 2^(p^2 −p) ,..., (p−1)^(p^2 −p) , p^2 −p+1} where p is any prime number Prove that for any value of p, however we split this set into two disjunctive sets, the arithmetic means of all elements of both sets cannot be equal to each other.


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Question Number 182203 by depressiveshrek last updated on 05/Dec/22
$Let A={1^(p^2 −p) , 2^(p^2 −p) ,..., (p−1)^(p^2 −p) , p^2 −p+1} where p is any prime number Prove that for any value of p, however we split this set into two disjunctive sets, the arithmetic means of all elements of both sets cannot be equal to each other.$
$${Let}\:{A}=\left\{\mathrm{1}^{{p}^{\mathrm{2}} −{p}} ,\:\mathrm{2}^{{p}^{\mathrm{2}} −{p}} ,...,\:\left({p}−\mathrm{1}\right)^{{p}^{\mathrm{2}} −{p}} ,\:{p}^{\mathrm{2}} −{p}+\mathrm{1}\right\} \\ $$$${where}\:{p}\:{is}\:{any}\:{prime}\:{number} \\ $$$${Prove}\:{that}\:{for}\:{any}\:{value}\:{of}\:{p}, \\ $$$${however}\:{we}\:{split}\:{this}\:{set}\:{into}\:{two} \\ $$$${disjunctive}\:{sets},\:{the}\:{arithmetic} \\ $$$${means}\:{of}\:{all}\:{elements}\:{of}\:{both}\:{sets} \\ $$$${cannot}\:{be}\:{equal}\:{to}\:{each}\:{other}. \\ $$
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