[Mathematical Analysis for small readers)

Set $ \ mathbb {p} $ to a number field. If $ P_1 (x), \ cdots, P_r (X) $ r$ on the number field $ \ mathbb {p} $ is an irrevocable polynomial with different Prime Minister coefficients of $1 $. Proof: $ f (x) = P_1 (X) \ cdots P_r (x) $ there is no duplicate root in the field $ \ mathbb {p} $.

 

Proof: Use the reverse verification method. if $ f (x) $ has $ K (\ geq 2) $ heavy root $ x = A $, then $ \ bee \ label {poly_div} f (x) = P_1 (x) \ cdots P_r (x) = (X-a) ^ 2G (x ). \ EEE $ order $ x = A $ \ Bex P_1 (a) \ cdots P_r (A) = 0. \ EEx $ and at least one $ I $ makes $ p_ I (A) = 0 $, so $ \ BEX (X-a) \ mid p_ I (X ). \ EEx $ since $ p_ I (x) $ is the first and not allowed, we have $ \ Bex p_ I (x) = x-. \ EEx $ replace the above formula into \ eqref {poly_div}, Which is simplified with $ \ Bex P_1 (x) \ cdots P _ {I-1} (X) P _ {I + 1} (x) \ cdots P_r (x) = (X-a) g (x ). \ EEx $ As demonstrated above, we can find that $ \ Bex \ exists \ J \ neq I, \ st p_j (x) = x-. \ EEx $ so $ p_ I (x), p_j (x) $ is the same. this contradiction is not true. therefore, there is a conclusion.