By Takashi Ono (auth.)

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Let p be a prime and a be an integer such that p f a. Let e = (n, p -1), n EN, and write p -1 = ef. Then, we have xn == a (p) has a solution ¢:> af == 1 (p ). When the equation has a solution, the number of solutions (modp) is e. PROOF. 20, to solve xn == a (p) is equivalent to solving ny == b (p - 1), a linear equation, where y = indr x, b = indr a. 8). If that is so, dividing both sides of ny == b (p -1) by (e), the equation n'y == b' (f), n = en', b = eb', has a unique solution Yo modulo f. Then e numbers Yo, Yo + f, ...

Then, we have xn == a (p) has a solution ¢:> af == 1 (p ). When the equation has a solution, the number of solutions (modp) is e. PROOF. 20, to solve xn == a (p) is equivalent to solving ny == b (p - 1), a linear equation, where y = indr x, b = indr a. 8). If that is so, dividing both sides of ny == b (p -1) by (e), the equation n'y == b' (f), n = en', b = eb', has a unique solution Yo modulo f. Then e numbers Yo, Yo + f, ... ,Yo + (e -l)f make up solutions of ny == b (p -1). Finally, we have a f == 1 (p) ¢:> rbf == 1 (p) ¢:> (p - 1) I bf ¢:>e lb.

Multiplying b3- 1 on both sides, we have + b1 b3- 1 a"-1 + .. ·bn _ 1b3- 1 1X + bn b3- 1 = (bolXt + b1(bolXt- 1 + ... + bn _ 1b3- Z(b olX) + b n b3-t, 0= b3a" which means that boa- E lL. D. 41. Prove that Q. lL c (J). (Here Q. lL = {aa-; a E 0, a- E lL}. 20. Let Yl, ... , Yl E ([: be complex numbers not all 0 and letM be the module in ([: generated by them: M = {Y = ~~=l CiYi; Ci ElL}. If a complex number a- E ([: has the property a-Mt: M, then a- is an algebraic integer: a- E lL. PROOF. By the assumption, we have a-Yi = ~;=l cijYj, cij ElL, 1 ~ i ~ l.