Lacunary Polynomials Over Finite Fields by L. Rédei

By L. Rédei

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3)) a JIix-oc) = + *). xd-l (9> II, § 9] THE SOLUTION OF PROBLEM I 31 and therefore a it follows that F(x) = h(Xy-(-\y. Because of this and of (8), (7) turns into the polynomial equation h(x)d = x\x o +h(x)) + (-iy (10) (with the unknown h(x) and the parameter κ). Instead of solving this equation merely by comparing coefficients, we compare the sinkings of both sides with one another. This will lead us to the result much more quickly. Since, as a result of d\q — 1, d is prime top, Proposition 4 implies that (h(x)dy = hK On the other hand, the sinking of the right-hand side of (10) is by (9) equal to d n ' where we have taken into consideration that h(x) has, from (8), at least two terms.

Since, on the other hand, d>\, we find that έ/=2. (13) Moreover, it can be also inferred that only = is possible in (12), instead of ^ , and consequently that h° = ^ and therefore that %-l. e. (16) We shall now very soon achieve our aim. From (14) and (16), we can write (17) Substituting this into (10) and taking (13) into account we obtain Simple comparison of coefficients gives the system of equations As 2 ^ 0 , the second equation implies and thus the first and the third equations give Since, accordingly, 2μ = (1 +μ) 2 , we also have: for suitable v.

P1k; a€F\0), where i, j \ k, a are uniquely determined. As a consequence of the full reducibility of g(x), it follows that the binomial xk—cc must possess the same property. On the other hand (taking into consideration that p{k), let us calculate the greatest common divisor (xk-oc, (x*-a)') = ( a * - a , Jbc*-1) = 1. Accordingly, xk — a has no multiple zeros. This, when combined with the previous result, implies that xk — (x\xq — x. Because even holds, and thus xk-oi\xq-1-\ q-l k\q-\, a k = 1.

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