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Solving equations over finite fields is an important problem from both theoretical and practice points of view. The problem of solving explicitly the equation P a ( X ) = 0 over the finite field F Q , where P a ( X ) : = X q + 1 + X + a , Q = p n , q = p k , a ∈ F Q ⁎ and p is a prime, arises in many different contexts including finite geometry, the inverse Galois problem [1] , the construction of difference sets with Singer parameters [9] , determining cross-correlation between m-sequences [10] and to construct error correcting codes [5] , cryptographic APN functions [6] , [7] , designs [21] , as well as to speed up the index calculus method for computing discrete logarithms on finite fields [11] , [12] and on algebraic curves [18] . In fact, the research on this specific problem has a long history of more than a half-century from the year 1967 when Berlekamp, Rumsey and Solomon [2] firstly considered a very particular case with k = 1 and p = 2 . In this article, we discuss the equation P a ( X ) = 0 without any restriction on p and gcd ⁡ ( n , k ) . In a very recent paper [15] , the authors have left open a problem that could definitely solve this equation. More specifically, for the cases of one or two F Q -zeros, explicit expressions for these rational zeros in terms of a were provided, but for the case of p gcd ⁡ ( n , k ) + 1 F Q − zeros it was remained open to compute explicitly the zeros. This paper solves the remained problem, thus now the equation X p k + 1 + X + a = 0 over F p n is completely solved for any prime p, any integers n and k.

We study the polynomial f(x)=x^q^+^1+ax+b over an arbitrary field F of characteristic p, where q is a power of p and ab 0. The polynomial has arisen recently in several different contexts, including the inverse Galois problem, difference sets, and Muller-Cohen-Matthews polynomials in characteristic 2. We prove f has exactly n rational roots, where [email protected]?{0,1,2,Q+1} and [email protected]?GF(q)=GF(Q). If F is finite then we find the exact number of a,[email protected]?F^x such that f has n rational roots, for each n. We also prove many arithmetic properties of f. For example, if F is finite and f has a rational root r, then f has exactly two rational roots if and only if N"F"/"G"F"("Q")(r-1) 1. The techniques rely on a detailed analysis of the splitting field and Galois group, together with frequent use of Hilbert's Theorem 90.