The known solutions to the equation τ(p) ≡ 0 (mod p) were p = 2, 3, 5, 7, and 2411. Here we present our method to compute the next solution, which is p = 7758337633. There are no other solutions up to 10.

Therefore 0 + l)* s P + 1 mod p\ Now put t=<r + vp, (0 < c r < p 1). Then ^ = cr̂ , 0 + l)*' s (er + 1)*" mod p. Therefore (er + l)* = a + 1 mod p, (0 < a < p 1). This is relation (7) of my previous note; from this follows (1) as in the earlier treatment. Hence (1) is satisfied by all primes of the form Qn + 1. Therefore the test can be useful only when the exponent p is 3 or is of the form 6n — 1.

A classical theorem conjectured by Jacobi asserts that for an odd prime p, the sum of the quadratic residues in the interval (0, p) is less than the sum of the quadratic nonresidues if and only if p ■ 3 (mod 4). We generalize Jacobi's problem to fcth powers (mod p), k > 2, and we consider in some detail a generalization of Jacobi's conjecture to quadratic residues and nonresidues (mod n), n an ...

(c0 + c1ζ + · · ·+ cp−2ζ) ≡ c0 + c1 + · · ·+ cp−2 mod p. The number p is not prime in Z[ζ], as (p) = (1 − ζ)p−1, so congruence mod p is much stronger than congruence mod 1− ζ, where all classes have integer representatives. Of course not every element of Z[ζ] that is congruent to a rational integer mod p is a pth power, but Kummer discovered a case when this converse statement is true, for cert...

A subset $S$ of vertex set $V(D)$ is an {em indpendent dominating set} of $D$ if $S$ is both an independent and a dominating set of $D$. The {em indpendent domination number}, $i(D)$ is the cardinality of the smallest independent dominating set of $D$. In this paper we calculate the independent domination number of the { em cartesian product} of two {em directed paths} $P_m$ and $P_n$ for arbi...

Here ∑∗ denotes summation over primitive characters χ (mod q), φ(q) denotes the number of primitive characters (mod q), and ω(q) denotes the number of distinct prime factors of q. Note that φ(q) is a multiplicative function given by φ(p) = p − 2 for primes p, and φ(p) = p(1 − 1/p) for k ≥ 2 (see Lemma 1 below). Also note that when q ≡ 2 (mod 4) there are no primitive characters (mod q), and so ...

Let p(n) denote the ordinary partition function. Subbarao conjectured that in every arithmetic progression r (mod t) there are infinitely many integers N ≡ r (mod t) for which p(N) is even, and infinitely many integers M ≡ r (mod t) for which p(M) is odd. In the even case the conjecture was settled by Ken Ono. In this paper we prove the odd part of the conjecture which together with Ono’s resul...

Let k be an algebraically closed field, let R be an associative kalgebra, and let F = {Mα : α ∈ I} be a family of orthogonal points in Mod(R) such that EndR(Mα) ∼= k for all α ∈ I. Then Mod(F), the minimal full subcategory of Mod(R) which contains F and is closed under extensions, is a full exact Abelian sub-category of Mod(R) and a length category in the sense of Gabriel [8]. In this paper, we...

Here are two typical results about the numbers mentioned in the title: If p is a prime such that p = 1 (mod 6) and p > 67, then there are exactly six numbers mod p , each of which has six sixth roots less than 2yf}p in absolute value. If p is a prime such that p = 1 (mod 8), then there is at least one number mod p which has eight eighth roots less than p3/4 in absolute value.

Abstruct -An implementation, called MOD-CHAR, of Char’s spanning tree enumeration algorithm [3] is discussed. Two complexity analyses of MOD-CHAR are presented. It is shown that MOD-CHAR leads to better complexity results for Char’s algorithm than what could be obtained using the straightforward implementation implied in Char’s original presentation 131. The class of graphs for which MOD-CHAR a...