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A276270
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Number of solutions to the congruence x^3 + y^2 + 1 == 0 (mod n).
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1
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1, 2, 3, 2, 5, 6, 3, 4, 15, 10, 11, 6, 11, 6, 15, 8, 17, 30, 27, 10, 9, 22, 23, 12, 25, 22, 45, 6, 29, 30, 27, 16, 33, 34, 15, 30, 47, 54, 33, 20, 41, 18, 51, 22, 75, 46, 47, 24, 21, 50, 51, 22, 53, 90, 55, 12, 81, 58, 59, 30
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OFFSET
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1,2
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COMMENTS
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Conjecture 1. If {k,m} is one of the pairs {3,2}, {7,2}, {11,5}, {11,7}, {13,2}, {13,11}, {17,2}, {17,5}, {17,10}, {17,11}, {19,2}, {19,3}, {19,7}, {19,9}, {19,13}, {19,14}, {23,17}, {23,19}, {25,11}, {29,5}, {29,7}, {29,13}, {29,23}, then the set {x^k + y^m: x and y are integers} contains a complete system of residues modulo any positive integer n, and also so does the set {x^k - y^m: x and y are integers}.
Conjecture 2. Let P(x) and Q(y) be non-constant polynomials with integer coefficients. Then there is a positive integer N(P,Q) depending only on P(x) and Q(y) such that for any prime p > N(P,Q) the set {P(x) + Q(y): x,y = 0,1,...,p-1} contains a complete system of residues modulo p. Moreover, when Q(y) = y^2, we may take N(P,Q) = 5 if P(x) = x^4, N(P,Q) = 11 if P(x) = x^5, and N(P,Q) = 13 if P(x) = x^6.
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LINKS
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EXAMPLE
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a(1) = 1 since 0^3 + 0^2 + 1 == 0 (mod 1).
a(2) = 2 since 0^3 + 1^2 + 1 == 0 (mod 2) and 1^3 + 0^2 + 1 == 0 (mod 2).
a(3) = 3 since 1^3 + 1^2 + 1 == 0 (mod 3), 1^3 + 2^2 + 1 == 0 (mod 3), and 2^3 + 0^2 + 1 == 0 (mod 3).
a(4) = 2 since 3^3 + 0^2 + 1 == 0 (mod 4) and 3^3 + 2^2 + 1 == 0 (mod 4).
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MATHEMATICA
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Do[r=0; Do[If[Mod[x^3+y^2+1, n]==0, r=r+1], {x, 0, n-1}, {y, 0, n-1}]; Print[n, " ", r]; Continue, {n, 1, 60}]
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PROG
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(PARI) a(n)={my(v=vector(n)); for(i=0, n-1, v[i^2 % n + 1]++); sum(i=0, n-1, v[(i^3 - 1)%n + 1])} \\ Andrew Howroyd, Aug 02 2018
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CROSSREFS
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KEYWORD
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nonn,mult
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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