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A260946
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Least positive integer k < prime(n) such that i^2 + j^2 = k^2 for some 0 < i < j with i*j*k a primitive root modulo prime(n), or 0 if no such k exists.
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3
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0, 0, 0, 0, 10, 0, 15, 5, 5, 5, 0, 17, 5, 15, 5, 10, 10, 26, 10, 17, 5, 5, 5, 5, 5, 13, 15, 5, 10, 15, 15, 10, 13, 13, 5, 17, 5, 10, 5, 10, 10, 10, 25, 61, 10, 13, 17, 25, 5, 50, 15, 13, 17, 10, 15, 5, 5, 10, 17, 10, 26, 10, 10, 17, 5, 10, 5, 5, 5, 13
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OFFSET
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1,5
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COMMENTS
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Conjecture: a(n) > 0 for all n > 11. In other words, for any prime p > 31 there are a,b,c among 1,...,p-1 with a^2 + b^2 = c^2 such that a*b*c is a primitive root modulo p.
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LINKS
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EXAMPLE
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a(5) = 10 since 10^2 = 6^2 + 8^2, and 6*8*10 = 480 is a primitive root modulo prime(5) = 11.
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MATHEMATICA
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SQ[n_]:=IntegerQ[Sqrt[n]]
Dv[n_]:=Divisors[Prime[n]-1]
Do[Do[If[SQ[k^2-j^2]==False, Goto[aa]]; Do[If[Mod[(Sqrt[k^2-j^2]j*k)^(Part[Dv[n], t]), Prime[n]]==1, Goto[aa]]; Continue, {t, 1, Length[Dv[n]]-1}]; Print[n, " ", k]; Goto[bb]; Label[aa]; Continue, {k, 1, Prime[n]-1}, {j, 1, k-1}]; Print[n, " ", 0]; Label[bb]; Continue, {n, 1, 70}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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