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A260960
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Least positive integer k < prime(n) such that there are 0 < i < j with i^2 + j^2 = k^2 for which (i*j)/2 is a primitive root modulo prime(n), or 0 if no such k exists.
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1
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0, 0, 0, 0, 5, 5, 5, 17, 13, 17, 10, 10, 5, 13, 13, 25, 5, 5, 39, 25, 17, 5, 5, 5, 17, 29, 5, 5, 5, 5, 5, 5, 5, 34, 17, 5, 5, 26, 13, 13, 5, 10, 29, 13, 13, 5, 34, 5, 5, 5, 5, 25, 25, 5, 5, 13, 17, 5, 5, 10, 29, 13, 13, 61, 17, 13, 17, 17, 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 any n > 4. In other words, for any prime p > 7, there exists a right triangle whose three sides are among 1,...,p-1 and whose area is a primitive root modulo p.
We have verified this for primes p < 10^5.
We also conjecture that for any prime p > 31, there exists a right triangle whose three sides are among 1,...,p-1, and whose perimeter and area are quadratic residues modulo p.
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LINKS
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EXAMPLE
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a(7) = 5 since 3^2 + 4^2 = 5^2, and (3*4)/2 = 6 is a primitive root modulo prime(7) = 17.
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MATHEMATICA
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SQ[n_]:=IntegerQ[Sqrt[n]]
Dv[n_]:=Divisors[Prime[n]-1]
Do[Do[Do[If[SQ[k^2-j^2]==False, Goto[cc]]; Do[If[Mod[(j*Sqrt[k^2-j^2]/2)^(Part[Dv[n], t]), Prime[n]]==1, Goto[cc]]; Continue, {t, 1, Length[Dv[n]]-1}];
Print[n, " ", k]; Goto[aa]; Label[cc]; Continue, {j, 1, k-1}]; Label[dd]; Continue, {k, 1, Prime[n]-1}]; Print[n, " ", 0]; Label[aa]; 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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