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A248173
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Primes equal to 3 or congruent to 2 mod 3 that satisfy (1+a^p) == (1+a)^p (mod p^2) for all a between (p-3)/2.
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0
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3, 5, 11, 17, 23, 29, 41, 47, 53, 71, 89, 101, 107, 113, 131, 137, 149, 167, 173, 191, 197, 233, 239, 251, 257, 263, 269, 281, 293, 311, 317, 347, 353, 359, 383, 389, 401, 431, 449, 461, 467, 479, 491, 503, 509, 521, 557, 563, 569, 587, 593, 599, 617, 641, 647
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OFFSET
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1,1
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COMMENTS
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Apart from 3, subsequence of A003627.
Gives an easily testable condition which allows occasionally to prove the first case of Fermat’s Last Theorem over number fields for a prime number p == 2 mod 3.
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LINKS
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MATHEMATICA
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selQ[p_] := p == 3 || Mod[p, 3] == 2 && AllTrue[Range[(p-3)/2], Mod[1+#^p, p^2] != Mod[(1+#)^p, p^2]&];
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PROG
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(PARI) isok(p) = {if ((p==3) || (p % 3) == 2, for (a=1, (p-3)/2, if (Mod(1+a^p, p^2) == Mod((1+a)^p, p^2), return (0)); ); return (1); ); return (0); }
lista(nn) = forprime(p=3, nn, if (isok(p), print1(p, ", ")));
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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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