

A235366


Smallest odd prime factor of 3^n  1.


9



13, 5, 11, 7, 1093, 5, 13, 11, 23, 5, 797161, 547, 11, 5, 1871, 7, 1597, 5, 13, 23, 47, 5, 11, 398581, 13, 5, 59, 7, 683, 5, 13, 103, 11, 5, 13097927, 1597, 13, 5, 83, 7, 431, 5, 11, 47, 1223, 5, 491, 11, 13, 5, 107, 7, 11, 5, 13, 59, 14425532687, 5, 603901, 683, 13, 5, 11, 7, 221101, 5, 13, 11
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OFFSET

3,1


COMMENTS

Levi Ben Gerson (12881344) proved that 3^n  1 = 2^m has no solution in integers if n > 2, by showing that 3^n  l has an odd prime factor. His proof uses remainders after division of powers of 3 by 8 and powers of 2 by 8; see the Lenstra and Peterson links. For an elegant short proof, see the Franklin link.  Sondow
One way to prove it is by the use of congruences. The powers of 3, modulo 80, are 3, 9, 27, 1, 3, 9, 27, 1, 3, 9, 27, 1, ... The powers of 2 are 2, 4, 8, 16, 32, 64, 48, 16, 32, 64, 48, 16, ...  Alonso del Arte, Jan 20 2014


REFERENCES

L. E. Dickson, History of the Theory of Numbers, Vol. II, Chelsea, NY 1992; see p. 731.


LINKS

P. Franklin, Problem 2927, Amer. Math. Monthly, 30 (1923), p. 81.
J. J. O'Connor and E. F. Robertson, Levi ben Gerson, The MacTutor History of Mathematics archive, 2009.


FORMULA

a(4n) = 5 as 3^(4n)1 = (3^4)^n1 = 81^n1 = (80+1)^n1 == 0 (mod 5).
a(6+12n) = 7 as 3^(6+12n)1 = (3^6)^(1+2n)1 = 729^(1+2n)1 = (728+1)^(1+2n)1 == 1^(1+2n)1 == 0 (mod 7), but 729^(1+2n)1 = (7301)^(1+2n)1 == (1)^(1+2n)1 == 2 (mod 5).


EXAMPLE

3^3  1 = 26 = 2 * 13, so a(3) = 13.
3^4  1 = 80 = 2^4 * 5, so a(4) = 5.
3^5  1 = 242 = 2 * 11^2, so a(5) = 11.


MATHEMATICA

Table[FactorInteger[3^n  1][[2, 1]], {n, 3, 50}]


PROG



CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



