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Numbers n such that 11^phi(n) == 1 (mod n^2), where phi(n) = A000010(n).
3

%I #19 Mar 15 2015 18:56:16

%S 71,142,284,355,497,710,994,1420,1491,1988,2485,2840,2982,3976,4970,

%T 5680,5964,7455,9940,11928,14910,19880,23856,29820,39760,59640,79520,

%U 119280,238560,477120

%N Numbers n such that 11^phi(n) == 1 (mod n^2), where phi(n) = A000010(n).

%C No further terms up to 10^9.

%C No more terms less than 10^10. - _Robert G. Wilson v_, Jan 18 2015

%C The first 30 terms are divisible by 71. Are there any terms not divisible by 71? - _Robert Israel_, Dec 30 2014

%C By Corollary 5.9 in Agoh, Dilcher, Skula (1997), if there are no further Wieferich primes to base 11 apart from 71, then the answer is no. - _Felix Fröhlich_, Dec 30 2014

%H T. Agoh, K. Dilcher and L. Skula, <a href="http://dx.doi.org/10.1006/jnth.1997.2162">Fermat Quotients for Composite Moduli</a>, J. Num. Theory, Vol. 66, Issue 1 (1997), 29-50.

%p select(t -> 11 &^ numtheory:-phi(t) mod t^2 = 1, [$1..10^6]); # _Robert Israel_, Dec 30 2014

%t a253016[n_] := Select[Range[n], PowerMod[11,EulerPhi[#], #^2] == 1 &]; a253016[500000] (* _Michael De Vlieger_, Dec 29 2014; modified by _Robert G. Wilson v_, Jan 18 2015 *)

%o (PARI) for(n=2, 1e9, if(Mod(11, n^2)^(eulerphi(n))==1, print1(n, ", ")))

%Y Cf. A077816, A242958, A242959, A242960, A245529, A241977, A241978.

%K nonn

%O 1,1

%A _Felix Fröhlich_, Dec 26 2014