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Numbers k such that k^2 + 1 is composite and phi(k^2 + 1) == 0 (mod k).
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%I #24 Jun 19 2022 02:19:22

%S 8,32,50,64,128,192,216,336,360,900,1000,1152,1250,1280,1344,1408,

%T 1800,2450,2880,4096,5440,6528,7564,9216,9800,11008,12544,13824,14450,

%U 14976,16576,18432,20000,21420,21440,24200,25675,36000,36288,43778,46656

%N Numbers k such that k^2 + 1 is composite and phi(k^2 + 1) == 0 (mod k).

%C If k^2 + 1 is prime, trivially phi(k^2 + 1) == 0 (mod k).

%H Amiram Eldar, <a href="/A067519/b067519.txt">Table of n, a(n) for n = 1..500</a> (terms 1..300 from Donovan Johnson)

%t n2Q[n_]:=Module[{c=n^2+1},CompositeQ[c]&&Divisible[EulerPhi[c],n]]; Select[Range[50000],n2Q] (* _Harvey P. Dale_, Apr 16 2015 *)

%o (PARI) for(n=1, 46656, m=n^2+1; if(isprime(m)==0, if(eulerphi(m)%n==0, print1(n ", ")))) \\ _Donovan Johnson_, Nov 14 2013

%Y Cf. A000010, A066820.

%K nonn

%O 1,1

%A _Benoit Cloitre_, Feb 22 2002