%I #11 Mar 12 2019 11:46:49
%S 33,36003,426747,220067817
%N Composite squarefree numbers k such that k^2+1 is divisible by p-1, where p are all the prime factors of k.
%C Tested up to 3*10^10. - _Giovanni Resta_, Mar 06 2019
%e Prime factors of 33 are 3, 11 and 33^2+1 = 1090, 1090/2 = 545, 1090/10 = 109.
%e Prime factors of 220067817 are 3, 59, 131, 9491 and 220067817^2+1 = 48429844079145490, 48429844079145490/2 = 24214922039572745, 48429844079145490/58 = 834997311709405, 48429844079145490/130 = 372537262147273, 48429844079145490/9490 = 5103250166401.
%p with(numtheory): P:=proc(q) local a,k,ok,n;
%p for n from 1 to q do if not isprime(n) and issqrfree(n) then a:=factorset(n); ok:=1; for k from 1 to nops(a) do if frac((n^2+1)/(a[k]+1))>0 then ok:=0; break; fi; od; if ok=1 then print(n); fi; fi; od; end: P(10^20);
%o (PARI) isok(n) = {if (issquarefree(n) && !isprime(n) && (n>1), my(f = factor(n)[,1], x = n^2+1); for (k=1, #f, if ((x % (f[k]-1)), return (0));); return (1);); return (0);} \\ _Michel Marcus_, Mar 12 2019
%Y Cf. A002997, A006972, A208728, A225711, A304291, A306685.
%K nonn,hard,more
%O 1,1
%A _Paolo P. Lava_, Mar 06 2019
%E a(4) from _Giovanni Resta_, Mar 06 2019