%I
%S 1,1,2,1,4,1,8,1,10,1,726,2,12,1,4,1,18,3,20,1,96,23,22,1,24,1,72,2,
%T 30,8,30,1,32,35,34,1,222,40,26,1,1312,43,42,46,360,44,48,2,588,1,50,
%U 2,5100,1,88,1,19152,60,8,16
%N Least positive integer m such that m*n divides phi(m^2+n^2), where phi(.) is Euler's totient function.
%C Conjecture: (i) a(n) exists for any n > 0.
%C (ii) For each n > 0, there is a positive integer m such that m*n divides sigma(m^2+n^2), where sigma(k) is the sum of all positive divisors of k.
%C Note that a(n) = 1 if n^2 + 1 is prime. When n^2 + (n+1)^2 is prime, n*(n+1) divides phi(n^2 + (n+1)^2) = n^2 + (n+1)^2  1 and hence a(n) <= n + 1.
%C If (n*q)^2 + 1 is prime for some q > 0, then for m = n^2*q the number phi(m^2+n^2) = phi(n^2)*phi((n*q)^2+1) = phi(n^2)*n^2 *q^2 is divisible by m*n = n^3*q.  _ZhiWei Sun_, Oct 03 2014
%H ZhiWei Sun, <a href="/A248058/b248058.txt">Table of n, a(n) for n = 1..1242</a>
%e a(5) = 4 since 4*5 divides phi(4^2 + 5^2) = phi(41) = 40.
%e a(919) = 37160684 since the product 919*37160684 = 34150668596 divides phi(919^2 + 37160684^2) = phi(1380916436192417) = 1379413805929632 = 40392*34150668596.
%t Do[m=1;Label[aa];If[Mod[EulerPhi[m^2+n^2],m*n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,60}]
%o (PARI)
%o a(n)=m=1;while(eulerphi(m^2+n^2)%(m*n),m++);m
%o vector(100,n,a(n)) \\ _Derek Orr_, Oct 01 2014
%Y Cf. A000010, A000203, A248035, A248036, A248054.
%K nonn
%O 1,3
%A _ZhiWei Sun_, Sep 30 2014
