OFFSET
1,3
COMMENTS
Conjecture: (i) a(n) exists for any n > 0.
(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.
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.
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. - Zhi-Wei Sun, Oct 03 2014
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..1242
EXAMPLE
a(5) = 4 since 4*5 divides phi(4^2 + 5^2) = phi(41) = 40.
a(919) = 37160684 since the product 919*37160684 = 34150668596 divides phi(919^2 + 37160684^2) = phi(1380916436192417) = 1379413805929632 = 40392*34150668596.
MATHEMATICA
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}]
PROG
(PARI)
a(n)=m=1; while(eulerphi(m^2+n^2)%(m*n), m++); m
vector(100, n, a(n)) \\ Derek Orr, Oct 01 2014
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Sep 30 2014
STATUS
approved