OFFSET
1,1
COMMENTS
Generalization of the questions proposed in Diophante problem A1885 (see links).
When p is prime, as a(p) is the smallest m such that tau(m^2 - 1) = 2*p, hence m^2 - 1 is of the form q * r^(p-1) with q > r primes, so we must solve the Diophantine equation (m-1)*(m+1) = q * r^(p-1) to get the smallest m, when only p is known.
Two cases must be checked:
-> r = 2: if 2^(p-3) + 1 is prime, then m = 2^(p-2) + 1, and
if 2^(p-3) - 1 is prime, then m = 2^(p-2) - 1.
If there is a solution m in the case r = 2, then a(p) is this smallest solution m (see example a(11)); if there is no solution m with r = 2, then try the 2nd case.
-> r odd prime: if r^(p-1) + 2 is prime, then m = r^(p-1) + 1, and
if r^(p-1) - 2 is prime, then m = r^(p-1) - 1 (example a(13)).
LINKS
Diophante, A1885, Cachés derrière leurs diviseurs (in French).
Adrian Dudek, On the Number of Divisors of n^2-1, arXiv:1507.08893 [math.NT], 2015.
EXAMPLE
tau(2^2 - 1) = 2 = 2*1, so a(1) = 2.
tau(3^2 - 1) = 4 = 2*2, so a(2) = 3.
tau(4^2 - 1) = 4 = 2*2, tau(5^2 - 1) = 8 = 2*4 so a(4) = 5.
For a(11): if r = 2, 2^8 + 1 = 257 is prime, while 2^8 - 1 is not prime, hence a(11) = 2^9 + 1 = 513.
For a(13):
if r = 2, 2^10 +- 1 are not prime, so not possible;
if r = 3, 3^12 +- 2 are not prime, so not possible;
if r = 5, 5^12 +- 2 are not prime, so not possible;
if r = 7, 7^12 - 2 = 13841287199 is prime, while 7^12 + 2 is not prime, then a(13) = 13841287199+1 = 13841287200.
CROSSREFS
KEYWORD
nonn
AUTHOR
Bernard Schott, Sep 17 2021
EXTENSIONS
a(31)-a(35) from Jinyuan Wang, Sep 23 2021
STATUS
approved