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Numbers k such that k^2 - 1 and k^2 + 1 have 8 divisors each.
1

%I #34 Jun 22 2024 00:24:43

%S 68,112,128,162,200,212,252,294,318,336,338,372,448,450,498,502,542,

%T 578,592,598,612,648,672,678,708,752,762,808,812,852,878,888,938,952,

%U 992,996,1012,1038,1098,1102,1116,1122,1188,1202,1212,1248,1258,1328,1362,1380

%N Numbers k such that k^2 - 1 and k^2 + 1 have 8 divisors each.

%C Among the first 10000 terms (from a(1) = 68 through a(10000) = 697578), k^2 - 1 and k^2 + 1 are each the product of three distinct primes, except for

%C 125 terms for which k^2 + 1 = 5^3 times a prime

%C 6 terms for which k^2 + 1 = 13^3 times a prime

%C 1 terms for which k^2 + 1 = 17^3 times a prime

%C 1 terms for which k^2 + 1 = 29^3 times a prime, and

%C 4 terms for which k^2 - 1 = p^3 * (p^3 +/- 2) (with p = 19, 29, 37, 83, respectively).

%C The first term for which both k^2 - 1 and k^2 + 1 are of the form p^3 * q is k = 41457661182: k^2 - 1 = 3461^3 * 41457661183, while k^2 + 1 = 5^3 * 13749901365452077097.

%F { k : tau(k^2 - 1) = tau(k^2 + 1) = 8}, where tau() is the number of divisors function, A000005.

%e 68 is a term: both 68^2 - 1 = 4623 = 3 * 23 * 67 and 68^2 + 1 = 4625 = 5^3 * 37 have 8 divisors.

%Y Cf. A000005, A002522, A005563, A069062, A108278, A193432, A347191.

%K nonn

%O 1,1

%A _Jon E. Schoenfield_, Jun 21 2024