Factors of generalized Fermat numbers

A search for prime factors of the generalized Fermat numbers Fn(a, b) = a2 n + b2 n has been carried out for all pairs (a, b) with a, b ≤ 12 and GCD(a, b) = 1. The search limit k on the factors, which all have the form p = k · 2m + 1, was k = 109 for m ≤ 100 and k = 3 · 106 for 101 ≤ m ≤ 1000. Many larger primes of this form have also been tried as factors of Fn(a, b). Several thousand new factors were found, which are given in our tables.—For the smaller of the numbers, i.e. for n ≤ 15, or, if a, b ≤ 8, for n ≤ 16, the cofactors, after removal of the factors found, were subjected to primality tests, and if composite with n ≤ 11, searched for larger factors by using the ECM, and in some cases the MPQS, PPMPQS, or SNFS. As a result all numbers with n ≤ 7 are now completely factored. 1. Generalized Fermat numbers Generalized Fermat numbers (GFNs) of the form Fn(a) = a 2 + 1 have been studied earlier by one of us, see [3], [4]. The access to fast computers has recently rekindled interest in these numbers [1]. Many of the properties of this particular type of numbers also hold for the slightly more general type Fn(a, b) = a 2 + b n . For obvious reasons we shall assume that a and b are lacking common divisors. 2. Properties of the factors of GFNs The well-known theorem on the prime factors of the ordinary Fermat numbers 2 n + 1 is in the more general case replaced by the following result. Theorem 2.1. Suppose that p = k · 2 + 1|Fn(a, b), with k odd, and that u ≡ a/b mod p is a 2-power residue but not a 2-power residue mod p. Then m = n+ t+ 1.—If u is not even a quadratic residue, we have to put t = 0. Proof. We know that a n +b n ≡ b2n(u2n+1) ≡ 0 mod p. Since u ≡ x2t for some x, −1 ≡ u2n ≡ (x2t)2n = x2n+t mod p. Suppose that x mod p belongs to the exponent d = (p − 1)/l = k · 2/l. Then x ≡ −1 mod p, and d/2 is also the smallest positive exponent, yielding −1. Therefore d = 2, l = k · 2/d = k · 2m−n−t−1, and m ≥ n + t + 1. Since we have presumed that u is a 2-power residue, but not a 2-power residue, x is a quadratic non-residue, and thus x(p−1)/2 ≡ −1 mod p. Thus, (p−1)/2 must be an odd multiple of d/2 and thus l = (p−1)/d is odd, which implies that m− n− t− 1 = 0. Thus m is in this case exactly n+ t + 1. Received by the editor May 6, 1996 and, in revised form, September 19, 1996. 1991 Mathematics Subject Classification. Primary 11-04, 11A51, 11Y05, 11Y11.