Lemma. Let u ≥ 3 be odd. Then A_{σ+γ}(3u) is never a power of 3.

Sketch of Proof: Suppose that A_{σ+γ}(d) is a power of 3 for odd d = 3u, u ≥ 3, i.e.,

A_{σ+γ}(d) = 3^{e}, (1)

for some integer e. Then, since A_{σ+γ}(d) is different from 3, 9, 27 (see Table 1), A_{σ+γ}(d) must be divisible by 81. There are nine d modulo 4212 with that property.

Next we consider (1) modulo m = 5, 8, 11, 13, 37, 41, 61. For each fixed m, the numbers A_{σ+γ}(d) modulo m are periodic (because A_{σ+γ}(d) satisfies the recurrence (4.32)), of length d(m) say, and the powers 3^{e} modulo m are periodic as well, of length e(m) say. Thus, for each fixed m, we obtain a number of solutions (d, e) to (1) mod m, the solutions (d, e) being determined modulo (d(m), e(m)) (which is short for: d is determined modulo d(m) and e is determined mod e(m)). Furthermore, if 3 | d(m), we may eliminate all solutions (d, e) with 3 |̸ d. (Recall that d = 3u.)

A computer search gives 84 solutions (d, e) to (1) mod 5, with each (d, e) being determined mod (624, 4), it gives 4 solutions (d, e) to (1) mod 8, with each (d, e) being determined mod (20, 2), it gives 13 solutions (d, e) to (1) mod 11, with each (d, e) being determined mod (240, 5), it gives 4 solutions (d, e) to (1) mod 13, with each (d, e) being determined mod (168, 3), it gives 233 solutions (d, e) to (1) mod 37, with each (d, e) being determined mod (2736, 18), it gives 21 solutions (d, e) to (1) mod 41, with each (d, e) being determined mod (480, 8), and it gives 12279 solutions (d, e) to (1) mod 61, with each (d, e) being determined mod (453960, 10). For example, the 4 solutions to (1) mod 13 are (d, e) = (3, 1), (21, 2), (63, 0), (105, 1) mod (168, 3).

We may now determine a combined solution (d, e), d determined modulo 2173560480 (the least common multiple of 4212, 624, 20, 240, 168, 2736, 480, 453960), e determined modulo 360 (the least common multiple of 4, 2, 5, 3, 18, 8, 10). But, as a computer calculation shows, there is no combined solution, a contradiction.