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The title says it all. For N less than 1000, if I've looked up the tables correctly, when the condition holds N can only be 17, 41, 73, 113, or 257.

Motivation:

Let N be an odd positive integer, f in Z[[q]] a modular form of level Gamma_0 (N), and p a prime not dividing 2N. I'll say N satisfies the mod 2 local nilpotence condition * if:

(*) For every choice of f and p, some power of the Hecke operator T_p takes f into 2Z[[q]].

Suppose in particular that N is prime. If N is 3,5 or 7 mod 8 and satisfies * then N is 3,5 or 7. Suppose however N is 1 mod 8. If h(-N) is not a power of 2 one may use an ideal-class-character whose order is not a power of 2 to construct first a weight 1 modular form of level 4N with quadratic character, and then from this form a mod 2 eigensystem of level N other than the identically zero eigensystem, thereby showing that * does not hold. A similar argument applies when h(-2N) is not a power of 2. So a necessary (but not sufficient) condition for * to hold is that h(-N) and h(-2N) are powers of 2. (The LMFDB database can be used to show that * fails when N is 73,113 or 257. I'm not sure what happens when N is 17 or 41)

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  • $\begingroup$ What is $h(-N)$? $\endgroup$
    – YCor
    Commented Jul 19, 2017 at 17:08
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    $\begingroup$ If we look at prime discriminants of size between $X$ and $2X$, there are about $X/\log X$ such discriminants, and the class numbers are of size $\sqrt{X}$ typically. So a random model would suggest that the chance of finding primes $N$ of size $X$ is about $1/\log X \times 1/\sqrt{X} \times 1/\sqrt{X} = 1/(X\log X)$. In other words, we might expect to find a prime of size $X$ about $1/\log X$ times. Summing this over $X=2^k$ gives a barely divergent series, and so ``Borel-Cantelli" suggests that there are infinitely many such $N$. Of course this is delicate, but we don't know any better. $\endgroup$
    – Lucia
    Commented Jul 19, 2017 at 17:34
  • $\begingroup$ The comment above suggests that there might be about $\log \log X$ primes $N$ below $X$ meeting the conditions of the question. $\endgroup$
    – Lucia
    Commented Jul 19, 2017 at 17:37
  • $\begingroup$ @YCor If N is a square-free integer other than 1, then h(N) is the class number of the quadratic field Q(root(N)). In the case at hand, h(-N) is the number of SL_2(Z)-classes of positive integral binary quadratic forms of discriminant 4N, and h(-2N) the number of classes of such forms of discriminant 8N. $\endgroup$
    – paul Monsky
    Commented Jul 19, 2017 at 19:58

1 Answer 1

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Up to more than 10^8, the list is

17 41 73 113 257 1153 1217 2657 4481 4993 5569 57649 1164817 140618353

I would guess this goes on.

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