Does there exist a half-integer weight theta function which is is equivalent to 1 modulo 4? Given a quadratic form $F$ in $n$ variables, there is an associated theta function $\theta_F(z) = \sum_{m \in \mathbb{Z}} q^{F(m)}$, which is a modular form of weight $n/2$.  Letting $F(m) = m^2$ gives a theta function $\theta_F(z) = 1 + 2 \sum_{n = 1}^\infty q^{n^2} \equiv 1 \pmod{2}$ with weight $1/2$.  
Does there exist a theta function $\theta_F(z) \equiv 1 \pmod{4}$ with half-integer weight?  
 A: This is nothing like a complete answer (EDIT-now it seems to be--see below), but it may suggest a fruitful attack, based on the comment (see below for a version incorporated into this answer) that I made earlier. Your question is related to my question 124243 on this site about a certain ring $M$ consisting of the mod $2$ reductions of elements of $\mathbb{Z}[[x]]$ that are the Fourier expansions of modular forms for $\gamma_0 (N)$. Suppose first that $N$ is an odd prime $p$.
Theorem(?) The element $g=x+x^4+x^9+x^{16}+\dots$ of $\mathbb{Z}/2[[x]]$ does not lie in $M$.
Proof(?)  In the notation of my question, and the other question of mine that it links to, let $h=g+g^2+r$. Then
$$ h(1+h)=(g+g^4)+(r+r^2)=f1+(f1+fp)=fp . $$
(Again, see below). Now I conjecture in my question that in the field of fractions of $M$, $fp$ has a zero of order $p$, a zero of order $1$ and $p+1$ poles counted with multiplicity. If this is true then both $h$ and $h+1$ have $\frac{p+1}{2}$ poles, counted with multiplicity. Furthermore one of them must have a zero of order $p$, giving a contradiction.
Now I'm reasonably sure that this conjecture is correct, is known to the experts, and follows from results proved by Igusa in the 1950's. Unfortunately no expert has yet responded to my question. I also think that similar techniques will show that $g$ is not the mod $2$ reduction of any modular form for $\gamma_0 (N)$ when $N$ is odd. But the theory, if I understand correctly, is much harder when the level is divisible by the characteristic, and if a proof is to be given based on the idea of my comment this case has to be considered.
Edited comment--Suppose that $F$ is a positive definite quadratic form over $\mathbb{Z}$ in $s$ variables where $s$ is odd. Here's an idea for showing that the power series $\theta_F$ in $\mathbb{Z}[[x]]$ attached to $F$ (I'll defy convention and write $x$ instead of $q$) cannot be congruent to $1$ mod $4$. Choose $S$ so that $s\times S=8m-1$ and let $G$ be a direct sum of $S$ copies of $F$ and one of the one variable 
form $z^2$. Then if $\theta_F$ is $1$ mod $4$, $\theta_G$ is $1+2x+2x^4+2x^9+2x^{16}+\dots$ mod $4$. Now (see Schoeneberg's paper from the 1930's for example) there is an $N$ such that $\theta_G$ is the "expansion at infinity" of a weight $4m$ modular form, $\phi$, for $\gamma_0 (N)$. Then $(1/2)(\phi-(E_4)^m)$ is a 
weight $4m$ modular form for $\gamma_0 (N)$ whose expansion at infinity lies in $\mathbb{Z}[[x]]$, and has mod $2$ reduction equal to $g=x+x^4+x^9+x^{16}+\dots$.  There are perhaps reasons to believe this can't happen--at any rate it appears to be an approachable problem of a mainstream variety.
One more remark. In my proof? given above I use the fact that the mod $2$ reduction $f_1$ of the expansion at infinity of the cusp form delta is $x+x^9+x^{25}+\dots$. This lovely if well-known fact comes from the infinite product formula for delta and Jacobi's triple product identity.
EDIT---MARCH 24
I now believe I have an argument showing that $g$ is not the mod $2$ reduction of a modular form for any $\gamma_0 (N)$, completing the answer to your question. I'll dispense with any ideas of Igusa, and instead use Hecke operators and classical results of Gauss on ternary quadratic forms. The idea is this. For each odd prime $q$ there is a "Hecke operator" $T_q$ on $\mathbb{Z}/2[[x]]$.
If $h$ is the reduction of a modular form for $\gamma_0 (N)$ then the $T_q(h)$ span a finite dimensional subspace of $\mathbb{Z}/2[[x]]$. Now let $h=g^{11}$. I'll use this last result to show that $h$ cannot be the reduction of a modular form.
To this end, fix $K$ and primes $p_1,...,p_K$ each of which is $5$ mod $8$. Then choose primes $q_1,...,q_K$ each of which is $7$ mod $8$ so that the Legendre symbol $(q_i/p_j)$ is $-1$ if $i=j$ and is $1$ otherwise. Now the coefficient of $x^{p_j}$ in $T_{q_i}(h)$ is the coefficient of $x^{p_j}(q_i)$ in $h$. By the last paragraph of the comments following my answer to MO question 28462-why are there usually an even number of representations as a sum of $11$ squares--(an answer based on Disquisitiones Arithmeticae)--, this coefficient is $1$ if $i=j$ and $0$ otherwise.
So the $T_{q_i} (h)$ are linearly independent. Since $K$ may be chosen arbitrarily large, the criterion of the last paragraph gives the result.
