A divisor sum congruence for 8n+6 Letting $d(m)$ be the number of divisors of $m$, is it the case that for $m=8n+6$,
$$ d(m) \equiv \sum_{k=1}^{m-1} d(k) d(m-k) \pmod{8}\ ?$$
It's easy to show that both sides are 0 mod 4: the left side since two primes appear to odd order in the factorization of $m$, and the right side since $m$ is neither a square nor the sum of two squares and so even products all appear twice.  
But they also seem to match mod 8 for small values.  Does this keep going, or does it fail somewhere?
 A: The congruence you state is true for all $m \equiv 6 \pmod{8}$. The proof I give below relies on the theory of modular forms. First, observe that
$$
\sum_{k=1}^{m-1} d(k) d(m-k) = 2 \sum_{k=1}^{\frac{m-2}{2}} d(k) d(m-k) + d\left(\frac{m}{2}\right)^{2}.
$$
Noting that $d(m) \equiv d\left(\frac{m}{2}\right)^{2} \pmod{8}$ if $m \equiv 6 \pmod{8}$, it suffices to prove that for every $m \equiv 6 \pmod{8}$ that $$\sum_{k=1}^{\frac{m-2}{2}} d(k) d(m-k)$$ 
is a multiple of $4$. The only terms in the sum that are not multiples of $4$ are those where $k$ is a perfect square, and $m-k = py^{2}$,
where $p$ is prime, and $p$ divides $y$ to an even power (or $k = py^{2}$ where $p$ divides $y$ to an even power, and $m-k$ is a square). It suffices therefore to show that if $m \equiv 6 \pmod{8}$, then $m$ has an even number of representations in the form $m = x^{2} + py^{2}$ with $x, y \in \mathbb{Z}$ with $x, y > 0$ and $p$ a prime number (and the $p$-adic valuation of $y$ is even).
If $m = x^{2} + py^{2}$, then either $x$ is even, which forces $p = 2$ and $y$ odd,
or $x$ is odd, in which case $y$ is odd and $p \equiv 5 \pmod{8}$.
The function $F(z) = \sum_{n=0}^{\infty} \sigma(2n+1) q^{2n+1}$, $q = e^{2 \pi i z}$ is a modular form of weight $2$ for $\Gamma_{0}(4)$. Here $\sigma(k)$ denotes the sum of the divisors function. A simple calculation shows that if $n \equiv 5 \pmod{8}$, then $\sigma(n) \equiv 2 \pmod{4}$ if and only if $n = py^{2}$ for some prime $p \equiv 5 \pmod{8}$ and a some perfect square $y$ (and moreover, the power of $p$ dividing $y$ is even, a condition which will remain in effect). It follows from this that
$$
  \frac{1}{2} \sum_{n \equiv 5 \pmod{8}} \sigma(n) q^{n} \equiv \sum_{\substack{p \equiv 5 \pmod{8} \\ y \geq 1}} q^{py^{2}} \pmod{2},
$$
where by this statement I mean that the power series on the left and right hand sides have integer coefficients and the coefficient of $q^{k}$ on the left side is congruent (modulo $2$) to the coefficient of $q^{k}$ on the right hand side.
By twisting $F(z)$ by Dirichlet charaters mod $8$, one can see that $G(z) = \frac{1}{2} \sum_{n \equiv 5 \pmod{8}} \sigma(n) q^{n}$ is a modular form of weight $2$ on $\Gamma_{0}(64)$. Now, observe that $F(z) \equiv \sum_{\substack{n \geq 1 \\ n \text{ odd }}} q^{n^{2}} \pmod{2}$. Now, we find that
$$
  F(z) G(z) + F(4z) F(2z) \equiv \sum_{x, y \geq 1, p \equiv 5 \pmod{8} \text{ prime }} q^{x^{2} + py^{2}} + \sum_{x,y} q^{x^{2} + 2y^{2}} \pmod{2}.
$$
where in the second term on the right hand side we have $x \equiv 2 \pmod{4}$ and $y \equiv 1 \pmod{2}$. 
The left hand side is now a modular form of weight $4$ on $\Gamma_{0}(64)$, and a computation shows that the first $1000$ Fourier coefficients are even. Indeed, Sturm's theorem proves that if the first $32$ coefficients are even, then all of them are. It follows that the number of representations of $m$ in the form $x^{2} + py^{2}$ when $m \equiv 6 \pmod{8}$ is always even, and hence the desired congruence is true.
