<s>[INST] congruence of Ramanujan tau function mod 2 The Ramanujan tau function, studied by Ramanujan (1916), is the function ${\displaystyle \tau :\mathbb {N} \to \mathbb {Z} }$ defined by the following identity:
${\displaystyle \sum _{n\geq 1}\tau (n)q^{n}=q\prod _{n\geq 1}(1-q^{n})^{24}=\eta (z)^{24}=\Delta (z)}$.
Wiki says $\tau(n)=\sigma_{11}(n) \operatorname{mod} 2$ for $n$ odd, is there a simple way to see this (without using modular form)?
 [/INST] Start with the Jacobi triple product
in the form
$$ \prod_{m=1}^\infty (1 - x^{2m}) (1 + x^{2m-1}y^2) (1 + x^{2m-1}/y^2) =
   \sum_{n=-\infty}^{\infty} x^{n^2} y^{2n}. \tag{1}$$
Let $\ y^2 = x t \ $ to get
$$ \Big(1 + \frac1t\Big) \prod_{m=1}^\infty (1 - x^{2m}) (1 + x^{2m}t) \Big(1 + \frac{x^{2m}}t\Big) =
   \sum_{n=-\infty}^\infty  x^{n^2+n} t^n . \tag{2}$$
Divide both sides by $\ 1 + 1/t \ $ to get
$$ \prod_{m=1}^\infty (1 - x^{2m}) (1 + x^{2m}t) 
\Big(1 + \frac{x^{2m}}t\Big) =
   \sum_{n=1}^\infty x^{n^2-n}\frac{ (1/t^n + t^{n-1})}{(1/t + 1)}. \tag{3}$$
Take the limit as $\ t \to -1 \ $ to get
$$ \prod_{m=1}^\infty (1 - x^{2m}) (1 - x^{2m}) (1 - x^{2m}) =
   \sum_{n=1}^\infty x^{n^2-n} (2n-1). \tag{4}$$
Define $\ E(q) := \prod_{n=1}^\infty (1 - q^n) \ $ and replace $\ x^2 \ $ with $\ q \ $ to get
$$ E(q)^3 = \sum_{n=1}^\infty (2n-1) q^{ n(n-1)/2}. \tag{5}$$
Taking modulo $2$ to get
$$ E(q)^3 \equiv \sum_{n=1}^\infty q^{ n(n-1)/2} \pmod{2}. \tag{6}$$
Raise both sides to the eighth power and using
 $\ (a + b)^{2^k} \equiv a^{2^k} + b^{2^k} \pmod{2} \ $ implies
$$  E(q)^{24} \equiv \sum_{n=1}^\infty q^{4n(n-1)} \pmod{2}. \tag{6}$$
Multiply both sides by $\ q \ $ and rewrite to get
$$ \sum_{n=1}^\infty \tau(n) q^n \equiv \sum_{n=1}^\infty q^{(2n-1)^2} \pmod{2}. \tag{7}$$
If $\ n \equiv 1 \pmod{2},\ $ then $\ \sigma_k(n) \equiv \sigma_0(n) \pmod{2}. \ $ Also
  $\ \sigma_0(n) \equiv 1 \pmod{2} \ $ iff $\ n \ $ is a square integer. This proves
that $\ \tau(n) \equiv \sigma_k(n) \ $ for $\ n \ $ odd.
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