20
$\begingroup$

The so-called "two squares theorem" can be proven by establishing the following identity:

$$\left(\sum_{n=-\infty}^\infty e^{\pi i \tau n^2}\right)^2 = \sum_{n=-\infty}^\infty \frac{1}{\cos(n \pi \tau)}$$

where $\Im \tau>0$.

Stein and Shakarchi give a lengthy proof in their complex analysis book. The proof is quite complicated and the motivation is unclear. I would be interested in a more intuitive (or at least concise) proof of the above identity.

$\endgroup$
  • $\begingroup$ What are $q$ and $\tau$ in the title of this question? $\endgroup$ – MPW Apr 1 '14 at 1:13
  • 3
    $\begingroup$ @MPW By convention, $q = e^{\pi i \tau}$ in the study of elliptic functions. $\endgroup$ – Argon Apr 1 '14 at 1:15
  • $\begingroup$ I think you would enjoy arxiv.org/abs/math/0611300 and the book Number Theory in the Spirit of Ramanujan by Bruce C. Berndt. Part of what you want is in the book, page 58; he applies this to two squares and some related quadratic forms. $\endgroup$ – Will Jagy Apr 1 '14 at 2:33
  • $\begingroup$ I'd stick to complex analysis at this point. Clearly enough, it seems like a work of pure residues by looking at the right side sum. I am not sure if clever theta-null identities can help, though. $\endgroup$ – Balarka Sen Apr 3 '14 at 9:41
17
+50
$\begingroup$

Below is the shortest proof I know. I would say this proof is motivated insomuch that it is a corollary--albeit not immediate--of the Jacobi triple product.

We proceed a la Ramanujan. The proof is quite elementary, but the algebraic manipulations are a bit tedious. The key ingredient of the proof is the Jacobi triple product: $$\sum_{n \in \mathbb{Z}} q^{n^2} x^n = \prod_{n = 1}^\infty (1 - q^{2n})(1 + q^{2n - 1}x)(1 + q^{2n - 1}x^{-1}), \quad |q| < 1, \quad x \ne 0.$$ Replacing $q$ with $\sqrt{q}$ and letting $x = \sqrt{q}y$ leads to $$\sum_{n \in \mathbb Z} q^{n(n + 1)/2} y^n = \frac{y + 1}{y}\prod_{n = 1}^\infty (1 - q^n)(1 + q^n y)(1 + q^n y^{-1}).$$ Now put $y = -z^2$ so that \begin{align*} \sum_{n \in \mathbb Z} (-1)^n q^{n(n + 1)/2} z^{2n + 1} &= \sum_{n \in \mathbb Z} q^{2n^2 + n} z^{4n + 1} - \sum_{n \in \mathbb Z} q^{2n^2 - n} z^{4n - 1}\\ &= (z - z^{-1})\prod_{n = 1}^\infty (1 - q^n)(1 - q^n z^2)(1 - q^n z^{-2}). \end{align*} Using the Jacobi triple product we can express the sums on the left-hand side as infinite products: \begin{align*} &z\prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n - 1} z^4)(1 + q^{4n - 3} z^{-4})\\ &\quad- z^{-1}\prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n - 3} z^4)(1 + q^{4n - 1} z^{-4})\\ &= (z - z^{-1})\prod_{n = 1}^\infty (1 - q^n)(1 - q^n z^2)(1 - q^n z^{-2}). \end{align*} If we differentiate both sides logarithmically with respect to $z$ and then set $z = 1$ we get $$1 + 4\sum_{n = 1}^\infty (-1)^n \frac{q^{2n - 1}}{1 + q^{2n - 1}} = \frac{\prod_{n = 1}^\infty (1 - q^n)^3}{\prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n - 1})(1 + q^{4n - 3})}.$$ Yet, $$\frac{\prod_{n = 1}^\infty (1 - q^n)^3}{\prod_{n = 1}^\infty (1 - q^{4n})(1 + q^{4n - 1})(1 + q^{4n - 3})} = \frac{\prod_{n = 1}^\infty (1 - q^n)^3}{\prod_{n = 1}^\infty (1 + q^n)(1 - q^{2n})} = \frac{\prod_{n = 1}^\infty (1 - q^n)^2}{\prod_{n = 1}^\infty (1 + q^n)^2} = \theta_4^2(q),$$ where the last equality follows from the Jacobi triple product (use $x = -1$) and Euler's identity: $$(1 + q)(1 + q^2)(1 + q^3)\cdots = \frac{1}{(1 - q)(1 - q^3)(1 - q^5)\cdots}.$$ To prove this identity write $$(1 + q)(1 + q^2)(1 + q^3)\cdots = \frac{(1 + q)(1 - q)(1 + q^2)(1 - q^2)(1 + q^3)(1 - q^3)\cdots}{(1 - q)(1 - q^2)(1 - q^3)\cdots}.$$ Evidently, all the terms of the form $(1 - q^{2n})$ cancel because $(1 + q)(1 - q) = (1 - q^2)$, $(1 + q^2)(1 - q^2) = (1 - q^4)$, etc.

Consequently, $$\theta_4^2(q) = 1 + 4\sum_{n = 1}^\infty (-1)^n \frac{q^{2n - 1}}{1 + q^{2n - 1}},$$ which is equivalent to $$\theta_3^2(q) = 1 + 4\sum_{n = 0}^\infty (-1)^n \frac{q^{2n + 1}}{1 - q^{2n + 1}}$$ as $\theta_4(-q) = \theta_3(q)$. However, $$\sum_{n = 0}^\infty (-1)^n \frac{q^{2n + 1}}{1 - q^{2n + 1}} = \sum_{n = 1}^\infty \frac{q^n}{1 + q^{2n}}$$ because \begin{align*} &(q + q^2 + q^3 + \cdots) - (q^3 + q^6 + q^9 + \cdots) + (q^5 + q^{10} + q^{15} + \cdots) - \cdots\\ &\quad= (q + q^2 + q^3 + \cdots) - (q^3 + q^6 + q^9 + \cdots) + (q^5 + q^{10} + q^{15} + \cdots) - \cdots\\ &\qquad+ (q^3 + q^7 + q^{11} + \cdots) - (q^3 + q^7 + q^{11} + \cdots)\\ &\quad\qquad= q(1 - q^2 + q^4 - \cdots) + q^2(1 - q^4 + q^8 - \cdots) + q^3(1 - q^6 + q^{12} - \cdots) + \cdots \end{align*} Thus $$\theta_3^2(q) = 1 + 4\sum_{n = 1}^\infty \frac{q^n}{1 + q^{2n}} = 1 + 4\sum_{n = 1}^\infty \frac{1}{q^n + q^{-n}} = 2\sum_{n \in \mathbb{Z}} \frac{1}{q^n + q^{-n}} = \sum_{n \in \mathbb{Z}} \frac{1}{\cos(n \pi \tau)}$$ for it is clear that $q^n + q^{-n} = e^{n\pi i \tau} + e^{-n\pi i \tau} = 2\cos(n\pi \tau).$

$\endgroup$
  • $\begingroup$ I would like to note that this whole thing becomes even more concise if one uses the q-Pochhammer symbol, as is something like standard in the subject. $\endgroup$ – user98602 Apr 3 '14 at 8:49
  • $\begingroup$ The subst has been $y=-z^2$, not $y=z^2$, it seems. $\endgroup$ – ccorn Apr 4 '14 at 23:05
  • $\begingroup$ @ccorn Yes, of course. It was a typo. Thanks for spotting it. $\endgroup$ – glebovg Apr 5 '14 at 1:24
  • $\begingroup$ Oh, and the last line should have all $\tau$ replaced with $n\tau$. Not spotted until now, although I have looked through the whole thing before. $\endgroup$ – ccorn Apr 5 '14 at 2:41
  • 1
    $\begingroup$ @ccorn Yes, thanks again. $\endgroup$ – glebovg Apr 5 '14 at 5:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.