Why are these sums equal?

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I've been looking at some pretty cool proofs of $\zeta(2)=\frac{\pi^2}{6}.$ recently, and the one proof that was the easiest to understand for me was how Euler originally presented it, by finding and manipulating the infinite series for $\frac{\sin(x)}{x}$. Through the the wild and mostly ridiculous thought process that leads me to most of my cooler mathematical experiments, I thought: "What if I made an infinite sum out of that infinite series?" Infinitiception. So, I ended up researching the sum in question: $$\sum_{x=-\infty}^{\infty}\left(\dfrac{\sin(x)}{x}\right)\;$$

and it turns out it is well known as $\pi$! This was awesome. I love pi, and while the connection here is quite obvious, it is still fun to see it pop up everywhere. But then, through some playing around, I found this:

$$\sum_{x=-\infty}^{\infty}\left(\dfrac{\sin(x)}{x}\right)\;= \sum_{x=-\infty}^{\infty}\left(\dfrac{\sin(x)}{x}\right)^{2}=\pi\;$$ This simply doesn't seem possible to me. How can a series equal itself squared? $\sin^2 \neq \sin$, and obviously $x^2 \neq x$, so how can this be true? Is it even true? I would appreciate an explanation with as little calculus as possible, unless it is needed or more elegant with calculus. Thanks!

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There's no paradox here. It's quite possible for the sum of a bunch of numbers to equal the sum of the squares of those numbers. This happens even in finite sums: $$ \frac13+\frac13+\frac43=\bigg(\frac13\bigg)^2+\bigg(\frac13\bigg)^2+\bigg(\frac43\bigg)^2 $$ $$ -\frac13+\frac12+c=\bigg({-}\frac13\bigg)^2+\bigg(\frac12\bigg)^2+c^2 \text{ for either } c = \frac{3\pm\sqrt2}6. $$