Sum of squared variables equals Chi-squared implies that the variables are standard normal? It is known that if iid $Y_1,...,Y_n \sim N(0,1)$ than $\sum_i Y_i^2 \sim \chi^2_n$.
However, if we know that (independent) $Y_1,...,Y_n$ have $\sum_i Y_i^2 \sim \chi^2_n$, can we say that $Y_1,...,Y_n \sim N(0,1)$?
 A: No. The sum of independent $\chi$-squares is $\chi$-squared, so $Y_i\sim\chi_1$ will work, and $\chi_1\neq\mathcal{N}\left(0,1\right)$.
A: One other interesting example is the skew-normal distribution.  Based on the standard normal distribution, but with an added parameter $\alpha$ to represent skewness, the density function is 
$$
   f(x)=2 \phi(x)\Phi(\alpha x), \quad \alpha \in \mathbb{R}.
$$
Its cdf (cumulative distribution function) an be written as 
$$
   F(x)= \Phi(x)-2 T(x,\alpha)
$$
where $T(x,\alpha)$ is the Owen T function and $\phi(x), \Phi(x)$ is the standard normal density and cdf. Using the property that $T(-x,\alpha)=T(x,\alpha)$ it is now an easy exercise to show that $X^2$ ($X$ distributed skewnormal $\alpha$) has the same distribution as $Z^2$ when $Z$ is standard normal (this is an important result in inference theory for skew normal models.)  
Another way to see it is using the comment of @whuber. If $X$ is a continuous random variable with density $f$, then the absolute value $\mid X \mid$ has density $f(x)+f(-x)$ for $x>0$. So we just have to look at 
\begin{align}
   2\phi(x)\Phi(\alpha x) + 2\phi(-x)\Phi(-\alpha x) &=& 2 \phi(x)\left\{ \Phi(\alpha x)+\Phi(-\alpha x)\right\} \\
&=& 2 \phi(x) \\
&=& \phi(x)+\phi(-x) \quad \text{for $x>0$.}
\end{align}
