Properties of $\chi^2(1)$ multiplied by a real value "$a$" Is it true that the $\chi^2$ distribution with $k=1$ (noted $\chi^2(1)$) multiplied by a real value $a$ is equal to $\chi^2(a)$ ?
If not, is there a particular distribution for $a\cdot\chi^2(1)$?
If yes, it doesn't make sense since the summing index is no longer integer, does it?
Positive value for $a$ is also mandatory, no?

TL;DR
$$ a>0, X\sim\chi^{2}(k)\implies aX\sim\Gamma\left( \frac{k}{2}, 2a\right) $$
The full story
The assertion is false. You can see it as follows.
Take $k\geq 1$, and consider $X\sim \chi^{2}(k)$ (read as: X is distributed as chi-square with k degrees of freedom). Take any constant $a>0$: what is the distribution of the random variable $aX$?
You can show that $aX\sim \Gamma\left( \frac{k}{2}, 2a\right)$.
Indeed,
$$ P(aX\leq t) = P\left(X \leq \frac{t}{a} \right) $$
(note that here we use the fact that $a>0$ otherwise the equality is not valid).
\begin{align*}
   P\left(X \leq \frac{t}{a}\right) &=
   \int_{0}^{ \frac{t}{a} } \frac{x^{k/2-1}e^{-x/2}}{2^{k/2}\Gamma(k/2)}dx \\
                                    &\text{use the substitution } x= \frac{l}{a} \\
                                    &=\int_{0}^{t} \frac{(l/a)^{k/2-1}e^{-l/(2a)}}{2^{k/2}\Gamma(k/2)} \frac{1}{a} dl \\
                                    &=\int_{0}^{t} \frac{l^{k/2-1}e^{-l/(2a)}}{(2^{k/2}a^{k/2-1})\Gamma(k/2)} \frac{1}{a} dl \\
                                    &=\int_{0}^{t} \frac{l^{k/2-1}e^{-l/(2a)}}{(2^{k/2}a^{k/2})\Gamma(k/2)} dl \\
                                    &=\int_{0}^{t} \frac{l^{k/2-1}e^{-l/(2a)}}{(2a)^{k/2}\Gamma(k/2)} dl \\
                                    &=P\left(Y \leq t\right) 
\end{align*}
where $Y$ is distributed as $\Gamma\left( \frac{k}{2}, 2a\right)$.
Thus, take $k=1$ and you have that
$$ a>0, X\sim\chi^{2}(1)\implies aX\sim\Gamma\left( \frac{1}{2}, 2a\right). $$
The latter has pdf equal to
$$ \text{pdf } \Gamma\left( \frac{1}{2}, 2a\right)=\frac{x^{1/2-1}e^{-x/(2a)}}{(2a)^{1/2}\Gamma(1/2)}$$
which is different from the pdf of $\chi^2(a)$, which is equal to
$$ \frac{x^{a/2-1}e^{-x/2}}{2^{a/2}\Gamma(a/2)}  $$
