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If $x_1(t)$ is periodic with period $T_1$ and $x_2(t) \neq −x_1(t)$ is periodic with period $T_2$, then $x_1(t)+x_2(t)$ will be periodic with period T if there exists integers n1 and n2 with no common factors such that $T = n_1T_1$ and $T = n_2T_2$.

I'm not understanding this. Why do $n_1$ and $n_2$ need to be integers? Assume $n_1$ = 4 and $n_2 = \pi$. Then $T$ repeats at every $4$ and at every $\pi$, so it simply is periodic at every $4$-multiple of $T_1$ or every $\pi$-multiple of $T_2$.

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    It has to do with the definition of the period. By definition, the period of a periodic function $f:D\subset\mathbb{R} \to \mathbb{R}$ is the smallest positive number $T$ such that $$\tag{1} f(x+T)=f(x) \quad \forall x\in D. $$ It is then clear that if $n$ is an integer, then for every $x\in D$ we have $$\tag{2} f(x+nT)=f(x) $$ To show that $n$ must be an integer for (2) to hold, consider the function $$ f(x)=\cos(x). $$ Its period is $T=2\pi$, and if you take $n=\pi$, you can check that (2) does not hold for any $x\in \mathbb{R}$. – 2017-01-13

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$x(t+T)=x_1(t+T)+x_2(t+T)$

we know for sure that $x_1(t+T)=x_1(t)$ if $T$ is a multiple of $T_1$

similarly $x_2(t+T)=x_2(t)$ if $T$ is a multiple of $T_2$

so if $T$ is a simultaneously a multiple of $T_1$ and $T_2$ then $x(t+T)=x(t)$.

So it is working for $T=T_1T_2$ but in fact the least T for which this is working for sure is $T=lcm(T1,T2)$. This may seems strange to speak about $lcm$ for reals, but it works merely the same than with integers.

This is why you have $T=n_1T_1=n_2T_2$ with $gcd(n_1,n_2)=1$, this is just a manner to describe the $lcm$.

So if you take for instance $f_1(t)=\sin(t)$ and $f_2(t)=\{t\}=t-\lfloor t\rfloor$, respectively of periods $2\pi$ and $1$, then the sum is not periodic because you cannot find integers such that $2\pi\times n_1=1\times n_2$ (since $\pi$ irrationnal).

But you can for $f_1(t)=sin(3t)$ and $f_2(t)=sin(\frac{22}{7}t)$ because both periods are rationnal multiples of $\pi$ and the sum is periodic with period $14\pi$.

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It does not say that $n_1$ and $n_2$ must be integer. It says that IF $T_1/T_2$ is rational then $x_1+x_2$ is periodic. Any integer multiple of a period of a function is also a period of it. So IF $T=n_1T_1 =n_2T_2$ with integers $n_1,n_2$ then $x_1+x_2$ is periodic with period $T.$ If $n'_1T_1=n'_2T_2\implies n'_2/n'_1=T_1/T_2,$ where $n'_1$ and $n'_2$ are integers, we can write $n'_2/n'_1=n_2/n_1$ where $n_2/n_1$ is in lowest terms.

If $T_1=4$ and $T_2=\pi$ and if $x_1(0)\ne x_1(\pi)$ then $x_1(\pi)+x_2(\pi)=x_1(\pi)+x_2(0)\ne x_1(0)+x_2(0).$

If $T_1/T_2$ is irrational then $x_1+x_2$ may fail to be periodic.

If a continuous real function $f$ with a non-zero period $P$ also has a period $Q$ where $P/Q$ is irrational then $f$ is constant. (Which is not obvious.) But an everywhere-discontinuous function can have periods whose ratio is irrational.