I've tried to prove this theorem, which is very simple, but is a kind of practice for me.
Let $a,b$ be two positive integers. Therefore, if $a+b$ is a composite number, $frac(\frac{a}{l}) + frac(\frac{b}{l}) = 0$ or $frac(\frac{a}{l}) + frac(\frac{b}{l}) = 1$, for some $l \in \mathbb{N}$, $l \neq 1$.
My proof:
If $a+b$ is composite, there's some $l,k \in \mathbb{N}, 1<l $ such that ${a+b \over l}=k$. If we suppose $l$ is a factor of both $a,b$, $frac({a \over l})+ frac({b \over l}) = 0$. But, if we denote $a=c_al+d_a$ and $b=c_bl+d_b$, for $d_a,d_b \in \mathbb{Z_+^*};|d_a|,|d_b|<l$, then ${a+b \over l}={c_al+d_a+c_bl+d_b \over l}= c_a+c_b+{d_a+d_b \over l}$. Since ${a+b \over l}$ is an integer, ${d_a+d_b \over l}$ must be an integer too. Here we have only two possibilites:${d_a+d_b \over l}=0$ or ${d_a+d_b \over l}= \pm 1$ (because $|d_a|,|d_b|<l$). From the first hypothesis, ${d_a+d_b \over l}=0 \iff {d_a \over l} = {-d_b \over l} \implies frac({d_a \over l}) = frac(-{d_b \over l}) \implies frac({d_a \over l})+frac({d_b \over l}) = 1 \implies frac(c_a+{d_a \over l})+frac(c_b+{d_b \over l}) = 1 \implies frac(a)+frac(b) = 1 $. From the second hypothesis, ${d_a+d_b \over l}= \pm 1 \iff {d_a \over l}= \pm 1 -{d_b \over l} \implies frac({d_a \over l})= frac(\pm 1 -{d_b \over l}) \implies frac({d_a \over l})= frac(-{d_b \over l}) \implies frac({d_a \over l})= 1-frac( {d_b \over l}) \implies frac({d_a \over l})+frac({d_b \over l}) = 1 \implies frac(c_a+{d_a \over l})+frac(c_b+{d_b \over l}) = 1 \implies frac(a)+frac(b) = 1$
Q.E.D.
Why $d_a,d_b \in \mathbb{Z_+^*}$ ?
$d_a,d_b \in \mathbb{N_+^*}$ is enough since $a,b \in \mathbb{N_+^*}$ and $c_a$ and $c_b$ can be equal to $0$.
Therefore $frac(\frac{d_a}{l}) + frac(\frac{d_b}{l}) = $$0$ or $1$ since $\frac{a+b}{l}$ is an integer and $0\leq d_a, d_b$
EDIT
Since $0\leq d_a, d_b< l$, $0\leq \frac{d_a}{l}< 1$ and $0\leq \frac{d_b}{l}< 1$ so $0\leq \frac{d_a}{l}+\frac{d_b}{l}< 2$ or $\frac{d_a+d_b}{l}=\frac{a+b}{l}-c_a-c_b$ is an integer so $\frac{d_a+d_b}{l}=0$ or $1$ and $frac(\frac{d_a}{l}) + frac(\frac{d_b}{l}) = $$0$ or $1$