Do there exist an infinite number of $S_m=\frac{m(m+3)}{2}$ such that $S_m=S_p+S_q=S_r-S_t=S_uS_v$? Is the following true?
"Letting $S_n=\frac{n(n+3)}{2}$ for $n\in\mathbb N$, there exist an infinite number of $S_m$ such that 
$$S_m=S_p+S_q=S_r-S_t=S_uS_v$$
where $S_i\gt 2\ (i=p,q,r,t,u,v).$"
Note that $S_i\gt 2\ (i=p,q,r,t,u,v)$. (otherwise this question is too easy.) 
Motivation : I've known that there exist an infinite number of the above examples in the triangular numbers, which can be represented as $\frac{n(n+1)}{2}$. This got me interested in the above. This expectation seems true by using computer.
Examples :
$$S_{45}=S_{15}+S_{42}=S_{52}-S_{25}=S_{5}S_{9}$$
$$S_{60}=S_{42}+S_{42}=S_{63}-S_{18}=S_{7}S_{9}$$
$$S_{63}=S_{36}+S_{51}=S_{64}-S_{10}=S_{6}S_{11}$$
$$\vdots$$
$$S_{1845}=S_{330}+S_{1815}=S_{1858}-S_{218}=S_{41}S_{60}$$
 A: Claim 1) There exist an infinite number of triples $(a,b,c)$ of odd numbers such that: $$(\spadesuit)\;(a^2-9)(b^2-9)=8(c^2-9).$$
For any odd value of $b_0$, $(3,b_0,3)$ is clearly a solution of $(\spadesuit)$. This implies that the Pell equation
$$ \frac{b_0^2-9}{8}a^2-c^2 = D = 9\frac{b_0^2-17}{8}$$
has a solution $(a,c)=(3,3)$, hence it has an infinite number of solutions. 
In particular, if $u,v$ satisfy
$$ u^2 - \frac{b_0^2-9}{8} v^2 = 1, $$
then 
$$ \left(ua+vc,b_0,uc+va\frac{b_0^2-9}{8}\right) $$
is another solution of $(\spadesuit)$. For example, starting from $(a,b,c)=(3,5,3)$ we have that $(15,5,21)$ is another solution of $(\spadesuit)$. 
However, an equivalent form of $(\spadesuit)$ is the following:
$$ S_{\frac{a-3}{2}} S_{\frac{b-3}{2}} = S_{\frac{c-3}{2}}, $$
so there exists an infinite number of "special numbers" $S_n$ that are product of two special numbers.
Claim 2) Every natural number $m\geq 3$ can be expressed as the difference of two special numbers.
This follows from the simple observation that $S_{n+1}-S_n = n+2$.
Claim 3) $m$ can be written as a sum of two special numbers iff $8m+18$ can be written as a sum of two squares both greater than $9$.
We have $8S_n=(2n+3)^2-9$, so $N=S_c+S_d$ implies $8N+18=(2c+3)^2+(2d+3)^2$. On the other hand, if $8N+18$ is the sum of two squares, it is the sum of two odd squares, since $8N+18\equiv 2\pmod{4}$, so $8N+18=C^2+D^2$ implies $N=S_{\frac{C-3}{2}}+S_{\frac{D-3}{2}}$.
Assume now that $(a,b,c)$ is a solution of $(\spadesuit)$. A simple inspection $\pmod{3}$ tells us that $3|c$, so $c=3d$, and $3$ divides at least one number between $a$ and $b$. Moreover:
$$8S_{\frac{c-3}{2}}+18 = c^2+9.$$
By choosing $b=5$, we have $c^2+9=2a^2=a^2+a^2$, so if $(a,5,c)$ is a solution of $(\spadesuit)$,
$$S_{\frac{c-3}{2}}$$
is the sum of two special numbers, and this is sufficient to prove the conjecture.
