$a=\frac{b×(3c-5)}{15}$ and $a,b,c$ are numbers from $S$, prove that S has infinitely many elements. This is a question from my exam and I don't really know how to solve it.
Let S be the set of positive integers such that for $a$ is any number from S, there exist two numbers $b$ and $c$ also from S such that:

$a=\frac{b×(3c-5)}{15}$

Prove that the set S has infinitely many elements.
And this is what I have done:
If $3c-5$ is divisible by $15$
Then $3c-5$ is divisible by $5$ and $3$
$ \rightarrow c$ is divisible by $5$
$ \rightarrow c = 5k \rightarrow 3c-5=5\times{3k-1}$ which is not divisible by $3$
So b must divisible by $15$
 A: I am going to find all the whole solutions of the following equation :
$15a=b(3c-5)$
Since $\;3\mid b(3c-5)\;$ and $\;3\not\mid(3c-5)\;,\;$ then $\;3\mid b\;,\;$ hence $\;b=3b_1\,.$
Consequently, we get that
$15a=3b_1(3c-5)\;\;,$
$5a=b_1(3c-5)\;.$
There are two possible cases : $\;5\mid b_1\;$ or $\;5\mid(3c-5)\;.$
First case :$\;5\mid b_1$
If $\;5\mid b_1\;,\;$ then $\;b_1=5b_2\;,$
$5a=5b_2(3c-5)\;\;,$
$a=b_2(3c-5)\;.$
Second case:$\;5\mid(3c-5)$
If $\;5\mid(3c-5)\;,\;$ then $\;5\mid c\;,\;$ hence $\;c=5c_1\;,$
$5a=b_1(3\cdot5c_1-5)\;\;,$
$5a=5b_1(3c_1-1)\;\;,$
$a=b_1(3c_1-1)\;.$
Therefore, all the whole solutions of the equation are the following ones :
$\begin{cases}
a=\lambda(3\mu-5)\\
b=15\lambda\\
c=\mu
\end{cases}\qquad\lor\qquad\begin{cases}
a=\lambda(3\mu-1)\\
b=3\lambda\\
c=5\mu
\end{cases}$
for any $\;\lambda,\mu\in\mathbb Z\;.$
For any $\;\lambda,\mu>1\;,\;$ we get infinitely many positive solutions of the equation.
Now I am going to prove that the set $\;S\;$ has infinitely many elements.
Since $\;3\mid b\;,\;$ there exist $\;n,\beta_1\in\mathbb Z^+\;$ such that $\;b_1=b=3^n\beta_1\in S\;$ and $\;3\not\mid\beta_1\;.$
Moreover,
$b_1=3^n\beta_1=\lambda(3\mu-5)\quad\lor\quad b_1=3^n\beta_1=\lambda(3\mu-1)\;.$
Since $\;3\not\mid(3\mu-5)\;$ and $\;3\not\mid(3\mu-1)\;,\;$ it follows that
$\lambda=3^n\lambda^*\;$ where $\;\lambda^*\in\mathbb Z^+\;$ and $\;3\not\mid\lambda^*\;.$
Consequently ,
$b_2=15\lambda=3^{n+1}\cdot5\lambda^*\in S\quad\lor\quad b_2=3\lambda=3^{n+1}\lambda^*\in S\;.$
In any case ,
$b_2=3^{n+1}\beta_2\in S\;$ where $\;\beta_2\in\mathbb Z^+\;$ and $\;3\not\mid\beta_2\;.$
By proceeding analogously, we get an infinite sequence of different elements of $\;S\;$ that is :
$b_k=3^{n+k-1}\beta_k\in S\quad$ for all $\;k\in\mathbb Z^+\;,$
where $\;\beta_k\in\mathbb Z^+\;$ and $\;3\not\mid\beta_k\quad$ for any $\;k\in\mathbb Z^+\;.$
Hence the set $\;S\;$ has infinitely many elements.
A: Suppose $S$ is finite, then there exists $m$ such that $3^m$ is the largest power of $3$ that divides any element of $S$.
Therefore there exists $a, b, c, d \in S$ such that $a = 3^m d$ and $15 a=b*(3c-5)$.
It follows that $3^{m+1}*5d=b*(3c-5)$ and since $3 \nmid (3c-5)$ we have $3^{m+1} \mid b$.
This contradicts the definition of $m$ hence $S$ cannot be finite.
Note strictly speaking $S$ could be the empty set. Also note $S$ is not unique. Any set than contains $2$ and is closed under multiplication by $15$, will have the desired property. You can show $S$ must contain $2$.
