<s>[INST] Prime Number Theory (Dividers) When $N = a^\alpha × b^β × c^γ$ is divided by $a$, the number of dividers decreases by $m$ units; when divided by $b$, decreases by $n$ and, dividing by $c$, decreases by $p$. Determine the exponents $α, β$ and $γ$.
Mine ended up giving in this system here:
$$
(α-a)(β+1)(γ+1)-(γ+1)(β+1) = (α+1)(β+1)(γ+1)-m; 
(β-b)(α+1)(γ+1)-(γ+1)(α+1) = (α+1)(β+1)(γ+1)-n;
(γ-b)(α+1)(β+1)-(α+1)(β+1) = (α+1)(β+1)(γ+1)-p
$$
 [/INST] When $N = a^\alpha \times b^{\beta} \times c^{\gamma}$ is divided by $a$, it becomes $\frac{N}{a} = a^{\alpha - 1} \times b^{\beta} \times c^{\gamma}$. Thus, as metamorphy's question comment indicates that if we can assume $a$, $b$ and $c$ are distinct primes, then since the number of divisors decreases by $m$, this means that
$$\begin{equation}\begin{aligned}
(\alpha - 1 + 1)(\beta + 1)(\gamma + 1) & = (\alpha + 1)(\beta + 1)(\gamma + 1) - m \\
(\alpha + 1)(\beta + 1)(\gamma + 1) - (\beta + 1)(\gamma + 1) & = (\alpha + 1)(\beta + 1)(\gamma + 1) - m \\
- (\beta + 1)(\gamma + 1) & = -m \\
(\beta + 1)(\gamma + 1) & = m
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Similarly, we also have
$$(\alpha + 1)(\gamma + 1) = n \tag{2}\label{eq2A}$$
$$(\alpha + 1)(\beta + 1) = p \tag{3}\label{eq3A}$$
Multiplying \eqref{eq1A}, \eqref{eq2A} and \eqref{eq3A} together gives
$$\begin{equation}\begin{aligned}
\left((\alpha + 1)(\beta + 1)(\gamma + 1)\right)^2 & = mnp \\
(\alpha + 1)(\beta + 1)(\gamma + 1) & = \sqrt{mnp}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Dividing \eqref{eq4A} by \eqref{eq1A} gives
$$\begin{equation}\begin{aligned}
\alpha + 1 & = \frac{\sqrt{mnp}}{m} \\
\alpha & = \sqrt{\frac{np}{m}} - 1
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
Similarly, you can also get that
$$\beta = \sqrt{\frac{mp}{n}} - 1 \tag{6}\label{eq6A}$$
$$\gamma = \sqrt{\frac{mn}{p}} - 1 \tag{7}\label{eq7A}$$
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