Exercise 4.1.11 (Divisors of $p^2 q^3$ )

Question

If $p$ and $q$ are distinct primes, prove that the divisors of $p^2 q^3$ coincide with the terms of $(1+p+p^2)(1 + q + q^2 +q^3)$ when the latter is multiplied out.

Richard Ganaye · Accepted Answer

\begin{proof} If $d \mid n = p^2 q^3$, where $p,q$ are distinct primes, then $d = p^i q^j$ where $0 \leq i \leq 2, 0 \leq j \leq 3$.  Therefore
\begin{align*}
(1+p+p^2)(1 + q + q^2 +q^3) &= \sum_{i = 0}^2 p^i \sum_{j = 0}^3 q^j\
&= \sum_{(i,j) \in [\![0 , 2]\!] 	imes [\![0 , 3]\!]} p^i q^j\
&= \sum_{d \mid n} d.
\end{align*}
So the divisors of $n = p^2 q^3$ coincide with the terms of $(1+p+p^2)(1 + q + q^2 +q^3)$ when the latter is multiplied out.
\end{proof}

Exercise 4.1.11 (Divisors of $p^2 q^3$ )

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Exercise 4.1.11 (Divisors of $p^2 q^3$ )

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