Question

Given that the divisors of $n=3^p{5}^q{7}^r$ are in the form of $4\lambda +1$,then $p+r$.

Answer 1

Step $1$: Find multiple of $3^p$in terms of $4\lambda +1$.

If $4\lambda +1$is the divisor of $n=3^p{5}^q{7}^r$,then $n$must be a multiple of $4\lambda +1$.

$3^p={(4-1)}^p$

$=4^p+.........+{(-1)}^p$

The terms in between are all factors of $4$, so taking $4$ common from them we get

$=4\lambda \text{'}+{(-1)}^p..........\left(1\right)$

where $\lambda \text{'}=$all the rest of the terms

Step $2$: Find multiple of $5^q$in terms of $4\lambda +1$.

$5^q={(4+1)}^q$

$$\begin{gathered} =4^q+......+1^q \\ =4\lambda \text{'}\text{'}+1........\left(2\right) \end{gathered}$$

Step $3$: Find multiple of $7^r$in terms of $4\lambda +1$.

$7^r={(8-1)}^r$

$=8^r+.........+{(-1)}^r$

$=8\left({\lambda }^0\right)+{(-1)}^r$

$=4\left(2{\lambda }^0\right)+{(-1)}^r$

$=4\lambda \text{'}\text{'}\text{'}+{(-1)}^r..........\left(3\right)$

Observing $\left(1\right),\left(2\right)$ and $\left(3\right)$ we get, $\left(2\right)$ is in the form of $4\lambda +1$ but $\left(1\right)$ and $\left(3\right)$ will be in the form of $4\lambda +1$ if $p$ and $r$ are even.

Hence, the results are:

1. $p+r$ is an even number.
2. $p+q+r$ can be even or odd, depending on $q$.
3. $q$can be any integer.