Proving $P$ for $N(n+N-1)$ Let $n$ be a non zero natural number.
We say that $n$ has the property $P$ if there exists a sequence of numbers  $a_k$,$a_1$,$a_2$...$a_m$ strictly positive rational numbers (not necessarily distinct), we get : $\sum_{k=1}^m a_k$ $=$ $n$ and $\sum_{k=1}^m \frac{1}{a_k} = 1$
Question : Suppose $n$ has the property $P$.
Show that  $N(n+N-1)$ too has the property $P$ for every non zero natural numbers $N$.
This probably is a very easy proof but i don't really know the correct answer to this because i'm completely new to Sums and Proofs.
Any hints/tips are appreciated.
I know that for the $P$ to work for $n$, $a_k$ has to be equal to $\sqrt n$ and $m$ has to be equal to $\sqrt n$. or maybe i'm wrong here too.
 A: Let $S$ be the set of positive integers $n$ such that there exists a finite sequence of positive rational numbers whose sum is $n$, and the sum of whose reciprocals is $1$.

Let $n \in S$, and let $N$ be a positive integer.

We want to show $N(n+N-1) \in S$.

Since $n \in S$, there exist  positive rational numbers $a_1,...,a_m$ such that
\begin{align*}
\sum_{i=1}^m a_i &= n\\[4pt]
\sum_{i=1}^m \frac{1}{a_i} &= 1\\[4pt]
\end{align*}
Let$\;q = m + N -1,\;$and let$\;b_1,...,b_q\;$be defined by
$$
b_i =
\begin{cases}
Na_i&\text{if$\;\,1\le i \le m$}\\[4pt]
N&\text{if$\;\,m < i \le q$}\\[4pt]
\end{cases}
\\[4pt]
$$
\begin{align*}
\text{Then}\;\;\sum_{i=1}^q  b_i
&=\sum_{i=1}^m b_i +\sum_{i=m+1}^q b_i\\[4pt]
&=\sum_{i=1}^m Na_i +\sum_{i=m+1}^q N\\[4pt]
&=N\left(\sum_{i=1}^m a_i\right) +N\left(\sum_{i=m+1}^{m+N-1} 1\right)\\[4pt]
&=N(n) + N(N-1)\\[4pt]
&=N(n+N-1)\\[14pt]
\text{and}\;\;\sum_{i=1}^q \frac{1}{b_i} 
&=\sum_{i=1}^m \frac{1}{b_i}+\sum_{i=m+1}^q \frac{1}{b_i}\\[4pt]
&=\sum_{i=1}^m \frac{1}{Na_i}+\sum_{i=m+1}^q \frac{1}{N}\\[4pt]
&=\frac{1}{N}\left(\sum_{i=1}^m \frac{1}{a_i}\right)+\frac{1}{N}\left(\sum_{i=m+1}^{m+N-1} 1\right)\\[4pt]
&=\frac{1}{N}(1) + \frac{1}{N}(N-1)\\[4pt]
&= 1\\[4pt]
\end{align*}
It follows that $N(n+N-1) \in S$, as was to be shown.
