Proving that almost all numbers are good.

Question

In this year's PreRMO examination(Leg 1), the question $12$ is as follows:

A natural number $k>1$ is called good if there exist natural numbers $a_{1}<a_{2}<.....a_{k}$ such that $ \frac{1}{\sqrt{a_1}}+\frac{1}{\sqrt{a_2}}+.....\frac{1}{\sqrt{a_k}}=1$. Let $f(n)$ be the sum of the first $n$ good numbers $n\geq 1$. Find the sum of all values of $n$ for which $\frac{f(n+5)}{f(n)}$ is an integer.

As I looked up video solutions, the instructor showed that both $k=3$ and $k=4$ are good and said that in a similar way, all numbers $k\geq 3$ are good. Is there any way to prove this? I tried using induction but I realised it was no good.

Answer 1

So suppose$$\frac 1{\sqrt {c_1}}+\frac 1{\sqrt {c_2}}+\frac 1{\sqrt {c_3}}=1$$

then $$\frac 1{\sqrt {c_1a_k}}+\frac 1{\sqrt {c_2a_k}}+\frac 1{\sqrt {c_3a_k}}=\frac 1{\sqrt {a_k}}$$

So from a solution for $n=3$ and a solution for $n=k$ you can manufacture a solution for $n=k+2$

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For the second part you need to make sure you include all the good numbers.

Answer 2

Consider $b_1=2,\,b_{n+1}=b_n(b_n-1)+1$ so $\sum_{j=1}^k\frac{1}{b_j-\delta_{jk}}=1$ (proof by induction on $k$ is an exercise). Now take $a_j=(b_j-\delta_{jk})^2$ to ensure $\sum_{j=1}^k\frac{1}{\sqrt{a_j}}=1$. The first "legal" sum this gives us is $\frac12+\frac13+\frac16=1$, so $k\ge 3$ are good.