Case study of observation

I am suprised that I could able to solve this question!

First, when I'm finding the pattern of $f^k (36)$ , I get this

$f^0 (36) =36$

$f (36) = \boxed{42}=36+6 = 6^2+6$

$f^2 (36) = 48 = 42+6$

$f^3 (36) = \boxed{54} = 48+6 =7^2+5$

$f^4 (36) = 61 = 54 + 7$

$f^5 (36) = \boxed{68} =61+7 =8^2+4$

$f^6 (36) = 76 =68+8$

$f^7 (36) = \boxed {84} =76+8=9^2 +3$

$f^8 (36) = 93 =84+9$

$f^9 (36) = \boxed {102} =93+9=10^2+2$

$f^{10} (36) = 112 = 102+10$

$f^{11} (36) = \boxed {122} = 112+10 =11^2 +1$

$f^{12} (36) = 133 = 122+11$

$f^{13} (36) = \boxed {144} = 12^2$

$k_1 =13$

I thought about what makes this consequence, every two row leads to something\s square and add a uniformly decreasing integer.

So I recall that the difference between $m^2$ and $(m+1)^2$ be $2m+1$ .

Then, I compare $f^1$ with $f^3$ and $f^3$ with $f^5$ , found that $f^3-f^1=(7^2+5)-(6^2+6)=(7^2-6^2)-1=(2m+1)-1=2 \times 6$

$f^5-f^3=(8^2+4)-(7^2+5)=(8^2-7^2)-1=(2m+1)-1=2 \times 7$

$f^7-f^5=(9^2+3)-(8^2+4)=(9^2-8^2)-1=(2m+1)-1=2 \times 8$

and so on.

So, I found a list to find for which $k$ satisfied $f^k (n)$ for any known $n$

1: Turns $n$ into $n=m^2+r$ which $m^2$ is smaller than $n$ and the nearest perfect square to $n$ . Notice that if there exists a $n'$ makes $n'=m^2+r'$ with $r'<r$ ,then $n$ shouldn't be considered, just as those numbers which didn't be boxed.

2: because of every two times of functioning $f$ , $r$ gets $1$ smaller, so first calculate how many time functioning to get $n$ , called $k_0$ , solution of $k$ will be $k=k_0+2r$ .

As the first question, $k_0=1$ leads to $n=42$ , $r=6$ , so leads to $k_1=1+2 \times 6=13$ . Ya, it works!

Continue to solve $k_2$ , we get $k_0=0$ , $r=4$ , so $k_2=8$ .

For $k_3$ , $k_0=1$ , $r=0$ , so $k_3=1$ .

For $k_4$ , $k_0=1$ , $r=5$ , so $k_4=11$ .

For $k_5$ , since $2072$ is quite large , $\sqrt{2072}=45.5192...$ , $f(2072)=2117=2116+1=46^2+1$ , so $2117$ is that $n$ , so $k_0=1$ , $r=1$ , then $k_5=3$ .

For $k_6$ , from above status we know $2115$ is exactly $1$ lesser than $46^2$ , so try to functioning once, we get $f(2115)=2115+45=2160=46^2+44$ , we get $k_0=1$ and $r=44$ , if we really expand steps of functioning this number we will get 89 row, it is a messy! So let us straightforward using the formula we get $k_6=1+2 \times 44=89$ .

$k_1+k_2+k_3+k_4+k_5+k_6-89= \boxed{36}$

first of all let $m^{2}\leq(n)<(m+1)^{2}$ .

we get $f(n)=n+m$ .

Then we try to find the value of $k$ s of which $f^{k}(n)=z^{2}$ where $z$ is an integer.

(here the interval of $n$ is $[m^{2},(m+1)^{2})$ .

We find the below pattern .

hence we get $k_{1}=13,k_{2}=8,k_{3}=1,k_{4}=11,k_{5}=3,k_{6}=89$ .

Proof of the pattern obtained

$n=m^{2}+p ; (1\leq p \leq m)$

$f(n)=f(m^{2}+p)=m^{2}+p+m$

$f^{2}(m^{2}+p)=m^{2}+p+2m=(m+1)^{2}+(p-1)$

$f^{2}((m+1)^{2}+(p-1))=(m+2)^{2}+(p-2)$

$f^{2}((m+2)^{2}+(p-2))=(m+3)^{2}+(p-3)$

$f^{2p}((m)^{2}+p)=(m+p)^{2}+(p-p)=(m+p)^{2}$

Simlarly we can prove for $m+1\leq p \leq 2m$