Question

Let
\[
f(n)=\left\lfloor \sqrt{n}\right\rfloor +n
\]
Show that for every integer m greater or equal to 1, the sequence
\[
f^{k}(m), \, k=0,1,2,3,\cdots
\]
has at least one square.

Answer 1

k=1,2,3
f^2(x)= f(f(x)
f^3(x)=f(f(f(x)))

Answer 2

yes
you have to show it for every n

Answer 3

No, @mahmit2012 , you're showing it for all \(m\), there exists a \(k\) such that \(f^k(m)\) is a square.

Answer 4

No the sequence f^k(m), k=1,2,3,4,... has a square for every m

Answer 5

Is that not saying the same thing as what I just said? \[\forall n,\ \exists k \in\mathbb N,\ f^k(n) \text{ is square}\] Also, is \(n\) restricted to particular values, such as \(\mathbb N\) or \(\mathbb Z\)?

Answer 6

@mahmit2012 In order to show that, you need to show that there exists some \(k\) such that \(f^k(n)\) is a square. If there didn't exist some \(k\), then there would not be at least one square in the sequence, which transcends all possible values of \(k \in \mathbb N\).

Answer 7

The sequence\[ (f^k(m))_{k \ge1} \] has a square for every \[m \ge 1\]

Answer 8

\[
f^8(7)= 36=4^2 \\
f^{10}(7+1)= 50\ne (4+1)^2 =25
\]

Answer 9

Ah, Putnam problems. They look easy, but they're deviously hard.

Answer 10

In this book where I've seen the problem, the author presented a conjecture that was not quite right:
If \(m=n^2+b \text{where } 0 \leq b <2n+1\), then \(f^2(m)=k^2+(b-1)\).
But he said that a similar conjecture is correct, and leads to the full solution. So the key is finding the correct conjecture.

Answer 11

*

Answer 12

if i suppose there is no m,k wich satisfies it and prove there is a solution,then for absurd i will prove it

Answer 13

@eliassaab what you think?

Answer 14

Sorry, k=0 is included and
\[
f^0(m)=m
\]

Answer 15

It is enough to prove it for all integers between two given squares.Fix n. It is enough to prove the statement for all m such that
\[
n^2 \le m \le (n+1)^2
\]

Answer 16

It is enough to prove it for all integers between two consecutive squares

Answer 17

You did not attend my solution that was right.
Check your solution again that is wrong and not generalized solution.

Answer 18

It's not true for r=3:
\[m=7=2^2+3 \text{ but } f^6(m)\neq (2+3)^2\]

Answer 19

Suppose that
\[

m=n^2 + r

\]
Let us distinguish two cases:

case I:

\[ 1\le r \le n\\
f(m)=n^2 +r + n \\
f^2(m)=n^2 + r + n +n, \text { since } n^2 + r + n < n^2 + 2n +1\\
f^2(m)=n^2 + 2n +1 -1 +r=(n+1)^2 + r-1
\]

If r =1, then \[ f^2(m)\] is a square.
In the general case \[f^{2 r}(m)\] is a square.


case II:

\[ n+1\le r \le 2n +1\\
f(m)=n^2 +r + n= \\
n^2 + r + n+n -n +1 -1=\\
n^2 + 2 n +1 + r-n -1=\\
(n+1)^2 + r-n-1
\]

If r =n+1, then \[ f(m)\] is a square.
If r> n+1, you can show that if k= 2(r-n-1)+1, then
\[
f^k(m)
\]
is a square.