Determine all postive integer k

[anemone]

Determine all positive integers $k$ for which $f(k)>f(k+1)$ where $f(k)=\left\lfloor{\dfrac{k}{\left\lfloor{\sqrt{k}}\right\rfloor}}\right\rfloor$ for $k\in \Bbb{Z}^*$.

 

[kaliprasad]

Determine all positive integers $k$ for which $f(k)>f(k+1)$ where $f(k)=\left\lfloor{\dfrac{k}{\left\lfloor{\sqrt{k}}\right\rfloor}}\right\rfloor$ for $k\in \Bbb{Z}^*$.

Spoiler

if k+1 is not a perfect square then the floor of square root of k and k+ 1 are same so
f(x) < f(x+1)
so we need to look at k+1 being a perfect square say $n^2$
$f(k) = \left\lfloor\dfrac{n^2-1}{n-1}\right\rfloor= n + 1$
$f(k+1) = \left\lfloor\dfrac{n^2}{n}\right\rfloor= n$
so k is of the form $n^2-1$ for $n\gt 1$

 

[anemone]

Spoiler

if k+1 is not a perfect square then the floor of square root of k and k+ 1 are same so
f(x) < f(x+1)
so we need to look at k+1 being a perfect square say $n^2$
$f(k) = \left\lfloor\dfrac{n^2-1}{n-1}\right\rfloor= n + 1$
$f(k+1) = \left\lfloor\dfrac{n^2}{n}\right\rfloor= n$
so k is of the form $n^2-1$ for $n\gt 1$

Hey kaliprasad, thanks for participating and your answer is correct! I think it might be necessary(?) to prove that for $k=n^2+a$ with $0<a<2n$,

$f(k)=\left\lfloor{\dfrac{n^2+a}{n}}\right\rfloor=n+\left\lfloor{\dfrac{a}{n}}\right\rfloor$ which is non-decreasing.

But then this is an easy proof, so, there is no big deal here.:)

 

[kaliprasad]

Hey kaliprasad, thanks for participating and your answer is correct! I think it might be necessary(?) to prove that for $k=n^2+a$ with $0<a<2n$,

$f(k)=\left\lfloor{\dfrac{n^2+a}{n}}\right\rfloor=n+\left\lfloor{\dfrac{a}{n}}\right\rfloor$ which is non-decreasing.

But then this is an easy proof, so, there is no big deal here.:)

Hello Anemone,

Spoiler

Both are effectively same as $k= n^2+ a$ with $0\lt a\lt2n$ is same as k+1 is not perfect square and I have mentioned that numerator is increasing and denominator is constant and I should have said $f(x) \le f(x+1)$ instead of $f(x) \lt f(x+1)$

 

[CellCoree]

To determine all positive integers $k$ for which $f(k)>f(k+1)$, we first need to understand what the function $f(k)$ represents. In this case, $f(k)$ is the floor function of the quotient of $k$ divided by the floor function of the square root of $k$. This means that $f(k)$ is equal to the largest integer that is less than or equal to the quotient of $k$ divided by the largest integer that is less than or equal to the square root of $k$.

Now, to determine the values of $k$ for which $f(k)>f(k+1)$, we can start by looking at the behavior of the function $f(k)$ for different values of $k$. For example, when $k=1$, we have $f(1)=1$ since the largest integer less than or equal to the square root of $1$ is $1$. When $k=2$, we have $f(2)=1$ again since the largest integer less than or equal to the square root of $2$ is also $1$. However, when $k=3$, we have $f(3)=2$ since the largest integer less than or equal to the square root of $3$ is $1$, and $3$ divided by $1$ is $3$, which is greater than $2$.

This pattern continues as we increase the value of $k$. For instance, when $k=4$, we have $f(4)=2$, and when $k=5$, we have $f(5)=2$ as well. However, when $k=6$, we have $f(6)=3$ since the largest integer less than or equal to the square root of $6$ is $2$, and $6$ divided by $2$ is $3$, which is greater than $2$.

From this pattern, we can see that the values of $k$ for which $f(k)>f(k+1)$ are those that are perfect squares or one less than a perfect square. This is because the floor function of a square root will always return the value of the square root itself, while the floor function of a non-perfect square will return a value that is one less than the square root.

Therefore, all positive integers $k$ for which $f(k)>f(k+1)$