MHB Determine all postive integer k

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The discussion centers on determining positive integers \( k \) for which the function \( f(k) = \left\lfloor{\dfrac{k}{\left\lfloor{\sqrt{k}}\right\rfloor}}\right\rfloor \) satisfies \( f(k) > f(k+1) \). A key point made is that for \( k = n^2 + a \) where \( 0 < a < 2n \), the function can be expressed as \( f(k) = n + \left\lfloor{\dfrac{a}{n}}\right\rfloor \), indicating that \( f(k) \) is non-decreasing. This suggests that the conditions for \( f(k) > f(k+1) \) may be limited. The conversation emphasizes the simplicity of proving this property, indicating that the analysis is straightforward. Overall, the focus is on the behavior of the function and the specific conditions under which it decreases.
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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 $k\in \Bbb{Z}^*$.
 
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anemone said:
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 $k\in \Bbb{Z}^*$.

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 $n\gt 1$
 
kaliprasad said:
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.:)
 
anemone said:
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,
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 $f(x) \lt f(x+1)$
 
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