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with the property that ![\[f(x-f(y))=f(f(x))-f(y)-1\]](http://latex.artofproblemsolving.com/f/2/5/f25cc1e8ae0be1fd02b347fd94be4fab88af1d46.png)
holds for all 
.
be the set of positive integers. Find all functions 
such that![\[ m^2 + f(n) \mid mf(m) +n \]](http://latex.artofproblemsolving.com/2/f/4/2f409d1de993f1af8fd839bb8e9f87a57e1b8608.png)
and 
.
board with a number written in each cell. Every two neighbour rows sum up to at least 
, and every two neighbour columns sum up to at most 
. Find the sum of all numbers on the board.
such that the equation in the unknown 
:
admits 
roots: 
s.t. 
.
be a function, two times derivable on 
for which there exist 
such that![\[ \frac { f(b)-f(a) }{b-a} \neq f'(c) ,\]](http://latex.artofproblemsolving.com/a/3/1/a31861ef909a6e17ed8b495b9b9b16d887ba573f.png)
for all 
.
.
, 
and 
.
be complex numbers with 
. Find the greatest possible value of the expression 
be a positive root of the equation 
. Prove that the following sequence converges: 
(0<x<2π)
with domain ![$[0,2025]$](http://latex.artofproblemsolving.com/d/1/7/d17ac41d54c4b1888fb8bb43cf11b372691694a0.png)
satisfying 
and 
is strictly increasing on Compute smallest real M such that all functions in this set ,
.
have local extrema at 
respectively. find 
. Here 
are constants .
be the region in the complex plane enclosed by curve 
for 
. Compute perimeter of .
be defined by![\[
a_1 = 1, \quad a_{n+1} = a_n + \frac{1}{\sqrt[2024]{a_n}} \quad \text{for } n \geq 1, \, n \in \mathbb{N}
\]](http://latex.artofproblemsolving.com/8/8/f/88f423d87c08b20dc552703fcf30f2a8e9585902.png)
Prove that![\[
a_n^{2025} >n^{2024}
\]](http://latex.artofproblemsolving.com/a/6/8/a6806c817edb7ce29a399981137c1a3e93f63c63.png)
for all positive integers 
.
![$x + \frac{1}{\sqrt[2024]{x}}$](http://latex.artofproblemsolving.com/f/6/1/f61a70fe2258941f9d52de50dc7b3faffa2e6381.png)
is an increasing function for 
by differentiating. Also note that by induction 
for all .
is true, assume true for 
note that ![$$a_{k+1} = a_k + \frac{1}{\sqrt[2024]{a_k}} > k^{2024/2025} + \frac{1}{\sqrt[2025]{k}} = \frac{k+1}{\sqrt[2025]{k}} > (k+1)^{2024/2025}$$](http://latex.artofproblemsolving.com/1/d/3/1d3da88ded0f98b07414bf5a63035d9cf6d59835.png)
is an increasing function for by differentiating. Also note that by induction for all . is true, assume true for note that
![$$a_k + \frac{1}{\sqrt[2024]{a_k}} > k^{2024/2025} + \frac{1}{\sqrt[2025]{k}} $$](http://latex.artofproblemsolving.com/1/f/e/1fe31ebd5dc9aaaf67435686a0b5ada57fe14d4f.png)
![$$ a_k+\frac{1}{\sqrt[2024]{a_k}}$$](http://latex.artofproblemsolving.com/8/d/a/8da8a08ee9c5d6ebb91bcee9ace966fe6e387067.png)
is a increasing function and 
![$\implies a_k+\frac{1}{\sqrt[2024]{a_k}} >k^{\frac{2024}{2025}}+\frac{1}{\sqrt[2025]{k}}=\frac{k+1}{\sqrt[2025]{k}}$](http://latex.artofproblemsolving.com/5/f/c/5fcb4f8ec708d782e354067758922b5c2e71023c.png)
![$\frac{1}{\sqrt[2025]k}>\frac{1}{\sqrt[2025]{k+1}} \implies \frac{k+1}{\sqrt[2025]k}>(k+1)^{\frac{2024}{2025}}$](http://latex.artofproblemsolving.com/d/8/b/d8b222a6ef426559502149f12bfd25514b6d1707.png)
, but I don't know why increasing helps here. Could you please elaborate more? Thanks! is increasing so by defination of increasing functions we can say that 
using we get
, but I don't know why increasing helps here. Could you please elaborate more? Thanks! is increasing so by defination of increasing functions we can say that using we get
is trivial as 
and 
. We assume the result holds for some 
and show it also does for 
. Note,![\begin{align*}
a_{n+1}^{2025} &= \left(a_n + a_n^{-\frac{1}{2024}}\right)^{2025}\\
&= a_n^{2025} + \binom{2025}{1}a_n^{2024}\cdot a_n^{-\frac{1}{2024}} + \dots + \binom{2025}{2024}a_n\cdot a_n^{-\frac{2024}{2024}}+ a_n^{-\frac{2025}{2024}}\\
&= a_n^{2025} + \left[\sum_{i=1}^{2024} \binom{2025}{i} a_n^{2025-i}\cdot a_n^{-\frac{i}{2024}} \right]+ a_n^{-\frac{2025}{2024}}
\end{align*}](http://latex.artofproblemsolving.com/3/3/c/33c0cb79ff4ee5dbc4b38792adf1cf6dd0d7ef13.png)
we have the inequality,![\[\binom{2025}{i} a_n^{2025-i}\cdot a_n^{-\frac{i}{2024}} \ge \binom{2024}{i}n^{2024-i}\]](http://latex.artofproblemsolving.com/8/8/9/889e53ac5d00ac1cdf5ab0e64e2ebe730047db56.png)
However, this is clear as for all ,![\[\frac{2025}{2025-i} a_n^{\frac{2025(2024-i)}{2024}} \ge a_n^{\frac{2025(2024-i)}{2024}} > n^{\frac{2024(2024-i)}{2024}} = n^{2024-i}\]](http://latex.artofproblemsolving.com/7/1/8/7188ca8804b3444632a63a2ee5976188f937b16d.png)
as desired.![\begin{align*}
a_{n+1}^{2025} & = a_n^{2025} + \left[\sum_{i=1}^{2024} \binom{2025}{i} a_n^{2025-i}\cdot a_n^{-\frac{i}{2024}} \right]+ a_n^{-\frac{2025}{2024}} \\
& > n^{2024} + \sum_{i=1}^{2024} \binom{2024}{i}n^{2024-i} + a_n^{-\frac{2025}{2024}} \\
& > (n+1)^{2024}
\end{align*}](http://latex.artofproblemsolving.com/7/8/8/78805143bc8623cb15238eee67b4d42c124cfbc3.png)
which completes the induction.
.
.
.
for all 
.![\[a_{k+1}^{{2025}/{2024}} = a_{k+1} a_{k+1}^{{1}/{2024}} > a_{k+1} a_k^{1/2024} = a_{k}^{{2025}/{2024}} + 1 > k + 1.\]](http://latex.artofproblemsolving.com/3/5/7/357e6a5f60c1d375fec7a2fa8f176a2e04bd2711.png)