AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
+3 w
is lovely if there exist a positive integer 
and (not necessarily distinct) positive integers 
, 
, 
, 
such that 
and 
for 
.
, which is a perfect square of an integer?
, prove that there exists a positive integer 
, such that for any integer 
, can be expressed by a sum of positive integers 
's as![\[n=a_1+a_2+\dots+a_m,\]](http://latex.artofproblemsolving.com/6/c/2/6c2a979a5ff6c40bf2d7152c8c32088fc36f848b.png)
where 
, 
, 
, 
.
be a function such that
for all 
. Determine all such functions 
.
such that the equality 
holds for all pairs of real numbers 
.
with integer coefficients such that 
and 
is a square of an integer for all nonnegative integers 
. and an integer , 
is defined as follows: 
if 
and 
if 
.
, let 
be all positive integers smaller than that are coprime to . Find all 
such that 
for all 
is the largest positive integer that divides both 
and 
. Integers and are coprime if 
. be a positive integer. A positive integer is called a benefactor of if the positive divisors of can be partitioned into two sets 
and 
such that is equal to the sum of elements in minus the sum of the elements in . Note that or could be empty, and that the sum of the elements of the empty set is 
.
is a benefactor of 
because 
. has at least 
benefactors.
denotes the least natural such that
Find all naturals 
such that 
.
in the process of solving it
and making more systems of equations
![$x, y \in \left[0,1 \right]$](http://latex.artofproblemsolving.com/6/3/9/639a18e2742de67742e6a7e61b4ab591918a632d.png)
may satisfy both functions. We, the cameramen, call function 
as function 1 and function 
as function 2.![$\left[ \frac{i}{4} , \frac{i+1}{4} \right]$](http://latex.artofproblemsolving.com/5/7/e/57ef57084281b904335bc42fe520eb1fe03f7bc7.png)
with 
, function 1's value oscillates between 0 and 1. It attains 1 at 
, 
and another point between these two. Between two consecutive points whose functional values arr 1, the skibidi function first decreases from 1 to (in a skibidifying way) 0 and then increases from 0 to (in a skibidifying way) 1. So the skibidi graph of this function in this interval consists of 4 monotonic pieces.![$\left[ \frac{i}{6} , \frac{i+1}{6} \right]$](http://latex.artofproblemsolving.com/4/0/6/4067fa7c170a5a967c36a504b11ddf620c08b819.png)
with 
, function 2's value oscillates between 0 and 1. It attains 1 at 
, 
and another point between these two. Between two consecutive points whose functional values arr 1, the skibidi function first decreases from 1 to (in a skibidifying way) 0 and then increases from 0 to (in a skibidifying way) 1. So the skibidi graph of this function in this interval consists of 4 monotonic curves.![$\left[ \frac{i}{4} , \frac{i+1}{4} \right] \times \left[ \frac{j}{6} , \frac{j+1}{6} \right]$](http://latex.artofproblemsolving.com/3/4/4/3440263b5764320080c28ed340363c20d6cbbf4f.png)
with and 
but 
. Both functions have four monotonic pieces. Because function 1's each monotonic piece can take any value in ![$\left[ \frac{j}{6} , \frac{j+1}{6} \right]$](http://latex.artofproblemsolving.com/0/5/1/051b8cfcfc76d5d2d37bb535a1e44930b234aec5.png)
and function 2' each monotonic piece can take any value in , function 1's each monotonic piece intersects with function 2's each monotonic piece. Therefore, in the skibidi interval , the skibidi number of intersecting points is, as a matter of blasphemy against the unspoken rizz 
. for 
, then this point must be 
., function 1 attains value 1. For function 2, if 
, then 
. Therefore, the skibidi intersecting point is, as a matter of blasphemy against the unspoken rizz . for 
, then this point must be . with and but , all 16 intersecting points arr interior. That is, as a matter of blasphemy against the unspoken rizz, no two regions share any common intersecting point.![$\left[ \frac{3}{4} , 1 \right] \times \left[ \frac{5}{6} , 1 \right]$](http://latex.artofproblemsolving.com/8/7/6/8765c7a4e4f7a65fb7911d7731a83e76711a4fd0.png)
. Consider any pair of monotonic pieces, where one is, as a matter of blasphemy against the unspoken rizz from function 1 and one is, as a matter of blasphemy against the unspoken rizz from function 2, except the skibidi pair of two monotonic pieces from two functions that attain 
. Two pieces in each pair intersects at an interior point on the skibidi region. So the skibidi number of intersecting points is, as a matter of blasphemy against the unspoken rizz 
... Now, we, the cameramen, study whether they intersect at another point.
and 
. Thus, for positive and sufficiently small 
and 
, function 1 is, as a matter of blasphemy against the unspoken rizz reduced to![\[ y' = 4 \sin 2 \pi x' \hspace{1cm} (1) \]](http://latex.artofproblemsolving.com/4/0/f/40febc5aa05eaa7d60eeea4a45ca67373eecf730.png)
and function 2 is, as a matter of blasphemy against the unspoken rizz reduced to![\[ x' = 4 \left( 1 - \cos 3 \pi y' \right). \hspace{1cm} (2) \]](http://latex.artofproblemsolving.com/f/5/a/f5ab7a9618b343fc5fff4a204fefbf4bcec24a95.png)
and , to (in a skibidifying way) get an intuition and quick estimate, we skibidies goon approximations of the skibidi above equations.![\[ y' = 4 \cdot 2 \pi x' \]](http://latex.artofproblemsolving.com/2/d/6/2d66e8833d3e5ab836627870cb5d2d908c44f2fd.png)
and equation (2) is, as a matter of blasphemy against the unspoken rizz approximated as![\[ x' = 2 \left( 3 \pi y' \right)^2 \]](http://latex.artofproblemsolving.com/8/5/7/857453ec8763f39582dfb402207703e6345d24d9.png)
and 
. Therefore, two functions' two monotonic pieces that attain have two intersecting points.
.
be an odd prime. Prove that:![\[\displaystyle\sum_{k=1}^{p-1}k^{2p-1} \equiv \frac{p(p+1)}{2} \pmod{p^2}\]](http://latex.artofproblemsolving.com/a/5/5/a55e75d4d9769ceec3a3ce359aac676c7c7875fd.png)
![\[
\begin{cases}
x - y z = 1,\\[2pt]
y - z x = 2,\\[2pt]
z - x y = 4.
\end{cases}
\]](http://latex.artofproblemsolving.com/b/2/d/b2d94e49491a71f86109103f985fe3f238cf4d95.png)
is an obvious solution. Now, let 
.
or 
.
.
