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.
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and 
be positive integers with 
. Let 
be a polynomial of degree with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers 
such that the polynomial 
divides , the product 
is zero. Prove that has a nonreal root.
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positive integers 
such that for every index 
, with 
, 
is a multiple of 
. Determine the greatest possible value of the largest of the numbers.
and 
be positive integers satisfying the equation ![$(a, b) + [a, b]=2021^c$](http://latex.artofproblemsolving.com/2/4/6/246ccbb22acecb7adc72219afb8afb34c9b05101.png)
. If 
is a prime number, prove that the number 
is composite.
be the set of real numbers. Determine all functions 
such that ![\[ f(x^2 - y^2) = x f(x) - y f(y) \]](http://latex.artofproblemsolving.com/6/8/f/68f78e0df627f6ac04052433c01627bf16f0af8e.png)
for all pairs of real numbers 
and 
.
into 
or 
can sometimes be quite complex. That's why I’ve written a program to assist with this process.
be real numbers such that 
Prove that 
Where 
with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of are either all odd or all even.
, 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 
, 
, 
, 
and 
.
equally spaced points on a circular track 
of circumference . The points are labeled 
in some order, each label used once. Initially, Bunbun the Bunny begins at 
. She hops along from to 
, then from to 
, until she reaches 
, after which she hops back to . When hopping from 
to 
, she always hops along the shorter of the two arcs 
of ; if 
is a diameter of , she moves along either semicircle. arcs which Bunbun traveled, over all possible labellings of the points.
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.
.
Prove that 
