AoPS Academy
such that![\[
x^2 + y^2 + z^2 = 2.
\]](http://latex.artofproblemsolving.com/a/8/d/a8db4ff2ec57b9e926f2fc4f4eede3d2d4b3c556.png)
Prove that![\[
x + y + z \leq xyz + 2.
\]](http://latex.artofproblemsolving.com/f/7/5/f75ae1a7eb4676271c869021a9457348f9893569.png)
, find the smallest real number 
such that no matter how each integer point in the coordinate plane 
is colored with one of different colors, there always exist three non-collinear points 
of the same color such that the area of 
is at most .
pebbles is placed in a vertical column. This configuration is modified according to the following rules. A pebble can be moved if it is at the top of a column which contains at least two more pebbles than the column immediately to its right. (If there are no pebbles to the right, think of this as a column with 0 pebbles.) At each stage, choose a pebble from among those that can be moved (if there are any) and place it at the top of the column to its right. If no pebbles can be moved, the configuration is called a final configuration. For each , show that, no matter what choices are made at each stage, the final configuration obtained is unique. Describe that configuration in terms of .
be pairwise distinct positive integers such that 
is an integer. Prove that 
is not a prime number.
be a convex hexagon such that 
and ![\[ \frac{AB}{BC} \cdot \frac{CD}{DE} \cdot \frac{EF}{FA} = 1. \]](http://latex.artofproblemsolving.com/9/9/b/99be4560967e7521ee7160d01ffa634d0922f37f.png)
Prove that ![\[ \frac{BC}{CA} \cdot \frac{AE}{EF} \cdot \frac{FD}{DB} = 1. \]](http://latex.artofproblemsolving.com/7/e/2/7e2a05831899978b85421d282ec38281cba280c9.png)
be an integer. There are cells on a circle, and each cell is assigned either 
or 
. There is a rooster on one of these cells, and it repeats the following operation:
If the rooster is on a cell assigned , it changes the assigned number to and moves to the next cell counterclockwise. If the rooster is on a cell assigned , it changes the assigned number to and moves to the cell after the next cell counterclockwise.
, then the rooster would go around the circle exactly three times, stopping again at . Moreover, every cell would be assigned the same number as it was assigned right before the rooster went around the circle three times.
such that 
.![\[
a^3 + b^3 + c^3 \geq a\sqrt{b+c} + b\sqrt{c+a} + c\sqrt{a+b}
\]](http://latex.artofproblemsolving.com/7/9/e/79e23cd5fdf9630a62070feb0f2c38ae1d11163d.png)
be 
positive integers such that the 
products![\[a_1 a_2 a_3 \cdots a_n, b_1 a_2 a_3 \cdots a_n, b_1 b_2 a_3 \cdots a_n, \dots, b_1 b_2 b_3 \cdots b_n\]](http://latex.artofproblemsolving.com/0/2/c/02cbff15c4d0118156a9d24be09360facb264d5a.png)
form a strictly increasing arithmetic progression in that order. Determine the smallest possible integer that could be the common difference of such an arithmetic progression.
and 
denote three distinct integers, and let 
denote a polynomial having integer coefficients. Show that it is impossible that 
and 
.
, and 
are distinct.![$ P[\mathbb{Z}]$](http://latex.artofproblemsolving.com/b/2/6/b266cafa5e8b2475c08709e1e8ac1d631ccbe914.png)
, suppose we have by the given conditions to be exisitng such that
.
, and are distinct., suppose we have by the given conditions to be exisitng such that
.
and 
, then 
.
But they're much harder nowadays. Also, by = I meant congruence, sorry, and I obviously used the parentheses for order. I think, however, that it's a good solution. Edit: If there is something flawed with my solution, or if it is correct, tell me. If it is correct, I'd love to add this brief solution to the AoPS wiki, but I don't know latex. 
. Then 
, a contradiction.
. Then 
, also a contradiction.
divides 
and 
divides 
.
we get
are integral.![\[\left(\frac{P(a)-P(b)}{a-b}\right)\left(\frac{P(c)-P(a)}{c-a}\right)=\frac{b-c}{c-a}=k_1k_3.\]](http://latex.artofproblemsolving.com/0/c/6/0c661d4ab20f210ad7148524ee84223faf03eade.png)
But 
, so the only way for 
to be an integer is if 
. Similarly we get that 
. If one of 
is -1, WLOG say 
, then 
, contradiction. Thus, are all equal to 1, and plugging these in it follows that 
, 
and 
. Solving this system yields 
, but 
are distinct integers by definition, a contradiction.
, and are distinct., suppose we have by the given conditions to be exisitng such that.
for integers 
and 
and a polynomial 
with integral coefficients. To see why, write out the expression 
and then refactor it so you get a lot of terms of the form 
for some 
. Then 
divides all of these terms, and is a linear combination of terms of this form, so .
are all integers and their product is equal to , all three of them are each either or 
.
. THen either both of the remaining fractions equal 1 or both equal -1, so we examine the two cases. Both cases give us that 
, contradiction.
and 
. 
is a polynomial with integer coefficients then where 
are integers.
. We have 
and 
. So 
. Notice that 
so it is easy to see ., 
, and 
. We know what 
and 
are as they were given in the problem so we substitute them in. Let 
, 
, and 
where 
are integers. Note that 
and because are integers it follows either all of them are or two of them are and the other is . Let's split this into two cases:
.
, 
, and 
. This gives us that but the problem said are distinct so this case is ruled out.
and 
. again, meaning that it is impossible for , , and 
.
. Let 
Note that 
Subtracting the second from the first, third from second, and first from third gives
By the RHS, note that 
so 
Similarly, 
and 
Hence, 
so 
Assume WLOG that 
so 
and 
From the first equation, we get 
and substituting this in the second gives 
Hence, , contradicting the uniqueness of 
. So, by this fact we get 
. Now we will consider this into two cases when 
, 
and and the other case will be when 
, 
and 
. So, by adding all the inequalities we get 
which is impossible. Now we will consider when , and 
. From this we get , which is clearly not possible, as the question says that are distinct integers. Hence, not possible.
. We have that 
, which is clearly divisible by 
. Hence, 
Therefore, 
. Letting 
, 
, and 
, we have that 
. 
, so we know that 
. In other words, 
. If 
, then , which is impossible. 
, and are distinct., suppose we have by the given conditions to be exisitng such that.
![\[P(a)-P(b)=P(b+(a-b))-P(b)\]](http://latex.artofproblemsolving.com/1/9/b/19b02dda5fc7f5f35382ab52af2991a809eb13b8.png)
is always a multiple of 
. From this we obtain 
, 
, and 
. Thus, ![\[|a-b|=|b-c|=|c-a|.\]](http://latex.artofproblemsolving.com/d/1/1/d1173fd71dcf4780b3c701c5c52e98b12e360bb3.png)
Clearly 
cannot hold, as that would imply 
. Therefore, 
. Similarly, we get 
. Subtracting this gives 
, a contradiction.
such that: 
Prove that :
and 
?
,
,
,
,
.
.
(the division is valid as none of 
,
,
are zero.) Since 
,
,
are integers. But since they are integers that multiply to one, they must all have absolute value one.
)
.
and 
,
,
,
,
,
.
,
,
.
,
,
.
,
.
.
,
and 
are integers and 
;
;
,
,
,
.
.
(which is valid since the condition is symmetric), this means that 
must equal 
,
![$P(x)\in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/3/f/e/3fee4d24dc1e172d6472b4122f8e17e9084571a4.png)
,
for 
.![$f(x)\in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/0/a/0/0a04571f5290305c8b91f732a34451f2457ed370.png)
.
, for some 
.
and 
It follows that 
By difference of 
for positive integral 
out of every term, to get 
for integral 
Note that if 
,
,
,
.
where 
and because 
are distinct, we can divide both sides by 
without any hassle to get 
Since 
,
We can rearrange these equations to get 
Dividing both sides by two on all these equations yields 
Subtracting the first two equations from each other yields 
Let 
.
or 
,
.
,
and 
so we have 
Thus by definition, 
Similarly, with the argument we used previously, we can say 
But, by 
which then defines 
We also know 
and thus this means 
This implies that 
hence a contradiction. 
so at least one of them is equal to 
Thus, 
and 
so 
or 
Similarly, 
Solving for 
Hence, 
which simplifies to 
a contradiction.
simultaneously.
and 
implying 
and 
and 
simultaneously as ![$\mathbb{Z[X]}$](http://latex.artofproblemsolving.com/f/1/5/f15695032af130ac55b3f0af6bcbdd6093efc880.png)
which gives 
which can be figured out by substracting any two subsequent equations
(contradiction)