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.
points in the plane, no three of them belonging to a same line. Each of these points is colored using one of four given colors. Prove that there is a color and at least 
scalene triangles with vertices of that color.
and 
.
?
be positive integers satisfying ![\[\frac{ab}{a+b}+\frac{cd}{c+d}=\frac{(a+b)(c+d)}{a+b+c+d}.\]](http://latex.artofproblemsolving.com/b/8/c/b8c3fe14640a2dcb30b755b2347721da5d080a78.png)
Determine all possible values of 
.
+1 w
users, some pairs of whom are friends. Whenever user 
is friends with user 
, user is also friends with user . Events of the following kind may happen repeatedly, one at a time:, , and 
such that is friends with both and , but and are not friends, change their friendship statuses such that and are now friends, but is no longer friends with , and no longer friends with . All other friendship statuses are unchanged.
users have 
friends each, and users have friends each. Prove that there exists a sequence of such events after which each user is friends with at most one other user.
and natural 
such that 
and 
, then find the minimum value of ![$\sqrt{ad^3 g^4} +\sqrt[3]{be^3 h^4} + \sqrt[8]{cf^3 k^4}$](http://latex.artofproblemsolving.com/b/f/a/bfa17e8bb4cc05a84ed6a757f506d6d0ed2b0081.png)
, satisfying the condition
and 
.
be an acute triangle with incentre 
and 
. Let lines 
and 
intersect the circumcircle of at 
and 
, respectively. Consider points 
and 
such that 
and 
are parallelograms (with 
, and 
). Let 
be the point of intersection of lines 
and 
. Prove that points 
, and are concyclic.
such that for all any 
be non-negative real numbers such that 
Prove that
satisfying![\[1=\frac{b_1}{1^2} > \frac{b_2}{2^2} > \frac{b_3}{3^2} > \frac{b_4}{4^2} > \dotsb\]](http://latex.artofproblemsolving.com/a/e/6/ae65e3d53931fe9d60d0de76dc79540ccf44c26e.png)
and let 
denote the largest real number satisfying 
for all positive integers 
. What are the possible values of across all possible choices of the sequence 
?
, 
and 
for any positive integer .
.
we clearly have 
so we must have 
or 
. If we let 
where is any positive integer then 
and we have our first pair .
.
Thus we must have 
which rearranges to 
. Note that in order for this to be true, we must have 
or else 
. Now we simply just check the cases where 
and 
.
, then 
which is equivalent to 
from which we extract 
or 
. This gives us the pairs and .
, then 
which is equivalent to 
from which we see there are no positive integer solutions for .
These clearly all work. Note that 
(at the second step we subtracted times the modulus.) The point of this is that now everything is linear in .
. Then, we have 
hence there are no solutions as the expression is positive and smaller than the modulus.
. Then, clearly 
, which clearly works.
. Then, we have 
We can rearrange this to 
, there are not that many cases. If , we have 
For 
and 
this gives 
and 
, and other values of 
up to 6 do not give integer . If , we have 
here only 
is allowed which does not give an integer anyway. Finally, 
is not possible, hence done.
Due to size reasons 
, if it equals 0 then the solution set is . If it's <0 we must have 
; easy calculation reduces it into 
divides 
., or that 
can not hold.
which can be seen to hold.
, then there are no solutions., it follows that 
so 
. We can check that 
and 
works., then 
which implies 
which has no solutions.
, then 
which has no solutions.
Now we have a few cases:
, then 
, which is obviously impossible. If 
, then 
which implies 
and 
. This yields the solutions 
and 
., then the entire curve of solutions 
can be checked to work.
;-; 
. Also, these edge cases are so hard for me to find without making sillies.
, and for any 
.
., then we get that 
which gives us 
. But then the discriminant must be 
, that is 
which clearly has no solution., then we get that 
. Then we get that 
. This means that the discriminant must be 
, that is 
. This gives only solution since the left side is strictly increasing and it exceeds 
for . For , from the problem statement we get that 
. This gives us that 
each of which are solutions., then clearly 
and so 
which is another solution.
, for 
.![\[ab^2 + b + 7 \mid a^2b + a + b \implies ab^2 + b + 7 \mid b(a^2b + a + b) - a(ab^2 + b + 7) = b^2 - 7a.\]](http://latex.artofproblemsolving.com/b/e/d/bedf88f0178ab546f0d6feb72de1eeecbc5e30f0.png)
We now split into two separate cases:
Then 
, so that 
, which in turn gives us 
, for some . Then we find the corresponding 
. 
First observe that 
, else 
, which is a clear contradiction for 
. Hence 
, so that we obtain ![\[ab^2 + b + 7 \leq 7a - b^2 \implies b^2(a + 1) + b + 7 \leq 7a \implies b^2 < 7.\]](http://latex.artofproblemsolving.com/4/f/f/4ff8f4a9b8de7c7005420dd68dc00d8779690198.png)
Hence we split into cases for 
:
Then ![\[\frac{7a - 1}{a + 8} \in \mathbb{Z} \implies 7 - \frac{57}{a + 8} \implies a + 8 = 1, 3, 19, 57 \implies \boxed{a = 11, 49}. \bigstar\]](http://latex.artofproblemsolving.com/2/7/0/2704b8d70627d8809f46b8ac143c2e78ec2536f9.png)
Then ![\[\frac{7a - 4}{4a + 9} \in \mathbb{Z} \implies 1 + \frac{3a - 13}{4a + 9} \implies a \leq -22, \]](http://latex.artofproblemsolving.com/e/a/8/ea834238044622767cc1849a72fa675dcf2b9e40.png)
which is impossible. Therefore no solution exists in this subcase. 
, for arbitrary , as claimed. 
, , 
for all 
.
.
, contradicting the divisibilty condition..
, let 
, so
.
which is always valid, hence these are always solutions..
we get:
This gives and .
which gives no solutions.
. Then, 
, so 
so 
.
or 
. This means that or . When , we get 
so 
so or ., we get 
so 
so 
so there are no solutions to .
. Then, 
so 
. Plugging this in means 
. Note that 
so 
. Plugging back in, we get 
, which is always true., , and for all positive integer .
, so 
. We can then divide into cases, if , then clearly we have no solutions by size. If , then write 
, we write 
, we can see all solutions of this form work. If , by size we still require 
, or equivalently 
. We then check , we require 
, equivalently 
, equivalently 
, so there are no solutions for by divisor analysis. We now check , we require 
, so we have 
, so we have 
. We check gives 
, 
, so this pair works. We then check , we get 
, so this pair works as well. The answers are then 
.
. Since both sides are positive: 
.
: if we set 
, then 
, We can easily check that it is a solution with being a positive integer.
: Then from earlier, 
LHS is positive so RHS should follow. Meaning
. Then we can manually check and see that 
is a solution.
Multiply the RHS by to get 
and then subtracting times the LHS, we get 
Thus, we have that 
we have that 
but this is absurd.
we have that 
Thus, 
If 
we have 
Clearly, the solutions are 
If 
we have that 
Multiplying by 
we have that 
which has no solutions.
Thus, we have that 
We have 
We clearly have that 
for some 
Thus, 
The LHS is very clearly relatively prime to 
so we multiply the RHS by and then apply Euclids, we have that 
If it is non-zero, this is clearly absurd. Thus, 
We get the curve 
and 
.
and 
.
Is there any generalizations?
of positive integers such that 
is divisible by 
in positive integer
<0 (contradition)
(m i positive integer)
(k in positive integer)
=
then 
in positive integer![\[k\equiv 0 (mod b)\]](http://latex.artofproblemsolving.com/f/9/2/f92d5e37c71ebbca9b1cffc5dc9c2e83d2c5c9c1.png)
so 
(b >= 3and q i positive integer)
=
+
(u in positive integer)
,
,
.
,
.
,
, or
.
. But then we have
,
. And also 
,
we know 
. So we have all solutions of the form 
.
so the RHS is positive. So we have 
.
. So we check 
.
,
, giving the solutions 
. Since
,
. But since 
, we must have 
, which does not have an integer solution.
for positive integers 
then 
from here
so 
and 
so
then n divides 
, too.
.
then 
from here we will have
.
then 
also 
then we have to have 
here 
is positive integer then 
or 
otherwise inequality will lead to contradiction
then the problem will become as 
then 
and 
we will have 
and 
is divisible by 
so 
is an integer, too.
those are 
or 
)
by simliar of previous we will show that there is no solution for 
then it is easy to see there are no solutions.
,
.
is divisible by 
,
,
and we have the solution 
or 
.
can't be more than 
,
to 
and 
works, this gives the solutions 
and 
.
then 
and we have no solution.
,
,
since if 
then the left side is less than the right side (we would have 
,
,
)
.
since 
.
,
.
and 
work, giving the solutions 
.
are positive, then 
.
or 
.
so we must have 
and we must have 
since otherwise we have:
.
or 
.
and so we write:
.
must be positive. We rewrite this as:
and since 
is positive, the denominator must be positive and so 
.
into this expression in turn shows that only 
and 
give integer values for 
and 
.
.
.
,
does not give integer 
.
.
which satisfies the problem must also satisfy:
.
.
which is an integer.
?
?
,
?
.
?
is not actually used for the algebraic manipulations, but to set aside a case that forces 
,
,
.
.
(division by 
on the RHS of 
to be an integer?
must divide 
.
and 
and 
for all 
because 
that is impossible for 
therefore 
if we prove all the divisors we have that 
no solutions here
![\[y(x^2y+x+y)-x(xy^2+y+7) = y^2-7x.\]](http://latex.artofproblemsolving.com/8/2/c/82c9525fecfa2fa1730f2246886d7780f282fa74.png)
.![\[|xy^2+y+7| > |y^2-7x|,\]](http://latex.artofproblemsolving.com/c/3/9/c39303ba228bd6baedb223bc571b80db96f57d24.png)
implying there are no solutions, unless 
where we get the pairs 
.
![\[xy^2+y+7 \mid y(x^2y+x+y)-x(xy^2+y+7)=y^2-7x \]](http://latex.artofproblemsolving.com/3/e/7/3e7266ae7abb0ab975b5796a73c8509c5929457c.png)
which implies that 
which gives 
or 
or 
which Convsersely always works. The edge cases give 
and 
as solutions.
.
After checking we find the solution 
.
for some