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.
be a real numbers such that 
Prove that
Where 
be a cyclic quadrilateral with 
and let 
and 
be the feet of altitudes from 
to the lines 
and 
,.Prove 
bisects 
.
is strictly increasing : 
such that:
, let 
be the set of positive integers that do not exceed and are coprime to . Define 
as the smallest positive integer that allows to be partitioned into disjoint subsets, each forming an arithmetic progression.
satisfying 
, 
, and 
.
be a real numbers such that 
Prove that
Where
be natural number and 
be the set of distinct natural numbers. Define function 
with 
. Prove that have at least distinct values.
and 
such that
is a prime number.
be the midpoint of the side 
of triangle 
. The bisector of the exterior angle of point 
intersects the side in 
. Let the circumcircle of triangle 
intersect the lines 
and 
in 
and 
respectively. If the midpoint of 
is 
, prove that 
.
and satisfying the equation
?
is not a perfect square, then there exists an odd prime s.t. is a quadratic non-residue mod .
be the orthocenter of the triangle . Let and be the midpoints of the sides and , respectively. Assume that lies inside the quadrilateral 
and that the circumcircles of triangles 
and 
are tangent to each other. The line through parallel to intersects the circumcircles of the triangles and in the points and 
, respectively. Let be the intersection point of 
and 
and let 
be the incenter of triangle 
. Prove that 
.
that satisfies 
for all real 
triangle with 
. We take , point on AC, AB respectively such that 
, 
. 
, 
lines intersect at point . If 
, find 
.
upon division by 
be 
. Then consider the numbers 
for 
. Suppose that not all pairwise sums are different then![\[202(a+b)+r_a+r_b=202(c+d)+r_c+r+d\]](http://latex.artofproblemsolving.com/b/b/8/bb883b0f628b01552bf9324c56b682fdd3b9dcf9.png)
for some 
, all distinct. Note that 
. Since 
for all 
, 
and 
is either 
or has absolute value at least 
. The latter is impossible so it is zero, meaning 
. Now, 
and 
implies 
and as a result, 
.
then adding 
gives 
, contradiction. If 
then adding gives 
, also contradiction. Therefpre, all pairwise sums are different, and the largest number is between 
and 
.
be a prime. Let be the set of vectors 
for 
. Then the pairwise sums of elements of are distinct.
and 
then 
, and by unique factorization of 
in 
we have 
. 
. All vectors lie in 
, so we can apply the mapping 
to , which will give integers in ![$[1, 2p^2+p+1]$](http://latex.artofproblemsolving.com/a/4/d/a4dbcf6cc9cc909010afeefc0f0ee4c94891a487.png)
, and preserve the "pairwise sums are distinct" property. 
positive integers not exceeding 
,
![$[n]_p$](http://latex.artofproblemsolving.com/4/c/e/4ce12a27e3995de190e85f6e406874b072388b4c.png)
denote the remainder when ![$f(n) = 2pn + [n^2]_p$](http://latex.artofproblemsolving.com/0/4/b/04b846ff6a7c3b904bf843946cf6d6327d2ec997.png)
.
satisfies the conditions of the problem.
,![$f(i) = 2pi + [i^2]_p < 20200 + 101 < 25000$](http://latex.artofproblemsolving.com/3/b/9/3b9328d0da5f115abc1c08972add54a63e21d78d.png)
.
,
implies ![$|2p(i-j)| = |[i^2]_p - [j^2]_p| < p$](http://latex.artofproblemsolving.com/e/7/9/e797e83b2284885db1ebc3643de8e330d8a3d4f9.png)
,
.
are distinct and less than 100.
for 
,
.![$2p((a+b)-(c+d)) = [c^2]_p + [d^2]_p - ([a^2]_p + [b^2]_p)$](http://latex.artofproblemsolving.com/6/d/3/6d39b1023d75a2febae71cebc184c523772f0f51.png)
.
and has absolute value less than 
.
,
.
as well, 
and 
must be the two roots of the quadratic 
mod 
.
.
,
,
.
numbers 
less than 
with their pairwise sums different (then we just plug in 
to get the result)
is an integer then define 
to be the remainder obtained when 
for some 
and 
between 
and 
and 
,
(
)
are multiples of 
and 
or 
and 
for stewing up a contradiction,hence we have exhausted all possibilities and the result is clear by substituting ![$\sqrt{n}+0.998\sqrt[4]{n}$](http://latex.artofproblemsolving.com/7/6/d/76d052c913d23a956c47e00225e4f9cd945bd4d3.png)