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 sequence of real numbers with 
. Define the score 
of a permutation 
of 
to be the minima of the sum 
over all real numbers 
. attains the maxima over all permutations 
, if and only if for all 
, 
be a nonempty set of positive integers. We say that a positive integer 
is clean if it has a unique representation as a sum of an odd number of distinct elements from . Prove that there exist infinitely many positive integers that are not clean.
such that for any positive integer 
, satisfies 
.
and 
lie respectively on sides 
and 
of a triangle 
and 
. Segments 
and 
cross at 
. Circumscribed circles of triangles 
and 
cross again at point different from . Prove that point lies on the bisector of angle 
.
satisfying 
and their sum is an even number, consider the quadratic polynomial:![\[
P(x) = x^2 - (m^2 - m + 1)x + (m^2 - n^2 - m)(n^2 + 1).
\]](http://latex.artofproblemsolving.com/d/5/1/d51aaddfddfe06cc91abfdd3476a155770a8a5fd.png)
are positive integers but are not perfect squares.
such that![\[
ad = bc, \quad a + d = 2^k, \quad b + c = 2^m
\]](http://latex.artofproblemsolving.com/1/1/d/11d16732e3a77af46ad17adecf104a442a5d7f67.png)
for some positive integers 
and 
.
be a triangle with incenter 
and intouch triangle 
. Let be the foot of the perpendicular from 
onto 
. Assume that 
, intersect the sides , in 
respectively. Finally, let the rays 
, 
meet the circumcircle of 
in , 
respectively. Prove that the tangent from to the incircle and the line 
meet on the circumcircle of .
be a triangle with circum-circle 
. Let 
be a point in the interior of triangle which is also on the bisector of 
. Let 
meet in 
respectively. Suppose 
is the point of intersection of 
with 
; and 
is the point of intersection of 
with 
. Prove that 
is parallel to 
.
such that 
is a perfect square but none of the integers 
are perfect squares.
be an acute-angled triangle with 
. The altitudes 
and 
intersect at 
. Let 
be the midpoint of 
. Point 
is chosen on the extension of 
beyond and point 
is chosen on the segment 
such that 
. Prove that points 
and are collinear. be an acute-angled triangle with 
. Let be the incentre of triangle , and let 
be the points where the incircle touches the sides 
respectively. Let 
meet the line at 
respectively. Further assume that both and 
are outside the triangle . Prove that
are concyclic.
is also the incentre of triangle 
.
and 
, prove
and 
with conditions,
is a prime number.
is a perfect square.
is also perfect square.
and 
leads to and are nonnegative integers.
so 
and 
is prime so 
or 
then 
and 
for some integer thus 
is a perfect square so 
. This gives us 
which it is a contradiction., since is a perfect square then both and are perfect squares. Let 
for non negative integers 
. Hence 
is a prime thus 
so 
and 
for some integer 
and in this case 
cannot be a perfect square.
then 
and in this case 
cannot be a perfect square.
For this problem, a useful property is a perfect square is congruent to 0 or 1 mod 4.
and 
we have that and are non-negative integers. We proceed by contradiction.
must be an integer and similarly 
must also be an integer. Thus for 2 non-negative integers and 
we must have
and 
. Thus 
and 
.
and 
.
. This implies that 
. But this doesn't satisfy the third equation. Thus, there are no solutions.
and we have that and are non-negative integers. We proceed by contradiction. must be an integer and similarly must also be an integer. Thus for 2 non-negative integers and we must have and . Thus and . and .. This implies that . But this doesn't satisfy the third equation. Thus, there are no solutions.
.
so we have 
and 
are coprime. is a perfect square. in the first condition and factorize one sentence has to be 
use quadric formula to get a contradiction here too.
such that![\[
f(xy) = f(x)f(y) \;-\; f(x + y) \;+\; 1,
\quad \forall x, y \in \mathbb{R}.
\]](http://latex.artofproblemsolving.com/7/a/5/7a51a7be2306d18f4e80db1f5aac6e9bbf25c25c.png)
.
bisects 
.
and 
.
and 
.
is a perfect square. This means 
or both 
and 
are perfect squares. (Note: 
denotes the greatest common divisor of 
.
.
,
are both perfect squares
,
and 
.
.
by difference of squares. However, we also have 
by difference of squares.
and 
are prime numbers. From 
,
,
.
.
- a contradiction.
and 
both don't work, so 
,
.
Thus, either 
and 
,
Since 
,
(how do u type the congruent symbol?)
,
must be a perfect square.
for all positive integers k (this part is left as an exercise to the reader 
), a contradiction!
,
,
must be a perfect square, so 
is a square
for all positive integers k (another exercise 
if we suppose otherwise we get that 
and 
(because 
)
which is false.
and 
.
we obtain 
so 
and 
.
is a prime 
and 
.
.
becomes 
,
.
.
is a prime which is false. Hence there are no such integers.