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.
by 
and the number of distinct prime divisors of by 
. Let 
be a positive integer. Prove that there exist infinitely many positive integers 
such that 
and 
does not divide 
for any positive integers 
satisfying 
.
such that for any 
:
such that 
for all 
be a triangle with circumcircle 
and let 
be the 
-excircle. Let 
and 
be the intersection points of and . Let 
and 
be the projections of onto the tangent lines to at and respectively. The tangent line at to the circumcircle of the triangle 
intersects the tangent line at to the circumcircle of the triangle 
at a point 
. Prove that 
.
board and made 
moves so that it visited all squares and returned to the starting square. At every moment, the distance from the center of the square the king was on to the center of the board was calculated. A move is called 
if this distance becomes smaller after the move. Find the maximum possible number of pleasant moves. (The chess king moves to a square adjacent either by side or by corner.)
is lovely if there exist a positive integer and (not necessarily distinct) positive integers 
, 
, 
, 
such that 
and 
for 
.
, which is a perfect square of an integer?
, 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 
, 
, 
, 
.
be a function such that
for all 
. Determine all such functions 
.
such that the equality 
holds for all pairs of real numbers 
. with integer coefficients such that 
and 
is a square of an integer for all nonnegative integers 
. and an integer , 
is defined as follows: 
if 
and 
if 
.
then we will permute it. So the ordered paid 
can be 
or 
or 
. Now there are total 
permutations of the set 
. So we have total 
options for i.e. 
options for
And options for
Hence for we have total sum as 
, for we have total sum as 
, for we have total sum as . So total up to 
.
.... 
![\[
\begin{cases}
x - y z = 1,\\[2pt]
y - z x = 2,\\[2pt]
z - x y = 4.
\end{cases}
\]](http://latex.artofproblemsolving.com/b/2/d/b2d94e49491a71f86109103f985fe3f238cf4d95.png)
and 
: Prove that :
be an odd prime. Prove that:![\[\displaystyle\sum_{k=1}^{p-1}k^{2p-1} \equiv \frac{p(p+1)}{2} \pmod{p^2}\]](http://latex.artofproblemsolving.com/a/5/5/a55e75d4d9769ceec3a3ce359aac676c7c7875fd.png)
is an obvious solution. Now, let 
.
or 
.
.
is a permutation of the elements of the set 
}