[/quote]
,
be positive integers.Define injective function 
from 
to itself such that 
.
divides 
.
board that can be colored so that for each square in the board, at most one square adjacent to it is colored.
such that 
.
that divides 
for the positive integer 
.
points are collinear in Euclidean plane, such that Shining cannot find a group of three points such that their circumcircle contains none of the other remaining points. Is he always able to do so?
of positive integers such that 
in all of this cases :
3-Every prime number less than 
can be represented in the form 
by taking an Appropriate pair 
and 
. The number 
is the least prime number that cannot be represented as a sum or an difference of a power of 
, 
, 
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2^a + 3^b + 1 = 6^c
\]](http://latex.artofproblemsolving.com/1/9/e/19eb5efa2ccdf54cd277339f8d3b99f75168779d.png)
be real positive numbers such that: 
.
,
such that 
is a perfect square? (Here 
denotes the integer part of the real number 
?
satisfying
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\text{Find all non-negative integers } (a, b, c, n) \text{ such that}
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![\[
2^a + 3^b + 5^c = n!
\]](http://latex.artofproblemsolving.com/b/c/6/bc6b0754dd874f646fcab124c91583b620db273c.png)
of injective functions from the set of real numbers to itself satisfying
and 
.
is injective if 
for any distinct real numbers 
and 
.
,
such that any graph with 
Y by HWenslawski, Kobayashi
Find all pairs of positive integers 
, such that in a 
table (with 
horizontal lines and 
vertical lines), a diagonal can be drawn in some unit squares (some unit squares may have no diagonals drawn, but two diagonals cannot be both drawn in a unit square), so that the obtained graph has an Eulerian cycle.
, such that in a 
table (with 
horizontal lines and 
vertical lines), a diagonal can be drawn in some unit squares (some unit squares may have no diagonals drawn, but two diagonals cannot be both drawn in a unit square), so that the obtained graph has an Eulerian cycle.This post has been edited 1 time. Last edited by JustPostChinaTST, Mar 28, 2022, 2:27 PM
only. Making the degree of all vertices even is the equivalent and sufficient condition to create an eulerian path.
in S such that 
such that there is a path from 
to 
with only new edges.
.
,
and 
must not be in the same square. Therefore, the only way they can intersect (geometrically) is in a 2x2 square, a path goes from top left corner to center to bottom right corner, and another goes from bottom left to center to top right. We can "adjust" the paths such that the two paths do not cross each other i.e. the edges of one side are always to one side of the other path.
as points of a circle, where two edges do not cross each other (i.e. one edge stays on one side of any other edge)
,
be the graph consisting only diagonal edges, and call a vertex 
if it is on a side of the grid but is not a corner. Then all strange vertices have degree 
in 
,
such that there exists a path from 
to 
for all 
,
paths are disjoint.
,
such that all edges incident to 
times.
as the path between 
,
be the magnitude of the smaller of these two sets.
or for all 
and there exists some 
and 
be the sets of strange vertices on either side of 
and 
are both odd. WLOG let 
and the path between 
which generalizes readily for all 
