[/quote]
,
be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
.
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
,
,
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
and 
satisfies the following recursion formulas: 
,
.
interesting if 
of 
and 
is also a divisor of 
be a triangle, and let 
and 
be points on segment 
symmetric with respect to the midpoint of 
denote the circle passing through 
at 
.
denote the circle passing through 
at 
.
)
.
the feet of the altitudes lying on 
respectively. One of the intersection points of the line 
and the circumcircle is 
The lines 
and 
meet at point 
Prove that 
and circumcenter 
.
and 
be points on the line segments 
is a parallelogram. Prove that 
.
airline companies in a country. Between any two cities, there is exactly one 
and circumscribed about the incircle 
,
.
,
,
,
,
,
,
,
(distinct from 
intersects 
(distinct from 
be the midpoint of the arc 
of 
at 
and 
,
at 
.
.
but doesn't tell it to anyone. Bernaldo thens pick an arbitrary positive integer 
(not necessarily in 
.
chosen by Arnaldo.
be a set of 2025 positive real numbers. For a subset 
,
as the median of 
when all elements of 
.![\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]](http://latex.artofproblemsolving.com/1/d/f/1df2ea8487289111939a53dcc49576fbe6fdef37.png)
Find the smallest real number 
such that for any set ![\[
P(A) - Q(A) \leq C \cdot \max(A),
\]](http://latex.artofproblemsolving.com/c/d/e/cdea343d11c722a423e857aeb01202863405773e.png)
where 
denotes the largest element in 
in 
colors, there exists an arithmetic progression of size 
.
from this?
.
be foots of perpendicular lines from 
.
is cyclic and 
.
bisects 
such that 
Y by Adventure10, Mango247
An immortal flea jumps on whole points of the number line, beginning with 
. The length of the first jump is 
, the second 
, the third 
, and so on. The length of 
jump is equal to 
. The flea decides whether to jump left or right on its own. Is it possible that sooner or later the flee will have been on every natural point, perhaps having visited some of the points more than once?
. The length of the first jump is 
, the second 
, the third 
, and so on. The length of 
jump is equal to 
. The flea decides whether to jump left or right on its own. Is it possible that sooner or later the flee will have been on every natural point, perhaps having visited some of the points more than once?This post has been edited 2 times. Last edited by djmathman, Apr 15, 2016, 5:59 PM
at its 
.
times to the right and the last 
-![\[x+(2^{k}+1)+(2^{k+1}+1)+\dots +(2^{k+m}+1) - (2^{k+m+1}+1)=x-2^{k}+m.\]](http://latex.artofproblemsolving.com/3/3/8/338277af5c4166b66f5125316b96cee09c925902.png)
Now, choosing 
,
and
thus if we keep adding these consecutive numbers, the gap between the sum and the next number will eventually become one. This motivates the following solution:
using some of the numbers 3, 5, 9, 17 etc in the order (i mean that we have used all the numbers 
with 1<= j <=k for some k: Then you keep summing consecutive terms beginning wit
completing the induction.
,
such that![\[ (2^k+1)+(2^{k+1}+1)+\cdots+(2^{n-1}+1) - (2^n+1) = 1.\]](http://latex.artofproblemsolving.com/f/f/5/ff57a01cd4679a2dc104e2485e1801043fcfc407.png)
Proof: Let 
be the above expression. We know 
.
for all 
Therefore, 
,
works. 
that we can reach all 
in a finite number of moves.
,
to the left, and finally send Kelvin 
to the right. I claim that this shifts Kelvin's position right by exactly 
.
which is exactly 
,
.
on the way.If we can prove we can reach 
from 
.
then since 
so we can jump from 
)![\[(2^a+1)+(2^{a+1}+1)+\cdots +(2^{b-1}+1)-(2^b+1)=1,\]](http://latex.artofproblemsolving.com/a/5/f/a5f62ddbd534754df7828c50d897e765cb4718c0.png)
which obviously proves the claim. Simply select 
;
as needed.
Again assume at the 
jump the flea jumps to it's left. So then position at the 
Setting 
we get that the position at the 
jump, jump to the right.
Proof. Indeed, this is equivalent to 
as 
and 
.
the flea moves right and otherwise it moves left. We achieve every natural, done.
steps to the right and 
edit: oops misread the problem
.
.
If we choose 
this sum becomes 
Hence, choosing 
and 
for 
we obtain disjoint sums that all have sum one. We claim by induction that the flea can visit all natural numbers. For the base case, the flea can get to one by 
For the inductive step, from 
the flea can use 
to get to 
the number of steps needed to go from 
jumps in the negative direction, and then the next jump in the positive direction: this yields a net difference of 
which means that Kelvin can reach 
if flea reached point 
t
jumps to left and one jump to right - it works, since the new coordinate is 
This is because 
so for example the first 4 terms are 
and so on so in general 
so we can employ an algorithm where we move forwards if the move number is of the form 
for some 
on move 
,
,![\[2^{n+1}+1=(2^n+1)+(2^{n-1}+1)+\dots (2+1)-(n-3).\]](http://latex.artofproblemsolving.com/8/1/3/813a95a199e43257fc02f626ac92f15c9c427bae.png)
Rearranging, we get![\[(2^n+1)+(2^{n-1}+1)+\dots +(2+1)-(2^{n+1}+1)=n-3\]](http://latex.artofproblemsolving.com/b/0/8/b08868e1c68b652c3d7161bf809bdccbe5a2d1a2.png)
so take 
,
moves, if the fly is on point 
it can make 
and end up on point 
Now, we can choose 
So, from any point 
In other words, Kelvin can visit all points of integer coordinates at least once.
then does a left jump of 
,
From here, simply choose 
which in total moves the flea to the next integer. Then repeat this process starting at 
.
number of 
s
Now, we add 
.![\[ n+2^j-(2^j) - 1 \]](http://latex.artofproblemsolving.com/4/2/8/4284ec8e85cf77591c3bad2d1f207e439a7a02a0.png)
![\[ = n-1 \]](http://latex.artofproblemsolving.com/1/e/6/1e698ecdf72b8670aca4e3349c991588789e5d4f.png)
This is what we wanted as now, I can go to any positive integer by making my 
and then going down to my desired integer.
there exists a ![\[2^k + 1 + 2^{k + 1} + 1+ \dots + 2^n + 1 - (2^{n + 1} + 1)= 1\]](http://latex.artofproblemsolving.com/3/d/7/3d734abf45b2c20ac565a4270717f6447b9c3886.png)
