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 scalene triangle with point 
and 
on the plane such that 
is an equilateral . Let 
intersect 
and 
at 
and 
respectively and 
intersect 
and 
at 
and 
respectively .
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of square numbers. Prove that: 
is the sum of square numbers
real number on the board such that each number is more than 
.
on the board and write 
.
and 
.
?
be triangle and point 
be 
altitude of 
.
be a point on and such that 
and 
. Point 
is the intersection of 
and 
. Point be intersection of 
and 
. Line 
intersect circumcircle of again at .
.
bisect .
, define a sequence of integers 
by letting 
and 
for 
. Let 
. Determine the largest possible 
such that, for some positive integer , the numbers 
are all prime.
cities in a country, where 
is an integer. Some pairs of cities are connected by direct (two-way) flights. For two cities 
and 
we define:
A 
between and as a sequence of distinct cities 
, 
, such that there are direct flights between 
and 
for every 
;
A 
between and as a path between and such that no other path between and has more cities;
A 
between and as a path between and such that no other path between and has fewer cities. and in the country, there exist a long path and a short path between them that have no cities in common (except and ). Let 
be the total number of pairs of cities in the country that are connected by direct flights. In terms of , find all possible values 
+1 w
such that 
for all real number 
and 
be a sequence of perfect squares such that 
can be obtained by concatenating a digit to the right of 
. Determine all such sequences that are of maximum length.
be unit vectors on a plane, where is an odd number. Prove that there exist 
for 
such that 
a
.
. Prove that
![\[ 3^x - 5^y = z^2.\]](http://latex.artofproblemsolving.com/6/4/d/64da66e4dcb59114cc7a3b967c84f82211e6f6de.png)
must be 3k+1, then y divise by 2.
,
so 3 is divise by 2 or 5 that wrong if m,n is positive interger.
is divise by 4, but right hand is not. So there is no pair (x,y,z) satify for this problem.
,
is even. Let 
,
and 
.
.
,
.
is even. Consider modulo 4, 
,
,
(since 
)
by Zsigmondy's theorem. So the only solution is 
to see that if 
then 
is divisible by 
.
is divisible by 
which is not the case of the 
.
,
,
and 
.
so 
and using the method of the infinite descent we get 
.
so 
and using the method of the infinite descent we get 
.
and 
.![\[ 5^m+1=2\cdot 3^{\alpha} \textrm{ and } 5^{2m}+5^m+1=3^{\beta} \Longrightarrow 5^m=4\cdot3^{2\alpha}-3^{\beta},\]](http://latex.artofproblemsolving.com/4/3/9/4397cb3ea9b78617da91c3b3afaadbdd1d15508d.png)
we conclude that 
and we get:
and since each of these 
summands is congruent 1 modulo 3 we get 
.
does not work,others work too,that depends on the way you have proceeded.Specially here 
is divisible by 
for 
,
Can you explain why??? I don't see... 
first we use 
and get 
and 
so 
but from 
we get 
is in the form 
.
because if 
then 
which is impossible so we can conclude that 
and 
by zygmondy theorem we cannot find 
because if for all 
and 
.
works
,
for positive integer ![\[ (3^a-z)(3^a+z)=5^y \]](http://latex.artofproblemsolving.com/7/0/6/706b11f26a1a6bb86f50a4f37ff5db4dba42ec2d.png)
Suppose 
where 
.
.
,![\[ 2 \cdot 3^a=5^y+1\]](http://latex.artofproblemsolving.com/4/8/9/4893eb09ef8fcc7d6a8f792fdac02fff379cc8b0.png)
We can finish through LTE now but here is a more elementary approach. Suppose 
,
since the order of 
is 
.
.
,
.![\[ \boxed{(x, y, z)=(2, 1, 2)} \]](http://latex.artofproblemsolving.com/3/1/c/31cd7f9169b297997303dc136245333ccfc13bc5.png)
,
,
;
is not a quadratic residue 
.
;
,
.
,
is a square we can apply difference of squares. The equation becomes 
It follows that both 
and 
are multiples of 
.
is the only solution; basically what we want is for 
to be the average of two different powers of 
,
yields the solution 
.
yields 
,
as 
and 
.
gives us 
.
gives us 
,
.
,![\[(-1)^x-1\equiv 0,1\pmod{4},\]](http://latex.artofproblemsolving.com/2/e/3/2e3407f862df4b26cebe01f79bad796857957e71.png)
so the only possibility is that 
are even. We actually don't need that 
.![\[3^{2b}-z^2=5^y,\]](http://latex.artofproblemsolving.com/c/8/3/c83dbdc2072b61f68b7d6f7246904329ee79f258.png)
so![\[(3^b-z)(3^b+z)=5^y.\]](http://latex.artofproblemsolving.com/7/b/9/7b90205bcd5391bc9b83be52e5772023ffeb614b.png)
Let the first factor be 
,
.
,
.
and 
,![\[2\cdot 3^b=5^y+1.\]](http://latex.artofproblemsolving.com/b/4/c/b4cd3f581545011e4747505a027e46bdcdc05302.png)
For the left side to be divisible by 
to be odd. Then, we see that 
,![\[\nu_3(5^y+1)=\nu_3(25^y-1)=1+\nu_3(y)\]](http://latex.artofproblemsolving.com/e/c/f/ecf7cfb32a1f8c6260ca7e01e63545e236575341.png)
by LTE. Thus, 
,![\[6\cdot 3^{\nu_3(y)}=5^y+1.\]](http://latex.artofproblemsolving.com/5/7/a/57a76dd32ccd4077d6188d8c404e6ab363c2757b.png)
But note that the LHS is at most 
,
,
.
,
.
,
,
& 
So, 
Since, 
is not divisible by 
.
If 
.
,
.
is 
,
does not divide 
,
,
and 
,
which can't be equal to 
.
.
Since the only factors of 
is powers of 
and 
as powers of 
.
,
.
,
,
where 
,
,
,
.
.
must be divisible by 
,
is 
.
.
.
where 
and 
.
we get that 
but left side is not divisible by 
.
.
by taking 
we get that 
.
and that 
,
which gives us contradiction when watching 
so 
and 
Putting 
we get that
Note that taking 
in out original equation we get that 
Plugging it back in we get
But notice that 
for 
(Rigorous proof : Induction)
Hence the solutions are 
prime 
is solution
is a unique solution 
furthermore from the fact that 
we get that 
,
furthermore 
implies that 
,
which implies that 
is even.
,
however notice that for 
,
.
so now we can use 
to finish up our problem.
,
now let 
,
,
we get that 
we see 
From 
we also see that 
Say 
We have 
If 
we have 
a contradiction. Thus 
so, 
If 
we have 
and from Euler's Theorem 
We have 
From sum of cubes we have 
If 
we have 
Since 
we have 
a contradiction. Thus 
For 
we have 
no solution. For 
we have 
and 
,
,
.
,
.