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.
st for every two subsets 
of 
, there exists a polynomial 
with integer coefficients st either 
or 
where the equations are considered mod n.
is the set of reminders of 
where 
mod n.
be a positive integer such that 
.
.
be Rutthee sequence if 
be a positive integer and
for all 
and there are some positive integer 
sastisfied 
such that 
and 
also find all possible value of in Rutthee sequence
be triangle with point 
and 
on 
and 
are on the same side of 
be midpoint of segment and 
be midpoint of segment 
be intersection of 
with 
and 
be two positive integers, such that 
. Find all the integer values that 
can take, where ![\[ f(a,b) = \frac { a^2+ab+b^2} { ab- 1} . \]](http://latex.artofproblemsolving.com/c/b/2/cb27ccbcde6d90f98e10f8f6876c36ae8972af7f.png)
be positive real integer such that 
Prove that
such that 
and 
real number on the board such that each number is more than 
.
on the board and write 
.
and 
.
?
of positive integers satisfying the equation 
be a natural number . Define 
be cut number at 
if
Prove that if 
be cut number at 
prove that 
.
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 cell with 
property if after remove that cell from the board The board can be tile with 
tetromino. cell 
Note: tetromino can be rotated but not flipped
first we use 
and get 
and 
is odd.
so 
but from 
we get 
is in the form 
. Note that 
because if 
then 
which is impossible so we can conclude that 
and 
by zygmondy theorem we cannot find 
because if for all RHS will get e new primitive prime so we conclude 
and 
. Only 
works
. Taking the given equation modulo 
, we learn that![\[(-1)^x-1\equiv 0,1\pmod{4},\]](http://latex.artofproblemsolving.com/2/e/3/2e3407f862df4b26cebe01f79bad796857957e71.png)
so the only possibility is that 
and 
are even. We actually don't need that is even, but we will et 
. Thus, we have![\[3^{2b}-z^2=5^y,\]](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 
, the second 
. We see that 
, which means that we must have 
. Thus, 
and 
, so![\[2\cdot 3^b=5^y+1.\]](http://latex.artofproblemsolving.com/b/4/c/b4cd3f581545011e4747505a027e46bdcdc05302.png)
For the left side to be divisible by 
, we need 
to be odd. Then, we see that 
, so![\[\nu_3(5^y+1)=\nu_3(25^y-1)=1+\nu_3(y)\]](http://latex.artofproblemsolving.com/e/c/f/ecf7cfb32a1f8c6260ca7e01e63545e236575341.png)
by LTE. Thus, 
, so![\[6\cdot 3^{\nu_3(y)}=5^y+1.\]](http://latex.artofproblemsolving.com/5/7/a/57a76dd32ccd4077d6188d8c404e6ab363c2757b.png)
But note that the LHS is at most 
, so we have 
, or 
. When , we get that 
, so 
. This means that 
, so 
, so the solution must be the one that we claimed. It is easy to see that it works, so we're done. 
.
, since 
and 
, we must have as even, or else 
which can't be equal to 
. Therefore, we denote 
.
Since the only factors of 
is powers of 
, we must have 
and 
as powers of . I claim that 
. If is a power of greater than 
, then 
. However, this implies that 
, so 
, which is clearly impossible.
where 
is a positive integer. I claim that for all 
, there are no solutions. This is because then 
, thus 
, which implies is a multiple of . Rearranging, we get 
. However, then because is a multiple of , we have 
. Thus 
must be divisible by 
, which is clearly impossible. Thus, the only solution is when 
, so our only solutions for 
is .
. We get that is even and let's denote it as 
. Then 
. Then it follows that 
and 
where 
and 
. If we sum 
we get that 
but left side is not divisible by so we obtain that 
. Now our equation becomes 
.If 
by taking 
we get that 
is divisible by and let 
. Now we get that 
and that 
, then it follows that 
which gives us contradiction when watching 
so 
. Then plugging back we get that , 
and . With this we are done!
we get that 
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 
![\[ 3^x - 5^y = z^2.\]](http://latex.artofproblemsolving.com/6/4/d/64da66e4dcb59114cc7a3b967c84f82211e6f6de.png)
in positive integers.
prime 
is solution
is a unique solution 
gives us 
furthermore from the fact that 
we get that 
, is odd., we get that 
furthermore 
implies that is even. (
)
, we get that 
which implies that 
is even.
, now notice that if we sum the equations we get 
however notice that for 
, 
must be equal to 
. Thus 
so now we can use 
to finish up our problem.
, furthermore by plugging in our result into our original equation we get 
now let 
, thus 
, furthermore form the equation 
we get that , thus by plugging into our original equation we get 
.
we see is a multiple of 
. Let 
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 as the only solution.
Prove that
Let 
Prove that
Prove that
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... 
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 a square we can apply difference of squares. The equation becomes 
It follows that both 
and 
are multiples of 
to be the average of two different powers of 
,
.
yields 
,
as 
and 
.
gives us 
.
gives us 
,
& 
So, 
Since, 
is not divisible by 
.
If 
.
,
.
is 
,
,
,
,
.
,
.