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 real numbers such that![\[ a^2+x^2=b^2+y^2=c^2+z^2=(a+b)^2+(x+y)^2=(b+c)^2+(y+z)^2=(c+a)^2+(z+x)^2 \]](http://latex.artofproblemsolving.com/7/e/b/7ebfb2f663d390714310d3c43b9c68e1b101d632.png)
.
be a triangle with incenter 
and circumcircle 
. Circles 
passing through 
and 
passing through 
are tangent at . Let meet minor arc 
of at 
and at 
, and let meet minor arc 
of at 
and at 
. Rays 
and 
meet at 
. Let 
be a point such that 
is tangent to and 
is tangent to .
are collinear.
be the circumcircle of acute triangle . Points 
and 
are on segments and respectively such that 
. The perpendicular bisectors of 
and 
intersect minor arcs and of at points 
and 
respectively. Prove that lines 
and 
are either parallel or they are the same line.
of them there were either 
who were all friends among each other or who weren't friend with each other. Prove that you can separate all the people at the party in two groups in such a way that in the first group everyone is friends with each other and that all the people in the second group are not friends to anyone else in second group. (Friendship is a mutual relation).
and play a game, given an integer 
, writes down 
first, then every player sees the last number written and if it is 
then in his turn he writes 
or 
, but his number cannot be bigger than . The player who writes wins. For which values of does win?
be the circumcircle of a scalene triangle . Let 
be a circle internally tangent to in . Tangents from touch in and , such that lies in the interior of 
. Similarly, tangents from touch in 
and 
, such that lies in the interior of .
and 
concur on the bisector of 
.
, let 
be all positive integers smaller than that are coprime to . Find all 
such that 
for all 
is the largest positive integer that divides both 
and 
. Integers and are coprime if 
.
, 
and 
are the midpoints of sides 
and respectively. Point lies inside the triangle. Let 
. Prove that 
.
, 
, and 
of a hexagon 
inscribed in a circle 
intersect at a point 
, and the acute angle between any two of them is 
. Let 
be the radius of the circle tangent to segments 
and 
and internally tangent to ; the radii 
, 
, 
, 
, and 
are defined similarly. Prove that![\[
r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}.
\]](http://latex.artofproblemsolving.com/0/d/6/0d66d6685bbc97c7553e9ed3ba14ea0dc3328b49.png)
, let 
be the circumcentre, 
be the orthocentre, and be the centroid.
satisfy the following condition:
is constant.
. Find .
such that numbers 
are both perfect squares.
of nonnegative integers for which ![\[m^2 + 2 \cdot 3^n = m\left(2^{n+1} - 1\right).\]](http://latex.artofproblemsolving.com/4/1/3/4133d1af365417fa5b6a8b2560dc80c1d28a6a18.png)
]
, we get 
. Since if is a solution, 
is also a solution, so WLOG assume 
. Then we have 
. If either 
or 
, it is easy to check that 
are the only solutions by simple bounding. So now assume both and 
are at least 3. Taking mod 27, we get 
. If 
, then 
. Taking mod 19, we get 
, so 
. Taking mod 7, 
. Contradiction! Now if 
, 
. Similarly, taking mod 19, we get 
and taking mod 7 gives 
. Contradiction! So 
are the only solutions for 
, so 
.
. Therefore we get 
.
and 
, we get
.So 
, and 
.
, and it is easy to check that there is no 
such that 
.That's contradiction and we get 
or 
.
. Therefore
.
so 
or 
..
we have that 
is even and is even.
so
so we have that 
which implies //
so 
.
:
so we have
but
so . we get 
.
, which work. for any fixed , we will show that 
do not work.![\[ t^2 = \left( 2^{n+1}-1 \right)^2 - 4 \cdot 2 \cdot 3^n \iff 8 \cdot 3^n = \left( 2^{n+1}-1 \right)^2 - t^2. \]](http://latex.artofproblemsolving.com/2/c/1/2c1c2addd98b074c162eca46fb4fd31d02cc13a9.png)
The right-hand side factors as a difference of squares, where both factors sum to 
. In particular, these factors have the same parity, and thus must be of the form 
and 
, for nonnegative integers , with 
. Rearranging, we obtain ![\[ 2^{a+b+1} - 1 = 3^a + 2 \cdot 3^b. \]](http://latex.artofproblemsolving.com/7/2/a/72a7cf96d506e9fecbe26a5ca0ab27e8f9347172.png)
We have solutions 
and 
, which corresponds to 
and 
above. Manual inspection reveals no other solutions with 
, so we will suppose 
and derive a contradiction.
, we find is even and is odd (hence unequal). The idea is that the exponent of is much too large in the above equation. Indeed, 
Intuitively, this is way too big. We write 
which is false for 
, as desired.
, we get . Since if is a solution, is also a solution, so WLOG assume . Then we have . If either or , it is easy to check that are the only solutions by simple bounding. So now assume both and are at least 3. Taking mod 27, we get . If , then . Taking mod 19, we get , so . Taking mod 7, . Contradiction! Now if , . Similarly, taking mod 19, we get and taking mod 7 gives . Contradiction! So are the only solutions for , so . ??? How about 
??? Because it said that 
, we get . Since if is a solution, is also a solution, so WLOG assume . Then we have . If either or , it is easy to check that are the only solutions by simple bounding. So now assume both and are at least 3. Taking mod 27, we get . If , then . Taking mod 19, we get , so . Taking mod 7, . Contradiction! Now if , . Similarly, taking mod 19, we get and taking mod 7 gives . Contradiction! So are the only solutions for , so . ??? How about ??? Because it said that ———> is a solution,we will get 
and 
,
is also a solution
![\[ m^2-(2^{n+1}-1)m + 2\cdot 3^n = 0. \]](http://latex.artofproblemsolving.com/7/6/5/765088df59117c6fe09a5d91bcf4a675e5b3d7ed.png)
So![\[ (2^{n+1}-1)^2 - 8\cdot 3^n = k^2 \]](http://latex.artofproblemsolving.com/7/e/a/7ea0b492cda13ad27c42d0175c5a29418c7f25b5.png)
for some 
. Then![\[ 8\cdot 3^n = (2^{n+1}-1-k)(2^{n+1}-1+k). \]](http://latex.artofproblemsolving.com/7/2/3/723021ac12914257faea63e1ee523440758cb423.png)
So for some 
with , we have![\[ \{ 2^{n+1}-1-k, \ 2^{n+1}-1 + k \} = \{2\cdot 3^a,\ 4\cdot 3^b \}.\]](http://latex.artofproblemsolving.com/6/d/1/6d1bd10bbe366dfb4b81518eeb11fd18b6bd371a.png)
We can't have 
because then the sum of the two would be odd, but it is clearly even. Now, adding gives![\[ 2^{n+2} - 2 = 2\cdot 3^a + 4\cdot 3^b \implies 2^{n+1} - 1 = 3^a + 2\cdot 3^b. \]](http://latex.artofproblemsolving.com/3/7/5/3750316f1a5f640dfdc00c0f412eee6ff08d34b1.png)
mod 8 gives is even and is odd, so 
is odd too. Hence is even, so![\[ \nu_3(2^{n+1}-1) = \nu_3(2^{n+1} - (-1)^{n+1}) = 1 + \nu_3(n+1) \ge \min(a,b). \]](http://latex.artofproblemsolving.com/9/b/9/9b91798ae850019b1db96fb77d227780b19d4e4b.png)
Then![\[ 1+\nu_3(a+b+1) \ge \min(a,b) \implies a+b+1 \ge 3^{\min(a,b)-1}. \]](http://latex.artofproblemsolving.com/3/d/d/3dd54b98efaed14d24781702ceaf9313eea382a1.png)
This is a strong bound. We also know 
. Intuitively, the RHS is way too big, using the above bound. Suppose 
, in order to make the RHS bigger. We will still use size to narrow down the possibilities. So 
. We know 
, so 
. Then![\[ 2^{a+b+1} = 1+3^a+2\cdot 3^b > 3^b \implies 2^{a+1} > 1.5^b \ge 1.5^{3^{a-1}-a-1}. \]](http://latex.artofproblemsolving.com/e/5/a/e5ae2a813a2a61ab955788686e043576276c7f54.png)
If 
, this is not true. So 
. Now manual case check. It turns out the answer is s 
.
; then by Vieta there must be integers 
with 
and 
. WLOG let 
be even, and let 
; then we want to solve the equation 
. Take 
of both sides to get by LTE that 
which implies that 
. For convenience, let 
and 
. Now note that 
so from the previous centered equation we have 
which is only true when 
. From here it is easy to manually check that the only solutions are 
.
of nonnegative integers for which 
after several definitions, both of these cases are sufficient to solve the following equation.
1)
in which case it is not difficult to solve.(mod8).
this leads to a boundary of , and it is easy to see the rest, just as 
.
. Also, if is a solution for a fixed then 
is also a solution. Therefore we may WLOG . Note that we then get 
Now if 
or 
we can check that no solutions exist which implies that 
or that is odd. Then we have![\[\nu_3(2^{n+1}-(-1)^{n+1})=1+\nu_3(n+1)=\text{min}(k,n-k)\]](http://latex.artofproblemsolving.com/a/8/1/a81c90b4c19743c4af2ca488001fba73c3efc2d5.png)
so now we split into cases. If 
then we have 
which implies that 
or that 
which means that![\[2^{n+1}-1=3^{n-k}+2\cdot 3^k>\frac{2\cdot 3^{n-1}}{n+1}\]](http://latex.artofproblemsolving.com/8/d/e/8de9349e9ddb3d7588eae6d50a6f6d10c8e86efd.png)
so that 
. Then 
which gives 
. Only 
works and gives the solutions 
. Now suppose that 
. Then we have 
so we get 
and 
. Bounding, we get 
, so that 
and 
. Only 
works so we get the final two solutions 
. We are done. 
, so 
where 
and 
and 
Consider 
the RHS isn't divisible by , so 
or 
Thus, 
or 
One of these equals so pluggin 
gives no solutions by bounding, For 
, take mod to get 
, absurd. For 
does not work again because we get 
so 
For 
does not work by bounding. In summary, no solutions.
so ![\[\min(i,j)=\nu_3(4^{\frac{n+1}{2}}-1)=1+\nu_3(\frac{n+1}{2})=1+\nu_3(i+j+1)\le 1+\log_3(i+j+1)\]](http://latex.artofproblemsolving.com/a/3/7/a37018b9fe27665e82c07a6adeb00fdc08f2a1be.png)
Thus, 
so 
Suppose 
then this is a contradiction.
to get our solutions.
and 
(by mod
)
or 
and mod
) and 
can not be true at same time , Work on 
and 
and do same things above
and 
. Since the discriminant is a square, we have![\[ (2^{n+1} - 1)^2 - 8\cdot 3^n \implies 8\cdot 3^n = (2^{n+1} - 1 + k)(2^{n+1} - 1 - k)\]](http://latex.artofproblemsolving.com/c/0/c/c0ca1c0c8ccad34c44153be1954017d2d24fcaf1.png)
Note that the factors must be of the form 
and 
, with 
. Allowing to be negative, we may assume WLOG that the first factor is and the second one is . Hence,![\[k = 3^x - 2\cdot 3^y \text{ and } 2^{n+1} - 1 = 3^x + 2\cdot 3^y\]](http://latex.artofproblemsolving.com/1/e/f/1ef161d05dee353645d2f33a0d8abf2cccc24fd4.png)
We now bound and 
. We have![\[3^x < 2^{n+1} \implies 3^y > \frac{3^n}{2^{n+1}} \text{ and } 3^y < 2^n \implies 3^x> \frac{3^n}{2^n}\]](http://latex.artofproblemsolving.com/1/7/f/17fd1bf928bb116b5b25b0a30270b68d1ea63ee0.png)
Hence,![\[\frac{3^n}{2^{n+1}} < 3^{\min (x, y)} < 3^{\nu_3 (2^{n+1} - 1)} = 3^{1 + \nu_3 ((n+1)/2)} \le 3\cdot \frac{n+1}{2}\]](http://latex.artofproblemsolving.com/a/5/c/a5c4b1ac57cd6d6e88eea187199277c28f6716e2.png)
This gives 
. We get the solutions 
.
, that is, let 
. Then, if is a solution, is a root of 
and so the second root must be 
which is actually a positive integer as 
by Vieta's formulas. Hence if is a solution, then 
has positive integer roots 
. Note that 
and 
. A shows that and lead to no solutions, which implies that 
and 
, so 
, so is even. WLOG let 
. Now 
whence![\[\nu_{3}(m_{1})\leq \nu_{3}(m_{1}+m_{2}) = \nu_{3}(2^{n+1}-1)=\nu_{3}(4^{\frac{n+1}{2}}-1)=1+\nu_{3}\left(\frac{n+1}{2}\right)\leq 1+\log_{3}\left(\frac{n+1}{2}\right) \Longrightarrow m_{1}\leq \frac{3(n+1)}{2}\]](http://latex.artofproblemsolving.com/4/f/9/4f909536666893991db86fc1dc4c8539ba5394cb.png)
by LTE. Note that 
is decreasing on the interval ![$(0,\sqrt{c}]$](http://latex.artofproblemsolving.com/5/f/c/5fc0891909f965092326aa1a6976a2df50d1df3a.png)
where 
is any positive real number as 
. Now assume that 
. we have that 
, so we have that:![\[2^{n+1}-1 = m_{1}+m_{2} = m_{1}+\frac{2\cdot 3^{n}}{m_{1}} \geq \frac{3(n+1)}{2}+\frac{4\cdot 3^{n-1}}{n+1} \Longrightarrow (n+1)2^{n+2}-2n-2-3(n+1)^2-8\cdot 3^{n-1}\geq 0\]](http://latex.artofproblemsolving.com/8/6/9/8693d1b868d1240e17536b40ea9819b359340928.png)
This is a contradiction as the last inequality doesn't hold for , here's a This implies that 
and we're left to check the remaining cases. In the beginning, we mentioned that must be odd, so 
. A direct check shows that 
doesn't lead to a solution whereas and lead to the solutions 
.
.
for some 
. Then, the equation simplifies to 
We take of both sides.
, then we must have 
. This reduces to 
, which is obviously impossible for size issues. must be odd, and set 
. Then if 
, then 
But on the other hand 
But this implies 
, or 
Finite case check yields 
and 
. The other case works out in the exact same way by symmetry.
, notice that the transformation 
bijects between solutions with and 
. Thus, the other two solutions are 
and 
.
, and if 
is a solution, then so is 
, WLOG, . We can rewrite the equation is 
. If 
or 
, we can easily check that 
are the only solutions. Taking mod 27, we have 
, but taking mod 73, we get 
but 
is either 
or 
(since the order of modulo 
is 12), and it is easy to check none of these yield solutions.
and was interesting, since the order of 3 was also nice. From that it was fairly obvious.
; in particular, we claim n is odd: If n were even note that 2^{n+1}-1 is not 0 mod 3. mod 3 we get 
, so there are two cases.
. If 
has no solutions. If 
also has no solutions.
as desired. 
to see that 
. We also know that must be odd because taking 
, thus 
or 
for some positive integer 
. We can divide the equation by 
to get ![\[3^a + 2 \cdot 3^{n-a} = 2^{n+1} -1\]](http://latex.artofproblemsolving.com/8/3/e/83e9e7a0079b17f7389792f6b70744d09e975b11.png)
We can see that 
(note the expression is not equal only when 
). We can see that the LHS is always divisible by (if 
then the LHS is even which is not possible). This means that we can assume that is even. Thus we can see taking 
that we can see that we need ![\[\min(a, n-a) \leq 1+v_3\left(\frac{n+1}{2}\right)\]](http://latex.artofproblemsolving.com/b/3/4/b34e6b4a907558432eefd8c57c79f8df2687557c.png)
Notice that the first value of to attain on the LHS is 
. This means that we need ![\[3^x + 2 \cdot 3^{2 \cdot 3^{x-1}-x - 1} \leq 2^{2 \cdot 3^{x-1}} - 1\]](http://latex.artofproblemsolving.com/d/2/c/d2c35dd5d4de0b353aa2043743dbe6be54793460.png)
or ![\[3^{2 \cdot 3^{x-1}-x - 1} + 2 \cdot 3^x \leq 2^{2 \cdot 3^{x-1}} - 1\]](http://latex.artofproblemsolving.com/1/5/6/156b8ad0be1ccecd767f55ddca6ad3cc31fafdf6.png)
The first equation obviously has size issues when 
. Checking those finite cases we get 
Similarly we can see that the second equation has size issues when . Checking finite cases we get 
.
then we can divide by to get 
This solution set is actually isomorphic to the solutions. Which give 
. 
. we have . Thus let 
with 
. Then we have,
Clearly 
and hence 
must be odd. Specifically and are of opposite parity. Reindex such that 
whence and must be of opposite parity. Then we have,
For 
we can find solutions 
, 
. Assume that 
. Taking modulo 
and modulo 
we find that 
and 
. Finally taking modulo 
we find that,
Now we claim that there are no solutions. Indeed looking at the residues leaves for powers congruent to modulo 
we see, 
hence we are done.
then
But this fails by 
and the equation becomes
We can easily check that nor 
, then must be odd of the equation and note that
so 
so 
Which fails for 
so we can manually check for 
. Expressing the equation as a quadratic in , we require![\[8 \cdot 3^n = (2^{n+1}-1)^2-x^2 = (2^{n+1}-1-x)(2^{n+1}-1+x)\]](http://latex.artofproblemsolving.com/5/d/1/5d180b8e40a691d2ec4c605c098ed6ab6a3dfafc.png)
. Parity tells us these factors correspond to 
. Modulo 8 further tells us 
, so assume 
(where the other case is analogous) to give![\[2^{n+1}-1 = 2 \cdot 3^y + 3^{n-y} = 3^y \left(3^{n-2y}+2\right).\]](http://latex.artofproblemsolving.com/d/2/4/d24a3ba2e79bef82fabb4a3ab42734f3b1947427.png)
or 
are at most 2, casework reveals the solutions above. Otherwise, must be even, so we use LTE to find![\[v_3(2^{n+1}-1) = v_3\left(\frac{n+1}{2}\right)+1 \ge y \implies n+1 \ge 2 \cdot 3^{y-1},\]](http://latex.artofproblemsolving.com/0/3/3/0334824b7d6db23ffc021bb5353d022dadc999a1.png)
. (One way to see this is by noting the RHS of our equation grows faster than the left, so we can isolate and find an upper bound in terms of which excludes the interval we just proved.) 
is a perfect square, using difference of squares we get 
, so one of the two terms on the left side is , the other is , so we get 
, so we get 
, with 
. If is even, the right side is not divisible by , so one of is 
. If , we get 
, which obviously has no solutions. If 
, we get 
., which also has no solutions by size. Thus is odd, so 
, and 
, so 
. Thus 
. Thus we can bound 
. Thus, we can start by eliminating all for which 
, or equivalently 
. We can observe 
, then taking derivatives of both sides, we see 
for 
as 
as 
. between and 
. We already forced odd, so it remains to check 
. Checking 
gives 
, which has no solutions, checking gives 
, which resolves to 
, so we have solutions 
, both of which clearly work. Checking gives 
, which resolves to 
, so we have solutions 
, both of which clearly work. We have exhausted all cases, so we are done.
![\[(m,n) = (6,3), (9,3), (9,5), (54,5), \]](http://latex.artofproblemsolving.com/f/7/d/f7dff86c9b05f9e96e5f59b77d158f473dfce149.png)
which clearly work. Now we show they are the only ones, then 
., then 
.
., by taking the discriminant, we find ![\[k^2 = (2^{n+1} - 1)^2 - 8\cdot 3^n \]](http://latex.artofproblemsolving.com/e/2/1/e21405892f52a1d864742ccdd42d8c65219b9372.png)
for some nonnegative integer , so ![\[8\cdot 3^n = (2^{n+1} - k - 1)(2^{n+1} + k - 1)\]](http://latex.artofproblemsolving.com/a/f/5/af5b37b0d6eae08d344d499ddd33ff89005d7042.png)
and 
in some order. So . The equation is now equivalent to ![\[2\cdot 3^a + 4\cdot 3^b = 2^{a+b+2}- 1,\]](http://latex.artofproblemsolving.com/1/c/9/1c910197468aef2b90cd46d9ead2a45038df217f.png)
so ![\[3^a + 2\cdot 3^b = 2^{a+b+1} - 1\]](http://latex.artofproblemsolving.com/7/0/0/7001d04a8da2939d8d6ddc4b32edb0c57c6a2f48.png)
, then 
, absurd. If 
, then 
, also absurd.
gives even and odd (so 
).
is odd, we have 
(this can be found by factoring and LTE).
and 
. of the equation, we get ![\[c-1 = \nu_3(a+b+1) .\]](http://latex.artofproblemsolving.com/3/f/9/3f9cfb14f7f76dcc7d69a57579a177168fd04214.png)
Since 
is even, we have ![\[a+b+1 \ge 2\cdot 3^{c - 1}\]](http://latex.artofproblemsolving.com/1/b/4/1b434478f67b830d30bb7d73a39706d91b2f1ffe.png)
![\[2^{a+b+1} > 3^{M}\]](http://latex.artofproblemsolving.com/4/0/6/4062ee265ce35d9403accbf82403769d37f7e495.png)
so 
![\[2M\ge a+b+1,\]](http://latex.artofproblemsolving.com/2/d/6/2d6cf9f9ef1bfb5d2afae57596c367d6bc946d84.png)
so ![\[2^{c+1} > 3^{c- 1} \implies c\le 4\]](http://latex.artofproblemsolving.com/4/6/5/465a00a12f32e0651c45afa798841b7898e2eec6.png)
![\[2^{M + 5} \le 2^{c + M + 1} >3^M, \]](http://latex.artofproblemsolving.com/3/5/9/359553f76cdc81bcdae96b3fc5454c7470ac1124.png)
so 
.
, then since is odd, we have 
. So ![\[81 + 2\cdot 3^b = 2^{b+5} - 1\]](http://latex.artofproblemsolving.com/8/d/d/8ddbc63a406a5f5d5e645253c914cd5160a776e0.png)
Its easy to check that no odd 
satisfies this.
, then since is even, we have 
. So ![\[3^a + 54 = 2^{a+4} - 1\]](http://latex.artofproblemsolving.com/5/7/e/57e9fe262a94df40e2aa302663b9929f21cd8ea3.png)
It's easy to check that all even 
do not satisfy this.
, then since is odd, we have 
. So ![\[9 + 2\cdot 3^b = 2^{b+3} - 1\]](http://latex.artofproblemsolving.com/3/4/8/348b29bab00d2cdb27db2d42872fce74684dd67e.png)
Here we can check that the only possibility is , since 
. This gives .
, then since is even, we have 
. So ![\[3^a + 6 = 2^{a+2} - 1\]](http://latex.artofproblemsolving.com/a/b/a/abaee24113f49e9514157f9f3e3e5fec2c5a24b4.png)
Here we must have , which gives ..
are 
, 
, 
and 
which can easily be seen to work. We now work on showing that they are the only ones. and look at the discriminant, which must be a perfect square. Note,![\[\Delta_m=(2^{n+1}-1)^2 - 4(2\cdot 3^n)= (2^{n+1}-1)^2-8\cdot 3^n \]](http://latex.artofproblemsolving.com/1/b/2/1b2de422f2d4b496b049b987104cec06b58c7ce7.png)
Setting this expression equal to a perfect square and factoring we have that,![\[(2^{n+1}-1-x)(2^{n+1}-1+x) = 8 \cdot 3^n\]](http://latex.artofproblemsolving.com/e/7/8/e789e0b526dc57a3ac1fa6567cbd74aa708e87ec.png)
for some positive integer . Clearly must be odd, and hence each of the above factors must be even. Thus, they are and for a pair of non-negative integers such that . Summing the expressions for each factor we have,
Say and 
. Then the equation rewrites to![\[2^{n+1}=3^b +3\]](http://latex.artofproblemsolving.com/7/f/0/7f01d735a00fb523eff75b5baacaf4d1c7e98ec2.png)
which is a contradiction since the right-hand side is divisible by 3 but not the left. Similarly if and 
the equation reduces to![\[2^n=3^a-1\]](http://latex.artofproblemsolving.com/1/7/5/175fe70e8a1df998ee64de39a4ad2a49cdadc6d4.png)
to which the only solution is 
and 
and and by Mihaelescu's Theorem.
. Let 
and 
.We now show the following.
we must have 
.
means that divides the right-hand side and hence so must the left, implying that is even. Thus by Lifting the Exponent Lemma we have,![\[ \log_3\left(\frac{a+b+1}{2}\right) +1 \le \nu_3(3)+\nu_3\left(\frac{a+b+1}{2}\right) = \nu_3(2\cdot 3^a + 3^b) = \min(a,b)\]](http://latex.artofproblemsolving.com/c/2/3/c23ee7d68bbcd0bf6aa0a53c1574b4c81089a7b9.png)
since 
as is odd. Now,
However,![\[ 2^{m+M+1}-1 =2^{a+b+1}-1 =2\cdot 3^a + 3^b > 3^M\]](http://latex.artofproblemsolving.com/6/4/3/64302e08ed9615f86064e9e82d7b0f441c8635b6.png)
Thus,
for all 
, since 
for all 
and 
for all . This is a clear contradiction and hence, as claimed.
. Since we must have 
so,![\[-1 \equiv 2^{a+b+1}-1 = 2\cdot 3^a + 3^b\]](http://latex.artofproblemsolving.com/6/c/d/6cdef7f19f0d969a2ecdd3bef735b7d4025077ce.png)
Now, if were odd , 
or 
which is clearly impossible. Thus, must be even and since is odd, is in turn odd. Thus, we must have or 
.. Then, the equation reduces to![\[2^{b+2}-1=6+3^b > 2^{b+2} > 2^{b+2}-1\]](http://latex.artofproblemsolving.com/1/d/5/1d5c6cbbb5523d8c1e292e22a8c352c2ac212a03.png)
for all 
. Thus, we must have . This implies that .. Then, the equation reduces to![\[2^{a+3}-1 = 2\cdot 3^a+9 > 2^{a+3} > 2^{a+3}-1\]](http://latex.artofproblemsolving.com/5/3/1/5313080cb4d75c712a78c55a394786b9ceb165a0.png)
for all 
. Thus, we must have or 
. This implies that or . or , which we shall now solve separately. When ,![\[(m-6)(m-9)=m^2 - 15m+54=0\]](http://latex.artofproblemsolving.com/4/4/6/4467a2431f2ec82a9673363cc54e2618ea9ebf7d.png)
to which the only possible solutions are 
and 
. When ,![\[(m-9)(m-54)=m^2-63m+486 = 0\]](http://latex.artofproblemsolving.com/1/b/a/1bad08566f1c8f661bfa93c951abf94404fce900.png)
to which the only solutions are and 
. Thus, all solutions to the original equation must take the claimed forms, and we are done.
and 
Prove that
Let 
Prove that
and 
.![$\sqrt[3]{ab-1}+\sqrt[3]{bc-1}+\sqrt[3]{ca-1} \le \sqrt[3]{(a+b+c)^2}$](http://latex.artofproblemsolving.com/9/6/3/9633b5849794285e44519c745c1f616eceb77711.png)
give no solutions. Also, 
for 
& 
.
do not yield solutions. Hence, 
where 
Clearly, 
Hence, 
is odd 
.
& 
,
If 
.
& 
.
Taking, 
.
Hence no solutions! If 
Clearly, 
are odd. Again,
And similarly, checking the remainders we again get no solutions! When 
,
If 
,
.
Again taking 
,
for all nonnegative integers 
.
,
,
.
.
for all 
which suffices. We compute![\[g'(x) = 8\ln 3\cdot 3^{x-2}-2x-2>8\cdot 3^{x-2}-2x-2\]](http://latex.artofproblemsolving.com/2/7/0/270aec164f9d2d7293423c14b84d89e6efef1fb8.png)
If 
,
,
is increasing for 
.
,
for all positive integers 
,![\[-(n+1)2^{n+2}+2n+2+3(n+1)^2+8\cdot 3^{n-1} = 8t(n) + 3(n+1)^2+2n+2\]](http://latex.artofproblemsolving.com/7/d/2/7d2e7dda73d5b6a94a5990fc2bbedfaf2321d529.png)
We can compute that 
as the base of our induction. Now if 
,
as well:![\[t(k+1) = 3^{k} - (k+2)2^{k} > 3(k+1)2^{k-1} - (k+2)2^{k}= 2^{k-1} (3k+3 - 2k-4) = (k-1)2^{k-1} > 0\]](http://latex.artofproblemsolving.com/7/c/5/7c58fbc34ec14a395d3554157f875b4169f8f880.png)
as required, so the induction step is complete and so 
for all ![\[(n+1)2^{n+2}-2n-2-3(n+1)^2-8\cdot 3^{n-1}<0\quad \forall n\geq 6\]](http://latex.artofproblemsolving.com/6/2/7/6274aafb8a754f194c561e6a605b933ee41102a3.png)
.
.
be the number such that 
.
.![\[3^k + 2 \cdot 3^{n-k} = 2^{n+1}+1.\]](http://latex.artofproblemsolving.com/e/8/3/e839b1b5efe61bc3119d369a9f25e03a69dcd58c.png)
are greater than 
.