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 non-negative real numbers fulfilling ![\[\sum_{k=1}^na_k^2+{\left(n^2-3n+1\right)}\prod_{k=1}^na_k\geqslant{\left(n-1\right)}^2\text.\]](http://latex.artofproblemsolving.com/b/0/4/b04e7a65130d8a03a82014fcac1c9b31dfb7da54.png)
Prove or disprove: ![\[\frac1{n-1}\sum_{1\leqslant i<j\leqslant n}{\left(a_i-a_j\right)}^2\geqslant{\left({\left(n^2-2n-1\right)}\sum_{k=1}^na_k-n{\left(n-1\right)}{\left(n-3\right)}\right)}{\left(n-\sum_{k=1}^na_k\right)}\textnormal.\]](http://latex.artofproblemsolving.com/6/a/8/6a84740ee93278e04b315d45ad5079e69aeac3c1.png)
be a bounded sequence satisfying![\[a_n < \displaystyle\sum_{k=a}^{2n+2006} \frac{a_k}{k+1} + \frac{1}{2n+2007} \quad \forall \quad n = 1, 2, 3, \ldots \]](http://latex.artofproblemsolving.com/9/5/7/957789ae4242b934e9001abd1f330bd2168a745c.png)
![\[a_n < \frac{1}{n} \quad \forall \quad n = 1, 2, 3, \ldots\]](http://latex.artofproblemsolving.com/8/d/6/8d676053328fae8f562edf2fc7862d8059d76d5d.png)
be an infinite set of points on a plane. Any 3 points in can be covered by a strip region with a width of 
. Prove that can be covered by a strip region with a width of 
.
lies inside triangle 
such that 
and 
. Point 
is the midpoint of segment 
. Point 
lies on segment 
with 
. Prove that 
.
be an acute triangle. Let 
be a circle whose centre 
lies on the side . Suppose that is tangent to 
at 
and at 
. Suppose also that the circumcentre 
of triangle lies on the shorter arc 
of . Prove that the circumcircle of and meet at two points.
be a triangle. Let 
, 
, 
be the midpoints of its sides 
, 
, 
, and 
, 
, 
the midpoints of the altitudes from 
, 
, 
. Show that the lines 
, 
, and 
meet at one point.
is the circumcircle of 
,
is the orthocenter of is an arbitrary point on is the reflection point of wrt ,
meet 
at .
is the reflection point of 
wrt 
.
is a point on such that 
is parallel to ,
is a point on such that 
is parallel to .
is the circumcenter of 
is a pair of isogonal lines wrt 
be a triangle with incenter 
, and suppose the incircle is tangent to 
and at and . Denote by 
and the reflections of and over . Let be the intersection of with 
, and let 
be the midpoint of . Prove that 
and 
are perpendicular.
be an acute triangle. Denote by 
the foot of the perpendicular line drawn from the point 
to the side 
, by the midpoint of , and by 
the orthocenter of . Let 
be the point of intersection of the circumcircle 
of the triangle and the half line 
, and 
be the point of intersection (other than ) of the line 
and the circle . Prove that 
must hold.
the length of the line segment .)
. Let be the circumcenter of this triangle . Let , , be the feet of the altitudes of triangle from the vertices , 
, 
, respectively. Denote by 
, 
, 
the midpoints of these altitudes 
, 
, 
, respectively. The incircle of triangle has center and touches the sides , , at the points , , , respectively. Prove that the four lines 
, 
, 
and 
are concurrent. (When the point concides with , we consider the line as an arbitrary line passing through .)
be a circle and 
be circles internally tangent to at and . Assume that 
and 
are also externally tangent at point 
. Prove that the line through perpendicular to 
meets line 
on .
and are internally tangent at point . Chord of is tangent to at , where is the midpoint of . Another circle, 
is tangent to 
and at 
and respectively. Rays 
and meet at . If is the midpoint of major arc , show that 
., points 
lie on sides 
, respectively, such that 
. Altitudes 
meet at orthocenter . Points and lie on segment 
such that 
and 
. Lines 
and 
intersect at point 
. Compute 
.
of prime numbers which 
and
is an integer.
and 
, so 
and 
. Then we have in the usual way that 
.
. In that case 
and 
are even. Let 
, meaning 
is even and 
divides the expression above. First, 
and since 
, it follows 
.
divides 
. However by cyclotomic arguments, all prime factors of are 
or 
.
and 
are in 
. Hence 
as desired. 
and 
cases.
, we arrive at ![\[ a^b b^a - 1 \mid 4(a+b) (a^2+b^2). \]](http://latex.artofproblemsolving.com/d/7/f/d7f8b95ec306474bd81d430fbb036942b1c03d7c.png)
If 
then 
, 
and so 
which is a contradiction. On the other hand 
is evidently a solution. we get ![\[ a^b b^a - 1 \mid 6(a+b)(a^4+a^2b^2+b^4) \]](http://latex.artofproblemsolving.com/7/2/e/72e911cca4c778e33cd66efefbcfc6b59504c4b9.png)
If 
then 
, 
and 
which is a contradiction. On the other hand neither 
nor 
are solutions, since:
then 
, which is divisible by 
. But 
is not divisible by 
.
then 
, which is divisible by 
(dividing 
). But 
, which is not divisible by .
which visibly works.
factors as 
." right?" "No one cares [insert person name]".
, then divides 
. If the problem condition holds then it also divides 
. Subtracting we find that
be any prime dividing . Then 
and 
. Clearly is relatively prime to both 
and 
because it divides . Then the previous congruence implies that
divides 
. We assume for the moment that and deal with this case at the end. Assume first that the order is not divisible by . Then it divides and therefore
, 
or 
. The first case is impossible because implies that and are even and is odd. Now, if then 
implies that
divides 
. Because it is at most 
it follows that the order divides and 
which is impossible as 
. then similarly we find that 
and the order of 
must divide , so 
and 
for some positive integer 
and some choice of sign. Because both 
and are odd, it follows that is even. Now recall that must divide 
. If 
then the exponent of in is greater than , and 
, contradicting that must divide this expression. Thus 
and the only possibility is that 
. We now deal with these two cases.
.
must divide 
, which implies that 
. However this is false for every because
.
divides 
. We must now have 
, so we must have 
. The inequality now reduces to
then 
and we must have 
which gives 
, a contradiction. If 
then 
which we easiy can prove doesn't satisfy the condition.
must be divisible by , implying that divides 
, and therefore every prime divisor of is congruent to modulo . Now, we have
because all of their prime divisors is, but this is impossible because they differ by and is not .
then 
must divide 
. This is impossible for size reasons if 
. If then 
divides 
which once again doesn't hold for size reasons. Finally if 
we find that
![$(p+q)^{p-q}(p-q)^{p+q}-1\mid [(p-q)(p+q)]^{p-q}[(p+q)^{2q}-(p-q)^{2q}]$](http://latex.artofproblemsolving.com/7/6/6/76607b104b6781a8e39614b81dc2e98b40b2d7f1.png)
. Now, if is a prime and divides 
, it cannot divide the LHS, so we get 
Thus, as a preliminary bound we get 
or 
. From here on, assume 
, so both 
are odd. is a prime that divides the LHS. Then divides the RHS and is relatively prime to , so we get 
. Now, if 
, we get 
, so 
. But cannot divide the LHS, so we are limited to 
or 
. Suppose ; then 
or 
. Thus 
, but this contradicts . Thus, the only possibility is . Otherwise, 
. Thus, every prime factor of the LHS is either or 
.
, then the LHS can be written as 
. On the other hand, every factor of is either 
or . Thus, 
. The only way this can happen for is if and 
. Now, since we merely need to test 
, or 
. It is easy to check these cases fail. This exhausts all the cases for when is odd., we need 
or 
. The only value of which works is 
, and we get the ordered pair 
.
.
as in the beginning of the first solution and also we do case work.
of and do some casework (excluding cases 
and 
), after what we get that 
. if of the form . But there is a brute force finish in this case. Let's look. has remainder modulo 
, so itself has remainder modulo .
means 
or 
, so 
with an integer 
.
means 
, hence 
, or, equivalently, 
. we get 
, so 
. is order of modulo then 
, so the only thing left to do is to check primes 
.
for now. Let 
. Then 
, so 
. But 
, so 
. Suppose a prime divides 
. Note that is odd since 
are even. Then 
, so the order of 
mod divides , so it is in the set 
.
, i.e. 
, so 
. However, if , then![\[ 0\equiv (p+q)^{p-q}(p-q)^{p+q}-1\equiv q^{p-q}\cdot (-q)^{p+q}-1 \equiv q^{2p}-1 \equiv q^2-1 \pmod{p} \]](http://latex.artofproblemsolving.com/5/d/6/5d6aa734fc67ee0016a78a93a72f5510e53ae98d.png)
which is not possible, since . If , then![\[ 0\equiv (p+q)^{p-q}(p-q)^{p+q}-1 \equiv p^{p-q}\cdot p^{p+q}-1 \equiv p^{2p}-1\pmod{q}.\]](http://latex.artofproblemsolving.com/0/7/d/07d01e14c4dce77520464e33a5251ae1277d35e6.png)
Now the order of divides , so it is in the set 
. It also divides 
, which is a contradiction if the order is or , which means the order of 
is 1 or 2. But then 
, so 
, contradiction. mod is or . But it also divides , so 
. Hence, 
, so every prime factor of is or 1 mod . But are even, so 
for some . So 
, which means every prime factor of 
are 0,1 mod , and in particular are both 0 or 1 mod . This is impossible unless or .
. A check on these both gives that the only solution is .
and 
, it is equivalent to note that ![\[\frac1{(p+q)^{p-q}(p-q)^{p-q}}\left(\frac{(p+q)^{p+q}(p-q)^{p-q}-1}{(p+q)^{p-q}(p-q)^{p+q}-1}-1\right)=\frac{(p+q)^{2q}-(p-q)^{2q}}{(p+q)^{p-q}(p-q)^{p+q}-1}\]](http://latex.artofproblemsolving.com/4/9/3/49308ab2da916736a00e9c9c494f98f73614cc8c.png)
is an integer.
. We can easily verify this is the only solution for 
by applying the following claim and finite-case checking:
divides 
. But if 
, then ![\[(p+q)^{p-q}(p-q)^{p+q}-1\ge(p+q)^{2p-2q}-1\ge(p+q)^{2q}-(p-q)^{2q},\]](http://latex.artofproblemsolving.com/c/2/6/c26aea1ecb0a8d9cfb8321851646647e334e7943.png)
absurd. 
. Then and are even, so define ![\[s:=(p+q)^{(p-q)/2}(p-q)^{(p+q)/2}-1.\]](http://latex.artofproblemsolving.com/d/9/b/d9bc4143f9f30828c47023fba02acdd1fc034a3f.png)
Let 
be a prime dividing 
, so divides 
.
or 
divides 
. divides , so ![\[\left(\frac{p+q}{p-q}\right)^{2q}\equiv1\pmod r.\]](http://latex.artofproblemsolving.com/1/c/a/1ca17a55ee7defee17130a227553992a5325dd82.png)
It follows that 
.
, then 
.
, and so ![\[r\mid\left(\frac{p+q}{p-q}\right)^2-1\mid(p+q)^2-(p-q)^2=4pq.\]](http://latex.artofproblemsolving.com/c/b/6/cb646df24cca9ef66099d528b57158603392d72b.png)
is odd, 
.
, then ![\[0\equiv s^2-1\equiv q^{2p}-1\equiv q^2-1\pmod p,\]](http://latex.artofproblemsolving.com/e/b/6/eb62f8b4527655aa41e385f22a612eb1323817aa.png)
so 
divides either 
or 
. This is absurd by 
..
and 
are either 0 or 1 modulo . This is impossible under our assumption that .
of prime numbers which andis an integer.
Claim 01. 
.
, and by divisibility criteria, we have![\[ (p + q)^{p - q} - 1 \le (p + q)^{p - q}(p - q)^{p + q} - 1 \le (p + q)^{2q} - (p - q)^{2q} \le (p + q)^{2q} - 1 \]](http://latex.artofproblemsolving.com/8/a/4/8a46d0b0777dba2e38fd0fb8e886c19bac8acffa.png)
Therefore, we conclude that 
, but since and are both primes, we conclude that .
, then the only solution is 
.
, so we just need to check and 
. Notice that obviously satisfies, and therefore is a solution.
are both odd, and therefore and are both even.
, then either or 
.![\[ r \mid (p + q)^{p - q} (p - q)^{p + q} - 1 \mid (p + q)^{2q} - (p - q)^{2q} \]](http://latex.artofproblemsolving.com/e/8/f/e8f131114775bee9407056ea14c504bbd38ace02.png)
Then, we have 
. Suppose that 
, then we have ![\[ r \mid (p + q)^2 - (p - q)^2 = 4pq \]](http://latex.artofproblemsolving.com/f/7/3/f738e3203716731970215fe1b1946567f8e2afb2.png)
For obvious reason, 
since 
is even. Now, we have or ., then![\[ p \mid (p + q)^{p - q} (p - q)^{p + q} - 1 \Rightarrow p \mid q^{2p} - 1 \]](http://latex.artofproblemsolving.com/7/5/1/7519112e7198a6611e564667d61c26e84df8788d.png)
By Fermat Little Theorem, 
, and therefore 
. However, we know that 
and 
. This is a contradiction.
such that![\[ r \mid (p + q)^{p - q} (p - q)^{p + q} - 1 \]](http://latex.artofproblemsolving.com/c/2/9/c292ad7870feb5276b12d3f49be295c173e060fd.png)
then 
.
, which is an integer since and are both even. Therefore,![\[ r \mid N^2 - 1 \Rightarrow r \mid (N - 1)(N + 1) \]](http://latex.artofproblemsolving.com/d/1/0/d106e366cf06c056554b382595d6ddae50866e37.png)
Now, we conclude that 
. Now, notice that 
. This pretty much gives us . Since we have 
, this forces .
, which means we just need to check 
and 
, both of which don't work. is either or modulo , then must be either or modulo . Once you realized that 
for some 
, we are pretty much done.
, which obviously works.
are both relatively prime with the denominator. As a result,
This implies
In particular, we have 
, therefore, 
, so if then the only solution is 
. Now supppose is odd.
.
, then from 
we have
Let 
, then
Notice that 
, therefore , from which we easily check that there is no solution. 
.
. Notice that the 
, therefore, it can only be . This implies
, the equation becomes
hence 
, but by easy induction we have 
for 
, contradiction.
is a prime such that 
then 
is odd, 
.
by assumption, moreover,
Let the order of 
modulo be 
, notice that cannot be or otherwise 
which implies 
, contradiction.
, hence 
as desired. 
are both even, let 
, 
. By LTE,
Therefore, if 
, then we have 
, hence since 
. Otherwise, one of is congruent to mod while the other is congruent to mod , hence 
or 
, the former is impossible and the latter is rejected in CASE I.
. Notice that the problem is equivalent to 
due to the fact that 
. And this implies that the order of the element 
is either 
modulo 
. Now, we have 
, which means that .
or 
. We now have that 
is the only possibility. Hence, the lemma follows.
. Let be a prime dividing . If the order is or , we have that 
, and hence is either 
. If we have that 
, so we just need to try the values. From now on, we assume 
, and hence is odd, ,so is either . can't be because and can't be because by the claim. or , we have that 
, which implies that every prime divisor of is of the form 
,which implies 
. But notice that is of the form 
for some 
, and hence 
, which is not possible in this case., which comes from and trying .
, which works. be an arbitrary prime divisor of . Then, we must have 
. Thus, 
If 
is or , then 
, so 
. Therefore, either is even, , or . Otherwise, 
, so 
. We have three cases.. Then, 
, so ![\[(p+2)^{p-2}(p-2)^{p+2}-1 \mid (p+2)^{p+2}(p-2)^{p-2}-(p+2)^{p-2}(p-2)^{p+2}.\]](http://latex.artofproblemsolving.com/7/8/1/781a5920bc3bb698178dabae22f5932a3292e146.png)
Thus, we have 
If , then we have a contradiction because of size. We obtain that gives the solution and 
fails due to size reasons. Now, assume henceforth that and are even. is divisible by . Then, we have 
Since and are even, we have 
, so 
. Thus, 
, so 
or 
, which fails because . is divisible by . If is divisible by , then, we have 
, so 
. Since , we see that 
. Thus, 
, so 
or 
. If , the prime factorization of must consist of 
primes. Thus, we 
. Since 
and both factors cannot be 
, we are done in this case. Otherwise . We can do a similar Euclidean algorithm as in Case 1 to obtain that . Now, we are done by bounding.
.
so 
are eve, if we substract from the integer and remove the factors (we can do this because are coprimes with the denominator). We have that if then:The key idea is the following: then 
Then 
is divisor of but also is divisor of . But if is or then is or then: 
But is odd so or . then: 
So by FLT: 
then 
but and are odd so 
(Contradiction!) or 
.
is 
but have divisors: 
Then exists two divisors of with difference and both are so the difference is 
then the unique possible value of is .
So we have: 
. If 
if 
obvious this is false.
then n the problem statement we replace: 
that is false. then if 
is false. that works.
.
are both odd primes. Then, let some prime 
. Then, using the divisibility condition and dividing, we have 
Thus, either 
or 
If , then 
hence 
, or . But this is obviously impossible as are both odd primes.
, which is impossible, or . This crucially means that 
is the product of and 
primes. On the other hand, 
so each of these factors is or mod . But this clearly implies ., we can reduce to a finite case check by showing that 
. However, this is evident by size reasons, as otherwise 
When 
, the former expression is always greater, but this contradicts divisibility., testing cases and using the previous conclusion, we obtain 
is the only solution.
and 
. We have
Thus, 
. Let be the smallest prime dividing then 
and 
.
. Clearly, . If 
then 
. If 
then 
. In the first case, 
, so . In the second, . is clearly impossible since ![\[a^bb^a-1\equiv q^{p-q}q^{p+q}-1\equiv q^{2p}-1\equiv q^2-1\pmod p\]](http://latex.artofproblemsolving.com/e/c/f/ecf19778dea62a34831ffe4276baa55fb9be4620.png)
and 
, 
are both coprime to since is a larger prime than . If 
or then . Thus, the only prime factors of are or . Thus, any factor of is also or 
. This is absurd, since is one less than a perfect square and thus factors as 
for some , and one of them must break the or rule.. We have and are even, and so ![\[a^bb^a-1\mid a^6-b^6\implies a^bb^a-1\mid (a/2)^6-(b/2)^6\]](http://latex.artofproblemsolving.com/e/9/4/e94082197277d8be75cb756cb626511957ab1f62.png)
If 
then 
so surely 
which implies 
, absurd. Thus, 
or . It can be verified that neither works.. We have and are both odd. We also have ![\[a^bb^a-1\mid a^4-b^4\]](http://latex.artofproblemsolving.com/3/3/2/3320670c93ef1a9e9c212d8cb4e9368d2c5081cd.png)
If 
then 
so 
and so , again absurd. Thus, is or 
, and it can easily be verified that is the only solution.
and . Observe that 
, from which the divisibility is equivalent to ![\[ (a^bb^a-1) \mid a^{2q}-b^{2q}. \]](http://latex.artofproblemsolving.com/e/0/0/e00c2597d767a752081dce7680271f7df9ddf328.png)
Let 
be an odd prime divisor of . Then ![\[ \left(\frac{a}{b}\right)^{2q} \equiv 1 \pmod{p_0}, \]](http://latex.artofproblemsolving.com/8/f/9/8f9cb718694e4a0e888993badf1ae1208c6bd9c4.png)
so the order of modulo is one of , , , and . We proceed to do casework on these possible orders.
, then 
. or , then in either case we have ![\[ (p+q)^2 \equiv (p-q)^2 \pmod{p_0}, \]](http://latex.artofproblemsolving.com/9/e/d/9edb90956490f039d9290a6602317bf0fd17b225.png)
from which 
or 
.
, then ![\[ q^{2p} \equiv 1 \pmod{p}, \]](http://latex.artofproblemsolving.com/6/1/8/6185dcaedd3eee7c1cdc1771b3379c6b0d07367d.png)
so 
or 
by FLT. Neither are possible since . Thus, each prime divisor of is one of and 
. In particular, 
is .![\[ p_0 \mid (a^{b/2}b^{a/2}-1)(a^{b/2}b^{a/2}+1) \]](http://latex.artofproblemsolving.com/0/9/2/092a0caa5f080c1fcda270bcb2fb6d565351f841.png)
and realize that must divide exactly one of the two factors. Since each factor is in 
, and their difference is , we must have or . To check which work along with these values of , we first bound .
.
from our initial discussion. Then show that the LHS is greater than the RHS when 
, so .
as the only solution.
factors as ." right?" "No one cares [insert person name]".
this also gives us the 
bound/ Let ) Every prime divisor of must be of form 
unless 
then get a contradiction.
to get the only solution 
is the only solution.
. Assume . Let 
. Then note that 
and 
, so 
.
and 
, 
, hence either 
or 
. Therefore, there exists a prime such that 
and 
. But we have 
, which means 
, so 
, or equivalently, 
, a contradiction.. We split the problem into 2 cases:.
. If , then 
, so 
, which forces and we can easily check that 
is a solution.., then 
, so 
, a contradiction. Hence equals or 
and one can check that 
and 
are not solutions. Hence we're done. 
. Note that 
is just a finite case check due to size issues, giving us our claimed solution. We now assume 
, and suppose is an arbitrary prime divisor of the denominator, which also must divide the numerator. We then require![\[(p+q)^{p+q} (p-q)^{p-q} \equiv (p+q)^{p-q} (p-q)^{p+q} \equiv 1 \pmod r\]](http://latex.artofproblemsolving.com/4/d/4/4d45d5a3fb0aca415b26fade10220b57a21e28e0.png)
![\[\implies \left(\frac{p+q}{p-q}\right)^{2q} \equiv 1 \pmod r \implies k := \operatorname{ord}_r \left(\frac{p+q}{p-q}\right) \mid \gcd(2q,r-1) \mid 2q.\]](http://latex.artofproblemsolving.com/2/a/1/2a1a1f0f8de4e83abb18075c8b9c9eef12c82436.png)
and both result in contradictions. Then
, contradiction. 
, which works. Now we prove everything else fails. Suppose isn't .
. Clearly 
.
. Hence![\[a^b b^a - 1 \mid a^a b^b - 1 - (a^b b^a - 1) = a^a b^b - a^b b^a\]](http://latex.artofproblemsolving.com/d/5/0/d501cebd77388707db49c1a5fe5dca769668fefb.png)
Since are relatively prime to 
, we can divide the RHS by 
, so![\[ a^b b^a - 1 \mid a^{a-b} - b^{a-b} = a^{2q} - b^{2q}\]](http://latex.artofproblemsolving.com/2/1/a/21a1e9d903cebeb49be9f5981f5ae1e3295cee89.png)
Since , we have that 
. Now since , 
, so 
, meaning 
. Hence 
, so .
. Then 
.
has some prime factor that is not . are . Then notice that are both even, so is a perfect square. Let 
, we have that 
, so 
and 
are , implying , contradiction. . be an odd prime dividing that is not . We have divides 
, so 
.
modulo divides . However, it can't be a multiple of as doesn't divide . Hence the order of modulo is either or . modulo is , then 
. But 
, so divides 
, meaning that . We have 
and 
, and both are even, so 
, so 
. But 
, so 
, meaning that 
. Since 
, this means 
, which is absurd since 
. Therefore, the order of modulo must be .
, so . Hence 
. However, we have 
, so 
. If the order of modulo is or , we have a contradiction as cannot divide 
(since 
). Since the order of mod divides , it must be or . Hence 
, so 
.
, 
. By parity, can't be 
or 
., then 
, 
, so 
divides 
.
must be less than 
, implying that 
.
absurd.. But then 
, so![\[(3q + 1)^{q+1} (q+1)^{3q + 1} - 1 \mid (3q + 1)^{2q} - (q+1)^{2q}\]](http://latex.artofproblemsolving.com/f/3/8/f387fef13e8f979a845d74638f5da32b57d9ec78.png)
Hence 
, so 
.
absurd.., then 
, so or 
. To see does not work, notice it implies 
and 
, and 
clearly does not divide 
(in fact it is much larger). Hence ., then 
, so 
. If , then 
, but 
doesn't divide 
(this can be proven by dividing the RHS by 
and getting a size contradiction). If 
, then 
, but 
doesn't divide 
(since 
). Hence is not possible.
, so we are done.
is integer. Suppose is a prime such that 
Hence, 
We obviously require 
due to size, or 
We also have the equivalent condition ![\[\left(\frac{p + q}{p - q}\right)^{2q} \equiv 1\pmod r.\]](http://latex.artofproblemsolving.com/b/5/3/b53fd17e091a111d754342340e48e66f06592b31.png)
then 
so then 
so 
then so then 
so 
This implies that 
Then, 
It isn't hard to verify that is a valid solution.
then 
are exactly or 
However, 
have distinct values modulo unless 
It is easy to verify 
by hand.
,![\[
\frac{1}{x_1} + \frac{x_1}{x_2} + \frac{x_1 x_2}{x_3} + \cdots + \frac{x_1 x_2 \cdots x_n}{x_{n+1}} \geq 4\left(1 - x_1 x_2 \cdots x_{n+1}\right).
\]](http://latex.artofproblemsolving.com/2/b/6/2b63775b306d0d8b59c7a34ddecf90f653877b79.png)
Thus we obtain the bound 
,
.
.
If 
,
so 
.
so 
.
so 
.
so 
.
.
so 
.
so 
implying 
,
,
Thus each factor must congruent to 
.
,
We have 
so 
,
,
Using some calculus, we can show that 
thus the left hand side is always bigger, this yields contradiction.
We will firstly resolve the following case.
.
On the other hand, one can check that for 
which is a contradiction.
We then have that 
meaning that 
Then we have that 
.
,
we therefore have two cases.
meaning 
,
and therefore 
,
Therefore 
,
as 
is not possible as 
meaning that 
,
or 
unless 
,
.
consequently 
and therefore 
If 
then 
.
,
,
meaning that 
is decreasing over 
.
and therefore 
,
.
which is a contradiction to the inequality derived prior to 
.
which means that 
Similar to the previous case, we have the following lemma.
for all 
Moreover 
and therefore we are left to prove that 
or in other words that 
which holds for all 
,
and also 
,
meaning that 
We therefore get that 
,
which is only possible if 
.
,
for some 
and let we have that 
for some 
,
that 
,
,
and 
meaning that 
,
.
,
,
that is not equal to 
for some 
,
we do not have solutions by Catalan's conjecture because 
are absurd due to RHS being small. Now, 
so 
.
then we have 
so 
,
and 
we have 
so 
a contradiction for 
.
so 
for an even integer 
and 
are two odd integers meaning that one is not 
,
.
due to 
being coprime to LHS. Notice that 
so 
and 
because 
since this implies 
,
Checking