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 an integer greater than zero. Then, the value of the sum of the reciprocals of the cubes of the roots of the equation![\[
mx^4 + 8x^3 - 139x^2 - 18x + 9 = 0
\]](http://latex.artofproblemsolving.com/a/3/a/a3a9bca412e927632e4921c51f2a6593409031d9.png)
is equal to:
and 
be on segment 
of an acute triangle 
such that 
and 
. Let 
and 
be the points on 
and 
, respectively, such that is the midpoint of 
and is the midpoint of 
. Prove that the intersection of 
and 
is on the circumference of triangle .
, then prove that the following inequality holds:![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
be triangle, inscribed in parabola. Tangents in points 
forms triangle . Prove that 
. (
is area of triangle 
).
such that the equality 
holds for all pairs of real numbers 
.
, in which 
: 
, but the proof is important.
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
be an acute triangle with circumcircle 
. Let 
be the midpoint of 
and let 
be the midpoint of 
. Let 
be the foot of the altitude from 
and let 
be the centroid of the triangle . Let 
be a circle through and that is tangent to the circle at a point 
. Prove that the points 
and 
are collinear. with integer coefficients such that ![$P\left(\sqrt[3]5+\sqrt[3]{25}\right)=5+\sqrt[3]5$](http://latex.artofproblemsolving.com/5/8/e/58ec2a921cd988bd5ae017a8653544ddd758225f.png)
.
be a 
-element subset of the set 
. Prove that there exist numbers 
, 
in 
such that the sets ![\[ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100 \]](http://latex.artofproblemsolving.com/d/5/c/d5c2a7b943e8c7200361698f3eb6f83700ad68d6.png)
are pairwise disjoint.
positive integers 
such that for every index 
, with 
, 
is a multiple of 
. Determine the greatest possible value of the largest of the numbers.
be a triangle. Let and be the points in which the median and the angle bisector, respectively, at meet the side . Let and be the points in which the perpendicular at to 
meets 
and 
, respectively. And 
the point in which the perpendicular at to meets produced.
is perpendicular to .
such that the numbers ![\[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \dots, (a_{2018})^{2018}+a_1 \]](http://latex.artofproblemsolving.com/6/1/a/61ad92b49bd9b4b8839e6f6c4e81c758a11835a3.png)
are all powers of 
). Propose a positive 
performing 
. By the presentation of , we perceive that 
proportionately partitions all parts of the progression. proportionately parts 
.
power of phive. Our party personally presents:
as preferred.
,
purporting 
. Provoking LTE, since 
, 
The partisanship proves 
, proving 
, a preposterous predicament, proving the proposition. 
for some 
. My claim is that do not devide any 
. devides some of .Also assume 
.
Is not a perfect power of which is a contradiction !!
.![\[\Rightarrow a_1^{2018^2} +a_3 \equiv a_{2}^{2018}+a_3 \pmod{5}\]](http://latex.artofproblemsolving.com/6/7/8/678562f143317850a039fbd656121e5fa851bfc2.png)
![\[ \Rightarrow a_1^{2018^{2018}} +a_1 = a_{2018}^{2018}+a_1\equiv 0 \pmod{5}\]](http://latex.artofproblemsolving.com/5/7/2/572ac277302871d81c04af696fd20e9b3e48c66d.png)
do not devide 
hence ![\[ 5\mid a_1^{2018^{2018}-1} +1\]](http://latex.artofproblemsolving.com/c/8/b/c8bc717ca63b4562c0d2024dbd6390ef4eb42e76.png)
which forces ![\[5\mid a_1^{2.2018^{2018} -2} -1\]](http://latex.artofproblemsolving.com/2/d/0/2d04ef0a624adef4cf4bb1c93ff938a2a1ee72d0.png)
it means 
. Clearly It is a contradiction.
. Note that if 
for any , we have an immediate contradiction, as then 
will be divisible by a prime other than . Hence, 
for all . Iterating this yields ![\[ \nu_5(a_1) = 2018\nu_5(a_{2018}) = 2018^2\nu_5(a_{2017}) = \cdots = 2018^{2018}\nu_5(a_1), \]](http://latex.artofproblemsolving.com/d/a/e/dae690c335bde48a9f33113dfbff67a08c771804.png)
so 
, implying that 
for all .
. Iterating this yields
Since 
, we have 
, so 
. This is enough to imply that 
for all .
is minimal. Then divides all of the other terms, so
Iterating this yields
where dividing by makes sense because 
(if 
, then would be divisible by a prime other than , as 
). Recalling that 
, we have from LTE that
However, 
, so 
. Therefore, since is a power of , we have
Hence, we must have 
, but 
and 
(since 
). This is a contradiction, so we are done. 
.
is maximal and positive, then 
, so cannot be a power of . 
denote the smallest of the powers of , so all elements are divisible by . Evidently, 
, and for every index , ![\[ a_{i+2} \equiv - a_i^{2018^2} \pmod{5^e} \]](http://latex.artofproblemsolving.com/7/7/8/77850a6652a997353c584a317c45684b35e5fed3.png)
for all .
. But if we iterate this 
times we obtain ![\[ x \equiv x^{2018^{2018}} \pmod{5^e} \]](http://latex.artofproblemsolving.com/d/2/f/d2fd0de1f66c3570c2bcbb885a26eb314fcdbf75.png)
Thus 
has order dividing 
. But since 
, we find 
, So this forces 
.
, and the same is true for any other term. But 
and we have a size contradiction.
for all .
be maximal. It is evident that , a contradiction.
and as is not divisible by , we must have 
for all . Now suppose 
is minimal, so 
Now we are almost done; note that 
so 
which is obviously false.
such that the numbers ![\[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \cdots, (a_{2018})^{2018}+a_1 \]](http://latex.artofproblemsolving.com/b/a/4/ba4ac7fb2a62792bf76b5914c85994e6111069ff.png)
are all powers of does not divide for any divides one of it divides all.Now consider the number 
where 
is chosen so that 
.
cant be a power of since 
which concludes the claim.
Substituting repeatedly we get 
.Thus 
.Thus 
.But this is a contradiction.
for all .
and 
, if 
is a power of , then 
. Therefore, since 
is a power of for each , we have 
. Iterating this, we have 
since the sequence cycles back, which means 
, and hence 
for all . Therefore, 
for all , as claimed. 
be the smallest power of in the list, WLOG it is 
. (We can WLOG since the variables are cyclically symmetric.) Then all of the elements of the list are multiples of . We have for all (indices cycle) that 
, so![\[-a_{i+1}\equiv (-a_i)^{2018} \pmod{5^m}.\]](http://latex.artofproblemsolving.com/9/9/2/9926d23e3d013808b5cef25106db0a5362fb62b8.png)
Iterating this, we have 
. Since 
, we have![\[(-a_1)^{2018^{2018}-1} \equiv 1 \pmod{5^m}. \qquad (\clubsuit) \]](http://latex.artofproblemsolving.com/c/f/9/cf93d8cf864be5045c475ea053bf882758aa5f95.png)
In particular, said equivalence also holds in 
. Since 
, this implies 
, so 
. Now, 
implies![\[ m\le \nu_5\left( a_1^{2018^{2018}-1}+1\right) = \nu_5(a_1+1) + \nu_5(2018^{2018}-1) = \nu_5(a_1+1)\]](http://latex.artofproblemsolving.com/2/9/8/298c001fa7c58fbe6501024b54e212427307119a.png)
by LTE. Therefore, 
, but 
, a size contradiction.
such that the numbers are all powers of 
we have 
then all the terms of sequence 
is divisible by . Then we will Prove that this is never possible. For it let's start with 
then observe as 
Similarly we can get 
and so on. But then 
will be the clear contradiction.
is the smallest possible power of in this then we will have 
and as 
So we must have 
now this is enough to prove the same for all 
and Hence we will get Size Contradiction by LTE. So we are done there doesn't exist such sequence of natural numbers. 
Since , we have , so . This is enough to imply that for all .
means ?
Since , we have , so . This is enough to imply that for all . means ?
we get 
.
. Let 
be the index such that 
.![\[ a_{i_0}^{2018} + a_{i_0+1} = 5^{k_{i_0}}, \]](http://latex.artofproblemsolving.com/d/f/4/df496af200a5a24bdb1361e4fa05dc3a1356e72d.png)
we know 
, thus 
, but because of size reasons this is a clear contradiction.
. Perform a change of variables 
.
such that 
then we are done because of size reasons. So, 
, which means all of them are equal, making another change of variables 
as otherwise we are done because of size reasons. Thus, ![\[c_2 \equiv -2c_1, c_3 \equiv 4c_1, \cdots, c_{2018} \equiv (-2)^{2017}c_1 \implies c_1 \equiv 2^{2018}c_1 \pmod{5}.\]](http://latex.artofproblemsolving.com/7/9/b/79b1b5c65b21ce878954ce2d099e4a6588223ded.png)
From which we get 
, a contradiction.
. Now if one of the is divisible by , then so are the rest, and if WLOG 
we have a contradiction with 
being a power of 
none of the 
's is divisible by .
is the smallest out of all of those powers of ,so it divides all of the other numbers. We get that 
thus 
. Continuing in this fashion by induction we can get that 
and thus 
so 
. If 
, then 
, so either way 
. This however is impossible if 
because 
, so we must have 
. But then 
, which isn't possible, so the system has indeed no solutions!
and now note that by FLT we have 
hence since 
is 
on mod 5 we get that all 
are on mod 5.
multiple of . be the one with 
maximun compared with the others hence let 
and now using 
we have that 
which is not possible since 
. such that 
and note that its a recursion and in some moment we will get 
which is not possible are 
in mod 5.
where 
is the smallest power of between 
. It suffices to show that 
and now note that by the conditions.
Hence proved!.
Setting 
we get that 
but that means 
is a power of which is a contradiction!! exists! (thus we are done 
)
should be divisible by , otherwise we 
is not a power of where 
. Then using FLT we get
So each is modulo . Write 
. Let 
, with 
. Write each 
. Then 
. By definition not all 
are divisible by . Now note each power of is at least 
(as 
). Hence,
Using 
repeatedly gives
But this means divides every , contradiction. 
, we have two cases:
so 
then 
this obvious is contradiction because 
so 
isn't a power of 
with 
. We rewrite 
is a power of 5 then by LTE 
so 
(otherwise the contradiction is the same like the case 1) but: 
If we divide all original expressions by 
we have that the new expressions are multiples of , but: 
with 
, this implies the problem. 
such that the numbers are all powers of we see 
for all (Just assuming randomly,you can see that choosing any will work).Since we get 
,so repeating this process on other terms continuously gives 
, which is impossible.
Practicing this we get 
,then 
and so on.....finally we derive 
which immediately implies 
so we get that 
for all . as the minnimum of all possible .So 
for all .
Thus we get 
But 
,thus there doesnt exist such numbers 
For someone interested in how this works,
such that the numbers are all powers of does not divide for any divides one of it divides all.Now consider the number where is chosen so that . cant be a power of since which concludes the claim.Substituting repeatedly we get .Thus .Thus .But this is a contradiction..
such that the numbers are all powers of does not divide for any divides one of it divides all.Now consider the number where is chosen so that . cant be a power of since which concludes the claim.Substituting repeatedly we get .Thus .Thus .But this is a contradiction..
is not a power of if there exists 
such that 
. Suppose otherwise; we first narrow down the value of 
.
for any . for any then for all . Thus pick 
such that 
is minimal. We have
but then 
cannot be a power of . is even we have 
for all , which means that we must have 
. But then this means 
, so 
for all .
be the least power of present in the sequence, and WLOG let . We have
so 
as 
. Since 
, by Lifting the Exponent
but since we must have 
(otherwise which is impossible), it follows that is not a power of —contradiction. 
.
. Since it is the minimum, we have 
![$5\nmid a_i\forall i\in [2018]$](http://latex.artofproblemsolving.com/e/9/0/e90f074e19568f67968692cb796eacc35ec4db6a.png)
be the lowest. Then 
Which is not possible.![$$a_1((a_1)^{2018^{2018}-1}+1)\equiv 0\mod 5^{\alpha}\implies a_1\equiv 4\mod 5\implies a_i\equiv 4\mod 5\forall i\in [2018].$$](http://latex.artofproblemsolving.com/b/b/9/bb941ada4dd337e634a86c162fa1d7851f84a24b.png)
is a power of , we get that 
But 
as 
.
such that the numbers are all powers of 
for all 
.
be the lowest power of . We have
Let 
. Then 
and 
imply 
or 
.
for all , which is a contradiction since 
.
claims. are divisible by . then all should be. Hence say 
is the highest then 
which is a size contradiction. 
.
in 
is either 
; 
; 
; 
and only the last one works as it is the only one which is cyclic. 
is the smallest number in the sequence then see that in 
we have ![\[a_1^{2018}= -a_2 \implies a_1^{2018^2} = a_2^{2018} = -a_3 \implies \text{ } \dots \text{ } \implies a_1^{2018^{2018}} = a_{2018}^{2018} = -a_1\]](http://latex.artofproblemsolving.com/6/1/e/61eb12506a259355b236f650e50f5d12da1523cb.png)
And so we have ![\[t \le \nu_5 \left(a_1^{2018^{2018}-1}+1 \right) \overset{\text{LTE}}= \nu_5(a_1+1)+\nu_5(2018^{2018}-1)=\nu_5(a_1+1)\]](http://latex.artofproblemsolving.com/2/d/b/2dbcd72661f8bbacabeeac56e39670f127511c56.png)
And so we have ![\[a_1^{2018}+a_2=5^t \le 5^{\nu_5(a_1+1)} \le a_1+1\]](http://latex.artofproblemsolving.com/6/f/9/6f95df4cd4a7827c1d62a5b73a9db544132786c8.png)
which is a size contradiction (as 
).
such that this implies that 
since is a power of 5 (and strictly greater than 1). Induction then shows that 
. However, letting 
for some 
we note,![\[\nu_5(a_M^{2018}+a_{M+1}) = \nu_5(a_{M+1})\]](http://latex.artofproblemsolving.com/6/6/8/6687064bdd21c2c89805a5a27488831dfc58b5af.png)
since 
. Thus,![\[5^{\nu_5(a_{M+1})} = a_M^{2018}+a_{M+1} > a_{M+1} \ge 5^{\nu_5(a_{M+1})}\]](http://latex.artofproblemsolving.com/b/e/e/bee4034d901482e3c586aa3340627730aafca7bb.png)
which is a clear contradiction. Thus, if there exists a set of positive integers satisfying the desired condition, none of them may be divisible by 5.
we have that,![\[a_3 \equiv -a_2^2 \equiv -a_1^4 \equiv -1 \pmod{5} \]](http://latex.artofproblemsolving.com/a/c/5/ac58de2293c2a5def2f60578695e494ac2a805ab.png)
Further,![\[a_4 \equiv -a_3^2 \equiv -1 \pmod{5}\]](http://latex.artofproblemsolving.com/d/6/4/d645fe5ecae0502177db4fad8213fe69790d2159.png)
and continuing similarly we have that for all . Now, let 
be the minimum power of five among the given expressions. Then, 
for all where indices are consider 
. WLOG assume that 
and note,![\[a_1 \equiv -a_{2018}^{2018} \equiv -a_{2017}^{2018^2} \equiv \cdots \equiv - a_2^{2018^{2017}} \equiv -a_1^{2018^{2018}} \pmod{5^e}\]](http://latex.artofproblemsolving.com/8/4/5/84500330bda30df1698ed578726e3fe3e329df84.png)
Thus, by the Lifting the Exponent Lemma we have,![\[e \le \nu_5(a_1^{2018^{2018}}+a_1) = \nu_5(a_1) + \nu_5(a_1^{2018^{2018}-1}+1) = \nu_5(a_1)+\nu_5(a_1+1)+\nu_5(2018^{2018}-1)=\nu_5(a_1+1)\]](http://latex.artofproblemsolving.com/4/c/3/4c3938dbf23761e05c55b1ff73f5cc090ac6603e.png)
So 
. However,![\[5^e=a_1^{2018}+a_2 > a_1^{2018} \ge (5^e-1)^{2018}>5^e\]](http://latex.artofproblemsolving.com/b/1/d/b1d546034ec1e5b011286e8289b76be81546ffae.png)
for all which is clear contradiction. Thus, there indeed do not exist such positive integers 
as desired.
is an obvious solution. Now, let 
.
or 
.
.
integers with these as powers of ![\[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \cdots, (a_{2019})^{2018}+a_1 \]](http://latex.artofproblemsolving.com/1/8/f/18f27b4c93ed5abbe6c0f9f96fc8021059bc3411.png)
. Tumon2001 found it too low lavil for his taste so the typo was fixed
denote the highest exponent of 
for all 
.
Then 
and hence 
,
.
for 
.
Let 
Then since all 
'
for all 
Note that 
.
Proceeding similarly, 
for 
;
divides 
.
,
.
denote the smallest natural number such that 
.
Also, since 
it follows that 
.
,
and therefore 
divides 
,
,
be positive integers such that 
is the perfect power of prime 
.
or(analogously) 
, you immediately get a contradiction.
and so 
.
,
.
,
.
and so 
.
and using the lemma we have 
which thanks to lifting the exponent, reduces to 
,
is a constant.
for some naturals 
(note that since 
,
)
,
and so 
,
.
and so we have 
which implies 
.
.
we must have 
,
Thus there exist no such numbers, as required.
,
If 
then it will divide all 
. Since 
,
,
to be of the least power of 
we get 
.
,
which is not possible as 
and we are done!