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.
and 
be elements of the set 
such that 
, 
is a divisor of 
, and 
is a divisor of 
. Determine the largest value that 
can take.
such that the following equation holds:![\[
7^a + 1 = 2b^2
\]](http://latex.artofproblemsolving.com/e/0/8/e08d3fc7feefd7b4acc4a0f9c7e8bcfa328b5791.png)
have the property that 
is a perfect square for infinitely many positive integers 
?
and 
respectively be the incentre, circumcentre and circumcircle of triangle 
. Points 
are chosen on , such that 
, and point 
is the foot of the altitude from 
to 
. If 
are similarly defined, prove that lines 
and 
concurr on 
.
that is not isosceles at 
, let 
be its incircle, which is tangent to 
at 
, respectively. The lines 
and 
intersect the line passing through and parallel to 
at 
and 
, respectively. The lines passing through 
and perpendicular to 
intersect 
and 
at 
and 
, respectively.
are collinear and that 
are concurrent.
positive integer numbers such that the least common multiple (LCM) of all these numbers is not a perfect square. Mykhailo consecutively hides one of these numbers and writes down the LCM of the remaining 
numbers that are not hidden. What is the maximum number of the written numbers that can be perfect squares?
is a prime number and 
are positive integers. Prove that there exist infinitely such that 
are divisible by 
, let 
if is not squarefree and 
if is a product of 
primes, and let 
be the sum of the divisors of . Prove that for all we have![\[\left|\sum_{d|n}\frac{\mu(d)\sigma(d)}{d}\right| \geq \frac{1}{n}, \]](http://latex.artofproblemsolving.com/8/e/b/8eb88de7cd7ea8e2853dbc80cae24bdc6b77f9be.png)
be a sequance of natural integers. Proof that, any number in the sequance can be written only as or 
for any 
and not necessarily distinct primes 
for 
. And and 
are same in some case.
has no integer solutions
and 
with real coefficients satisfying
as the solution set of 
. From the statement we know that 
.
, we have 
and 
.
for polynomials 
, we get that 
.
, 
.
and 
.
and 
are pairwise distinct. Now look at 
. This must have as a root with multiplicity 
, and 
as a root with multiplicity 
. This implies that 
Rearranging gives the desired lemma.
, an obvious contradiction.
work, so 
and 
for some . Then we have: 
![\[ P' \cdot P^8 \cdot \left( 10P + 9 \right) = Q' \cdot Q^{19} \cdot \left( 21Q + 20 \right) \neq 0. \]](http://latex.artofproblemsolving.com/2/b/0/2b007bc7a7313d2fee06e81185be092842719f6f.png)
Thus 
should divide 
but ![\[ \deg(10P+9) = 21n > 20n-1 = \deg(Q' \cdot (21Q+20)). \]](http://latex.artofproblemsolving.com/a/0/2/a02800e5a576930474824d3332601eafcef123c5.png)
. is disjoint with that of 
. Similarly for 
. Note that 
, so the number of roots of 
, counting multiplicity must not be less than the sum of multiplicities of the roots of 
decreased by 1, so the number of roots of and is at least 
. But the number of roots of 
doesn't exceed 
, which is a contradiction.
and 
, the equation rewrites as 
. It is clear that the polynomial is indecomposable. Then according to Corollary 2.18 in http://dept.math.lsa.umich.edu/~zieve/papers/peter.pdf (which is proven using monodromy groups) there exists a linear polynomial 
such that 
or 
, where 
is the 
th Chebyshev polynomial. The first case is impossible since contains a nontrivial coefficient of 
but not of 
, and the second case is impossible since 
but 
.
be two coprime integers. Prove that there are no non-constant polynomials and with real coefficients such that 
the problem is true.
the problem is true.
.
the problem is true..
be the set of the roots of not counting multiplicity.Is it possible that 
and 
?
and differentiating to kill the one on the LHS.
is same as IMC 2000 day 2, prob 3.
, where 
and 
as shown in rkm0959's solution. Now, we look at the Lemma given below (it is basically a generalization of the Lemma in post #5), after which we can finish in a similar fashion as in rkm0959's solution.![$F \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/d/3/d/d3d45964209a0ff752736cab92d5a43b4afbfb1d.png)
be a non-constant real valued polynomial of degree 
. Consider the 
distinct real numbers 
where 
. Then the total number of complex solutions to the equation 
for all 
is at least 
. (The previous Lemma follows on taking 
and 
)
to denote the integers from 
to . First consider a general polynomial ![$A \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/c/0/1/c0197711a982fd1feffd53aaa95e1a7ea416795b.png)
. Suppose that 
has degree 
, where is the first derivative of 
. Then there exist roots of (not necessarily distinct) which are roots of also, which means that the remaining roots of are distinct and correspond to the elements of 
(Just use the fact that roots of with multiplicity have multiplicity 
in ). So we have 
. Return to our Lemma. Letting 
denote the total number of solutions to the given equations, and using the fact that 
, we have 
Then we wish to show that 
, which is equivalent to showing that 
Consider the polynomial 
. As all these terms in product are pairwise co-prime, so we get that 
is 
is well-known (maybe not exactly in that form). Similarly, the second half of the proof of the Lemma (introducing 
) is well motivated if one catches hold of the fact that degrees get added on multiplying pairwise co-prime polynomials.
to work with.
, we find that the degree of is 
for some 
and the degree of is 
. divides 
. Additionally, notice that 
hence we know that 
divides 
. But this is cap; the degree of the left is and the degree of the right is 
, smaller. Thus we are done.
work, so and for some . Then we have: Thus should divide but 
is an arbitrary root of either side of the equation (we will refer to this as a "root of the equation"), and let 
denote 
. We consider four cases—note that must fall in exactly one of these. is a root of , then it is a root of 
with multiplicity divisible by 
, and a root of 
with multiplicity divisible by 
. Thus is a root of the equation, it must have multiplicity divisible by 
. Thus we can say that there are distinct roots of the equation which contribute a total multiplicity of 
. is a root of 
, then it is a root of the equation with multiplicity divisible by , so we can say there are 
distinct roots with total multiplicity 
. is a root of 
, then we can say there are 
distinct roots with total multiplicity 
. is a root of 
, then we can say there are 
distinct roots with total multiplicity .
and similarly for the other pairs of variables. will have exactly 
roots with multiplicity (since each root contributes nine times) and will have exactly 
roots with multiplicity (since each root contributes once). Likewise, will have exactly 
roots and 
will have exactly 
roots. with multiplicity 
is a root of 
with multiplicity 
. Thus, has at least 
roots (with multiplicity). Thus,
and similarly 
(from looking at instead of ), so
On the other hand, this means that
which is absurd, hence no such polynomials exist. 
denote the set of roots, without multiplicity, of a polynomial . Then 
and 
for some positive integer .
by the lemma. On the other hand, 
This yields a contradiction. 
and its derivative 
. The former implies that any root of is a root of or . In particular, does not share any roots with , so 
. However this is absurd since 
, so no such polynomials exist.
and 
for some . Clearly we have that all the roots of and all the roots of 
or 
. Let be the set of distinct roots of , 
be the set of distinct roots of and and 
similarly. We have that 
by Putnam 1956 B7, thus we also get that as ,
, so we have a contradiction and thus no such polynomials exist.
and in question by mistake and Im too lazy to change, sorry.
.
and 
for some 
. Differentiate it once and see that we have the equations 
Divide the two equations and we get ![\[(Q+1)QP'(21P+20)=(P+1)PQ'(10Q+9]\]](http://latex.artofproblemsolving.com/c/5/a/c5a8d5b269e476db4902b34fc17e43899699e594.png)
But see that 
anfd so we have ![\[(Q+1)Q \mid (P+1)PQ' \implies 42d \le 20d+21d=41d\]](http://latex.artofproblemsolving.com/7/8/8/7887b454413432a90b6091bef92bd6845db2843c.png)
Which is a contradiction.
and let 
and 
be tangency points of the incircle of 
,
and 
respectively. Let the circle with centre 
and 
.
,
and 
,
.
and 
be feet of perpendiculars from 
and 
respectively. Prove that 
,
and 
are concurrent.
satisfying: 
.
![\[2^{xy}f(xy-1)+2^{x+y+1}f(x)f(y)=4xy-2,\forall x,y\in\mathbb{R}\]](http://latex.artofproblemsolving.com/d/1/d/d1de7196fd3ba60732c14fa2b156ec880b4829d0.png)
.![$X,Y,Z \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/8/a/5/8a5243f7f1c3c100a2636d38d6ad7244a0351170.png)
such that 
and 
have no common complex roots, and for some positive integer 
.![$X_1,X_2 \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/7/0/3/7039ca2ddf2d4dd047717a2782741a4ca75c475d.png)
such that 
,
,
holds.![\[ X= \prod_{i=1}^{n} X_i^{e_i}\]](http://latex.artofproblemsolving.com/9/e/0/9e073860988da37db0d2ae49c74ebb905f63717f.png)
.
,
divides 
,
divides 
,
.
and 
,![$Q_1,Q_2 \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/e/0/e/e0e47c04af3fde7bf35109842640f73dbe2da8f1.png)
such that 
,
,
.![$P_1,P_2 \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/0/9/6/0960572f4142d9415ffcb52005b80eeb3e260c7b.png)
such that 
, 
,
.
.
so that they all have positive leading coefficients.
and 
.
,
,
for some positive real 
are all positive)
for all sufficiently large 
for each fixed constant 
.
for all sufficiently large 
for each fixed constant 
.
,
such that
,
.
such that 
,
,
a contradiction.
at the original equation, we get ![\[P_2^9(P+1)=cQ_2^{20}(Q+1)\]](http://latex.artofproblemsolving.com/1/c/5/1c50d5a0ff054b004c847ef8aea8f5774f36ecdf.png)
.
such that
.
such that 
.
,
,
, a contradiction.
as 
goes to 
.
such that 
for every real 
,