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.
, where 
is rational for 
. A vector 
is called a rational point in 
-dimensional space. Denote the set of all such vectors as 
. For 
and 
in , define the distance between points 
and 
as 
. We say that point can move to point if and only if there is a unit distance between two points in .
, there exists a point that cannot be reached from the origin via a finite number of moves.
, any point in can be reached from any other point via moves.
be reals such that 
.Prove that
Where 
and integers 
, prove that for any positive numbers 
satisfying 
for 
(with the convention 
), the following inequality holds:![\[
\sum_{j=1}^n a_j^{2m-1} \leq \frac{mt}{2} \left( \sum_{j=1}^n a_j^{m-1} \right)^2.
\]](http://latex.artofproblemsolving.com/5/c/d/5cd52e9ce307563bde96f697a75079229d4bfcc2.png)
is inscribed in circle 
. The interior angle bisector of angle 
intersects side 
and at 
and 
(other than ), respectively. Let 
be the midpoint of side . The circumcircle of triangle 
intersects sides 
and 
again at 
and 
(other than ), respectively. Let 
be the midpoint of segment 
, and let 
be the foot of the perpendicular from to line 
. Prove that line 
is tangent to the circumcircle of triangle 
.
prove that
be an acute angled triangle with 
and incircle 
. Let touch the sides 
and at 
and 
respectively. Let 
and 
be points outside 
satisfying ![\[\angle BDX = \angle XEA = \angle YDC = \angle AFY = 45^{\circ}.\]](http://latex.artofproblemsolving.com/4/7/0/470b0486fa1342c4b6444ec1e3731c23bbb842fb.png)
Prove that the circumcircles of 
and meet at a point 
.
be a fixed integer. The cells of an 
table are filled with the integers from 
to 
with each number appearing exactly once. Let be the number of unordered quadruples of cells on this board which form an axis-aligned rectangle, with the two smaller integers being on opposite vertices of this rectangle. Find the largest possible value of .
of 
touch at , 
, let 
denote the line tangent to through 
. Define 
such that 
. Prove that the circumcenter 
of lies on 
.
of one variable with integer coefficients such that if 
and 
are natural numbers such that 
is a perfect square, then 
is also a perfect square.
so i'm assuming the only solutions are 
is not an odd polynomial there exists an so that 
is not divisible by 
so 
and 
also 
can not hold as an identity so 
and the polynomial is a constant function 
and 
. how can we show that these last two equations imply some contradiction if the degree of is too large?...
is not an odd polynomial there exists an so that 
is not divisible by 
, a nonzero constant (otherwise the conclusion is obvious...). Then obviously and 
are relatively prime and have the same degree. Differentiate the relation to obtain 
, thus divides 
and thus must be a constant times this thing, by the remark on the degree. This gives you a differential equation on and allows finding all solutions, but unfortunately it may very well happen that there are solutions of arbitrarily large degree (and it happens, actually...), but this imposes however very strong conditions on . Now, by choosing some other value 
for 
, I think we can conclude. I'm still puzzled by this apparently extremely easy problem, which however does not seem to have a decent solution.
but also vice versa so
and particularly
by putting different choices So 
holds for some fixed 
is constant so is linear and we can easily conclude
we don't even need the differential equations.
is not an odd polynomial there exists an so that is not divisible by , is diviseble by then 
is zero for all ....and P is odd...
imply 
, then i think we can conclude that the degree of is 0
being odd
do not need to be with integer coefficients. For example 
. is even. In this case we let 
and tha cofficent of the 
power is 
.
is a constant.
such that 
and 
(This is true due to Newton's formula for symmitric sums.)
for all 
. Then 
. Therefore is a integer, and for large enough , we have 
. Note that the sequence 
has the fomula 
where 
for large enough 
. So we conclude 
for large enough and therefore for all 
. And it's obvious that ![$ f(x) \in Q[x]$](http://latex.artofproblemsolving.com/4/b/1/4b13ba257b17b446018c823e9e1cfcf8d5ea08cd.png)
. is odd, let 
we do the same thing, and it's easy to claim that such 
don't exist.
do not need to be with integer coefficients. For example 
. takes square values for all integers ,
where is a polynomial with integer coefficients.
is a monic polynomial.
is a monic polynomial.
be a integer polynomial.We have 
such that 
.So by above theorem we have for any integer polynomial which has not any double root ,there are infinitely many prime 
which there exist some 
such that 
but 
.now suppose 
.(step 1)if has not any double root ,we have but a contraction.(step 2)if has some double root, we have 
such that 
has not any double root .
is a perfect square for any positive integer 
so 
is a perfect square for all 
and we have a contraction same way to step 1.
be a non constant polynomial with integer such that 
is an integer for all positive integers . Then there exists a polynomial 
with integer coefficients such that 
.
for some natural number . If is sufficiently large, we can separate complex roots of 
and 
and therefore and is relatively prime. By assumption, is even degree and the maximal coefficient is square. Therefore, 
for some ![$h(x) \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/5/e/1/5e1261f96cc148d56d09e638155055eb4f940bfd.png)
. Since and is prime, you can easily see that for some ![$g(x) \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/b/c/8/bc81caa0d160fc5ca1e3fb7719fc6271f495cd09.png)
.
. By assumption, 
. is odd.
. There are infinitely many 
such that 
is perfect square. Therefore, infinitely many such that 
is perfect square. However, there are also such that 
is not a perfect square and contradiction. is even, but 
is not a perfect square.
. And same as above. Here we used the fact that pell equation 
has infinitely many solution. is even and is perfect square and for some .
, with 
non zero.Assume non-constant.Then assuming, is not a square ,write it as a square times a bunch of irreducibles.Consider such a irreducible ,say 
,and then by Schur ,choose a very large prime factor of the polynomial 
.Then using Bezout,we can conclude that if 
,then does not divide any other irreducible factor of evaluated at 
.Then,again using bezout but this time with 
,we conclude that if ,is large enough ,then also doesn't divide 
,thus from Hensel,we can uniquely lift the root corresponding to mod to mod 
.Due to this unique lift ,we can find such an ,such that 
mod but does not divide 
.Thus we have that 
is odd.Now choose a 
such that 
is even and quite large.As is large (forgot to mention doesn't divide ,so doesn't divide 
). is a quadratic residue mod .So by hensel's on the polynomial 
,we see that the derivative is non zero mod .Thus,we can find a solution to this poly mod 
,such that 
is exactly 
.Thus we have found 
is a perfect square but 
is odd which is a contradiction.Thus,
is a perfect square.
is even(Basically choosing 
,such that 
has a large 
,which is obv to have small odd ,and obv doesn't divide .
being a perfect square is ensured by the very similar hensel argument as above).
as 
.So if we fix 
,we see that 
is a part of infinitely many Pythagorean triples,which is impossible.
is always a square for a fixed pair 
with being a square, so 
for some integer polynomial . In addition, we consider only the case 
.
(
is the leading coefficient of ), which shall be a square. If is even, plug 
for some prime to see that 
are QR's modulo for all primes , so 
is a QR for all primes , absurd, since 
. Now we can finish the other case with 
: 
is a square, hence is a square, so 
is a square, so 
and thus 
and 
, contradiction.
. If is constant so 
must be a square, so 
for some integer 
. Indeed all of these work. is positive. If it's negative then 
is negative for big , but is must also be twice a perfect square, contradiction.
such that the equation 
has a solution.'s satisfying this (perfect squares for instance).
, we can find infinetely many solutions of the form
Just so I'll write quicker, set 
, we'll use it at lot.
is the square of an integer, 
must be the square of an even integer, say, by definition, 
. So 
.
is a perfect square, so 
, for some integer 
. But, by definition, 
, so 
and 
must be equal. So we have that 
.
are pell's solutions for 
. This means that 
for some minimal solution 
and some integer 
. As there are only finitely many minimal solutions for pell's equation, by the pigeon hole principle one of them must appear infinetely many times, varying . WLOG call it .
so
So this is one estimate for 
.
and 
the leading coefficient. Now we also know that
As 
Therefore the ratio between the estimates converges to . So
So
So, a constant times an integer power of 
converges to . Due to the fact the the sequence 
is discrete, if it converges to , it's eventually . So there exists an integer 
such that
Rearranging things and substituting back to 
, we have
Now we are gonna make use of the Norm function defined by 
, where 
are rationals. One can easily prove that this function is well defined and is multiplicative. So we have
So
As the leading coefficient of is positive, is eventually positive, so we can take the squareroot of both sides to conclude that
But notice that this is true for infinetely many big integers 's, so it must be true that
But remember that is the leading coefficient of , so 
, so 
, so must be , and 
. is a square, so 
is a square, so itself is a perfect square. Given that, clearly is a square if and only if ac+bc is a square.
and 
, where is an arbitrary integer.
![$\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/4/b/7/4b7d768fffbb7a1a6163af392ed348f2d49d8783.png)
being positive integers, the Diophantine 
has a unique solution, which is 
., we have 
is a perfect square and 
is also a perfect square.![$A(x),B(x)\in\mathbb{Z}[x]$](http://latex.artofproblemsolving.com/5/2/e/52eb2fe1e8bedb81296067b07d58fe59d30f0351.png)
such that![\[2P(2x^2)=A(x)^2,\ \ P(x^2)+P(3x^2)=B(x)^2\]](http://latex.artofproblemsolving.com/d/9/c/d9ccf43e9b56f5d260f61e520489abe720c95dcf.png)
Let be the leading coefficient of and 
.
and 
for some integers 
. is even then 
for some 
, which means 
for some 
, which is a contradiction 
. is odd. But then is a perfect square from first equation, which means 
is a perfect square. From theorem 2, 
is forced. only 
works for some and for 
only 
works. These are the only solutions.
be a polynomial such that is a perfect square for all sufficiently large positive integers 
. Then there exists a polynomial 
such that 
identically., we have the numerical equality ![\[P\left(n^2 - k\right) + P(k) = P_k(n)^2\]](http://latex.artofproblemsolving.com/6/a/9/6a9ca915fb21cccd2d5ffc171f83b55d0fa65a0b.png)
for some function 
for all sufficiently large . Thus the polynomial 
in takes only perfect square values, and by the lemma, it follows that it must be identically equal to 
for some polynomial 
. is either even or odd. (Otherwise, it is easy to construct an such that 
.) We make the following claim:
is even for all 
or odd for all . such that was odd and 
was even. Then we can write
for some polynomials 
and 
. So in particular, we have the equality ![\[Q_{k+1}(x+1)^2 - P(k+1) = P(x-k) = xQ_k(x)^2 - P(k).\]](http://latex.artofproblemsolving.com/c/9/d/c9dde6b7becc7b2dcde35e81125cb6b6290994e6.png)
However, taking to be a sufficiently large negative number yields a negative RHS and positive LHS, hence contradiction. The symmetric case works out analogously. 
is even for all . Then the same equation in the previous display reads ![\[Q_{k+1}(x+1)^2 - P(k+1) = Q_k(x)^2 - P(k).\]](http://latex.artofproblemsolving.com/7/f/2/7f22fa132e2830e1b1758c2d4c998b65701167e8.png)
So for 
and , the coefficients of the squares of these polynomials agree everywhere except the constant term. It follows easily that their constant terms must also agree, thus 
for all positive integers , and is constant. We can verify 
in this case. is odd for all . Then by setting 
in the polynomial equality ![\[P(x-k) + P(k) = xQ_1(x)^2\]](http://latex.artofproblemsolving.com/4/4/7/4471eba06a27ff4bb972afe2abe225092ee54c73.png)
for all positive integers , we get that 
, i.e. is odd. Then setting 
in this sequence of equalities, note that 
is always a perfect square. So reading as a polynomial, 
is the perfect square of a polynomial. By applying the same argument with 
, we obtain that 
for some polynomial 
, as the numerical equality holds for all sufficiently large . [EDIT: the rest of this solution ended up being a fakesolve, so I would greatly appreciate anyone who could finish it off from here.]
, which can be manually confirmed to be the only valid polynomials with degree less than . Let 
so that it has degree .
and 
gives![\[2P(2x^2) = 2a_0+4a_1x^2+\dots+2^{d+1}a_dx^{2d} = [Q(x)]^2\]](http://latex.artofproblemsolving.com/a/4/8/a4808477660743756f30ecce79f42a67eb774fb9.png)
![\[P(x^2)+P(3x^2) = 2a_0+4a_1x^2+\dots+[1+3^d]a_dx^{2d} = [R(x)]^2,\]](http://latex.artofproblemsolving.com/d/7/c/d7c542205b1f1e1ee633ea410189a00b72746a8b.png)
![$Q, R \in \mathbb{Z}[x]$](http://latex.artofproblemsolving.com/b/4/8/b483d8bda474a298233767bf4ad2d071e388854d.png)
and both have degree . Let the leading coefficients of and 
be and , respectively. Then,![\[m^2 = 2^{d+1}a_d\]](http://latex.artofproblemsolving.com/2/4/a/24a5afcf3e5c016e4e48c051abcfa6099e5734d2.png)
![\[n^2 = a_d(1+3^d),\]](http://latex.artofproblemsolving.com/7/7/c/77cd19dc4be754dc76d148d0bfae07dd2e1d58d2.png)
and are positive integers. is even, then 
for some integer , so![\[\left(\frac{n}{c} \right)^2 = 2(1+3^d) \equiv 2 \pmod{3},\]](http://latex.artofproblemsolving.com/2/f/d/2fd357f82a3c33fac24168dfa3735f6151e6cf5e.png)
. Thus, there are no solutions for even except for . is odd, then 
for some integer , so![\[\left(\frac{n}{c} \right)^2 - 3^d = 1,\]](http://latex.artofproblemsolving.com/8/c/e/8ce43a82e4ebd251ebd0a28edb3ac3643f1f3d16.png)
as a solution. Hence, there are no solutions when 
.
Prove that
the equation 
and 
.![\[\sum_{i=1}^n (a_i-a_{i+1})^3\leq \frac{1}{2\sqrt{3}} (\sum_{i=1}^n a_i )^3,\]](http://latex.artofproblemsolving.com/8/1/0/810f07e3394370ce06aec1e7025bcf5cc5433d60.png)
where 
.
prove 
has degree=4n,(if P has deg.=n). and 
.
s.t. p+q=a perfect square.. so 
=
.
is also a perfect square for all 
.
P(1) being a perfect square.
for some integer 
.
.
and so, we may assume that 
where 
are primes and 
.
then let 
and 
where 
is a quadratic non residue modulo 
.
then let 
and 
.
for 
.
is changed to 
by our choice of 
;
where 
are pairwise distinct irreducible divisors of 
have integer coefficients. If the product 
is a constant then it must be a square and we may conclude. Else, there exists a non-constant polynomial in the set 
,
.
being irreducible has no common factor with the product 
(which might as well be empty or constant) and 
(the derivative of 
with integer coefficients and integers 
such that 
and 
for all 
is infinite, hence, we may choose a prime number 
and an integer 
.
,
.
where 
is an integer we shall choose later. By the Taylor's Expansion of polynomials, we have 
In particular, we let 
and we can select 
such that 
since the polynomials 
will yield 
,
.
and 
.
is an odd number, a contradiction to the fact that it is a square. The result holds.
is a perfect square then 
is also a perfect square. Define the polynomial 
where 
(not a polynomial/function etc) be the leading coefficient of 
for all 
where 
.
then we may write 
where, for obvious reasons of divisibility, 
is an integer. Let 
,
and so 
giving that 
and we get the requested contradiction. If 
.
which is again a contradiction! Thus, we must have 
and if 
then we see that the equation 
is satisfied for some integer 
.
is a perfect square then we get that 
.
for some integer 
