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.
such that![\[
f\left(\frac{x^2 - f(x)}{f(x) - 1}\right) = x
\]](http://latex.artofproblemsolving.com/8/0/0/8009a4709ffd6708023d67db2b544e605d1fbe68.png)
for all real numbers 
satisfying 
.
such that 
for all 
?
be a sequence of integers, with the property that every consecutive group of 
's averages to a perfect square. More precisely, for every positive integers 
and 
, the quantity ![\[\frac{a_n+a_{n+1}+\cdots+a_{n+k-1}}{k}\]](http://latex.artofproblemsolving.com/a/7/b/a7b72c85a43bcba1ebbc642e075c7443ccb6d23c.png)
is always the square of an integer. Prove that the sequence must be constant (all are equal to the same perfect square).
+1 w
be an acute triangle with 
. Let 
be its circumcircle, 
its orthocenter, and 
the foot of the altitude from 
. Let 
be the midpoint of 
. Let 
be the point on 
such that 
and let 
be the point on such that 
. Assume that the points , 
, 
, and are all different and lie on in this order.
and 
are tangent to each other.
(so 
is a function from the positive real numbers) such that![\[ \frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}
\]](http://latex.artofproblemsolving.com/0/4/b/04b955fb570ad384fb7f0e85bf0dacccf4c982a1.png)
satisfying 
be the intersection of the diagonals 
and 
of cyclic quadrilateral 
. If 
, 
, and 
, what is the circumradius of the quadrilateral?
such that:
for all 
.
be an integer. Ivan writes the numbers 
each on different cards. He then shuffles these 
cards, and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.
be six distinct real numbers satisfying the following condition:
,![\[
x_i^6 + (-1)^{i+1} x_i = -x_1 x_2 x_3 x_4 x_5 x_6.
\]](http://latex.artofproblemsolving.com/3/6/f/36f886654cf5f7af956e2475be94acbe62bb8153.png)
.
| 
with 
are positive integers and 
is prime (
), we get: 
, 
is the midpoint of 
. 
is an arbitrary point on 
. Let 
be the intersection of 
and 
. The line 
intersects with the circumcircle of 
, for the second time at 
. 
is the circumcenter of triangle 
. Prove that 
.
which 
is prime then 
, and , are even. Thus we can consider the expression mod 
, to assert that is of form 
for positive integers 
. will suffice. This is well known. So done
is prime then 
must be odd for 
. Thus we can consider the expression mod 
, to assert that is of form 
for positive integers . will suffice. This is well known. So done
is prime then 
must be odd for . Thus we can consider the expression mod , to assert that 
is of form 
for positive integers . will suffice. This is well known. So done factors!
, let be the minimal root of 
. I claim that this has 
, which will finish. Indeed, clearly 
, so let 
. Now
Putting this back in terms of gives
which implies the bound for sufficiently large .
be some arbitrarily large prime that is 
. Then let be the smallest that satisfies 
. Notice that 
since there are at most two distinct solutions to that add up to 
(you can prove this with primitive roots). So then let 
.
gives 
. again gives 
. Expanding gives 
which is enough as 
for sufficiently large .
be a prime equivalent to 
. Then 
has solutions. Note that if there are solutions for that are less than , then they come in pairs adding up to 
. Let 
then let the two smallest solutions be 
. Note that 
so 
, which implies 
. Solving the quadratic equation gives![\[p=\frac{4n+1+\sqrt{8n-15}}{2}\]](http://latex.artofproblemsolving.com/a/0/3/a038c6473ec9d00e54dbf399c2c2d8d485c43fd5.png)
Now, 
if and only if 
which is true for sufficiently large . The reason why sufficiently large ever appear in our method is because for any fixed , only finitely many divides 
.
which are 
. (By Dirichlet's, there are infinitely many.)
mod are 
and 
, with 
. Then 
, so 
. Since , we have 
, so 
. Now 
, so 
, and we're done by setting 
. Since can be arbitrarily large and 
, can also be arbitrarily large, so there are infinitely many such .
there exists with 
and 
This implies the problem, since no works for multiple as can have at most one divisor 
In fact we show a much stronger statement: this is true for all 
and 
Then choose that satisfies 
and 
Next, we can see that 
is equivalent to 
by manipulating. Thus we just want to prove this bound.
From 
it follows 
is a positive odd integer. Now ![\[n^2=\left(\frac{p-c}2\right)^2\equiv -1\pmod p,\]](http://latex.artofproblemsolving.com/f/2/d/f2d5bd653fd9ee305f23ecc3d5e0790ee4291499.png)
so this becomes 
In particular, 
and 
But 
so our claim is true.
be a prime number. Then there exist two solutions to the equation 
. Let 
be these solutions. Since 
, we have 
. That is, 
. But if 
, we would have:![\[
0 \equiv X_2^2 + 1 \equiv (2K + 1)^2 + 1 \equiv 2K(2K + 1) + 3K + 2 \equiv 3K + 2 \pmod{P}.
\]](http://latex.artofproblemsolving.com/5/5/b/55b88975aa2b877c5c773e6679b13e370d70f302.png)
Which is not possible, since 
we have 
, hence 
. Similarly, if 
we would have:![\[
0 \equiv X_2^2 + 1 \equiv (2K + 2)^2 + 1 \equiv 4K(K+1)(K+1) + 3K + 4 \equiv 3K + 4 \pmod{P}.
\]](http://latex.artofproblemsolving.com/4/c/b/4cbd871ef44e53b05dd3421bc78cab2e0acb7e4a.png)
Which is not possible, since 
then 
. From the above, we have 
. It follows that
Finally, we observe that:![\[
(P - 2X_1)^2 + 4 \equiv P^2 - 4P X_1 + 4X_1^2 + 4 \equiv 4X_1^2 + 4 \equiv 0 \pmod{P}.
\]](http://latex.artofproblemsolving.com/4/8/7/487da519a05edcfc4943bd0b30364c2f89749b8e.png)
From which 
, that is, 
. We conclude that:![\[
P \geq 2X_1 + \sqrt{P - 4} > 2X_1 + \sqrt{2X_1}.
\]](http://latex.artofproblemsolving.com/8/e/f/8ef59de561f330ffaa741fb695df079a7da3dc9e.png)
So for each there exists an integer such that 
and 
, thus we are done.
, think about it in terms of choosing a prime and considering the smallest for which 
. Clearly 
.
is not very strong, in fact, it says that only has to be slightly less than 
. Since the two solutions are symmetric, the only concern is that both solutions may be very close to . The following claim dispels this. is sufficently large and 
is a positive integer such that 
then 
. We have 
In particular, 
.
by symmetry, there exist 
such that .
, we have 
as desired.
. We will prove that there exists some such that 
and 
which is stronger than given bound. There are two where and 
, let 
be the smaller one. Suppose that 
. We have 
or 
. If 
, then 
which implies 
. However 
cannot be 
. Thus, 
and this gives as desired.
, we can see that for 
sufficient large, there exists a number 
such that the divisor of 
is greater than 
( is a constant). Since 
for very large , we're done.
and 
.
?
such that 
has a prime divisor greater than 
.
multiple of 
very large. Then we know there is 
and of course we may assume that 
and even 
,
also divides 
.
and observe that 
,
and so 
is at least 
.
if 
.
,
and one is greater. Let 
be the smaller one 
)
.
.
is equivalent to 
.
,
since the parity is different. Therefore 
,
and it's clear that the desired inequality is satisfied.
multiple of 
and of course we may assume that 
and even 
.
and observe that 
,
and so 
.
if 
has a prime divisor larger than 
divides some 
could solve the problem like harazi did because all one needs is to have infinitely many primes that divide 
and the classical Euclid's proof of the infinitude of primes can be trivially ajusted to give this statement. The opening sentence in TomciO's solution, on the other hand, would tempt me to reduce a few points immediately if I were one of the graders because it is quite doubtful that the contestant could prove it if asked.
. Let's see what the second day will bring 
mod 
and 
.
,
works easily because 
is not a perfect square.
,
as well. This still leaves infinitely many, but proving that is hard compared to the Euclid-style argument that there are infinitely many primes 
.
,
such that 
.
.
,
.![\[ p > p-1 - 2\alpha + \sqrt{p-1-2\alpha} \Leftrightarrow 2\alpha + 1 > \sqrt{p-1-2\alpha} \Leftrightarrow 4\alpha^2 + 6\alpha + 2-p >0.\]](http://latex.artofproblemsolving.com/f/4/0/f40faad5fd560d26a4eeee3354984eb9543b9c59.png)
Solving this quadratic inequality, since we assumed 
,![\[ \alpha > \dfrac{-6+\sqrt{36+16(p-2)}}{8} = \dfrac{-3+\sqrt{4p+1}}{4}.\]](http://latex.artofproblemsolving.com/a/1/0/a10cfbcfd205c76a23c59a0c85e591e755a80c03.png)
,
.
.
.
,
,
.
,
such that 
and 
we will done. So p
+
{
}
let us define set of positive integers 
such that 
and 
- the minimal of them.
and also 
else we can take 
instead of 
be 
.
,
.
.
and 
since 
and 
- contr; if 
and 
-contr.
,
or 
.
then there are infinitely many different integers 
{
}
isn’t square of integer and hence 
.
: 
directly from 
is immediate but not what we want, so we expect this is only sharp for finitely many 
which gives the idea to generalize to 
pretty easily.