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.
the set of all positive real numbers.
which have the property:![\[f(x)f(y)=2f(x+yf(x))\]](http://latex.artofproblemsolving.com/4/f/f/4ffbde35ff082585a188c1a4368d39338d1b07a4.png)
and 
.
such that :
for which there exists a sequence 
of positive integers satisfying the following conditions:[list]
for all integers 
for all integers 
the sequence 
is a permutation of
[/list]
such that 
is a positive integer.
be positive real numbers such that 
. Prove that
.
questions, each with 
answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?
, each of our twin children also begins to have twins every year. If this pattern continues, how many years will it take for my family to grow large enough to have the required number of members to guarantee winning the quiz gift?
, 
, and 
of a hexagon 
inscribed in a circle 
intersect at a point 
, and the acute angle between any two of them is 
. Let 
be the radius of the circle tangent to segments 
and 
and internally tangent to ; the radii 
, 
, 
, 
, and 
are defined similarly. Prove that![\[
r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}.
\]](http://latex.artofproblemsolving.com/0/d/6/0d66d6685bbc97c7553e9ed3ba14ea0dc3328b49.png)
such that 
for every 
.
and 
be polynomials with integer coefficients. Suppose that the integers 
and 
are roots of , and that 
. Prove that the equation 
has no integer solutions.
. Find .
be three positive real numbers with 
. Find the minimum value of the expression 
is strictly increasing : 
such that:
such that 
has a prime divisor greater than 
.
multiple of 
very large. Then we know there is such that 
and of course we may assume that 
and even 
, simply because 
also divides 
. Now, write 
and observe that divides 
, thus divides 
and so 
is at least 
. This ismmediately implies 
if is large enough, that is is large enough.
of the form 
. There are two solutions to the congruence 
, one of them is less then 
and one is greater. Let 
be the smaller one (
). Now we want find a lower bound for the , let's see what do we need. Take for simplicity 
. We want to have:
.
is equivalent to 
. But 
, so it's not a square. Of course 
since the parity is different. Therefore 
, i.e. 
and it's clear that the desired inequality is satisfied.
multiple of very large. Then we know there is such that 
and of course we may assume that 
and even , simply because also divides 
. Now, write 
and observe that divides 
, thus divides 
and so is at least 
. This ismmediately implies 
if is large enough, that is is large enough.
divides some 
could solve the problem like harazi did because all one needs is to have infinitely many primes that divide for some 
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 appears exactly once as the divisor here, except for 
and 
. Harazi's "large enough" threshold is 
, and 
works easily because 
is not a perfect square. were changed to 
, we would need to exclude all primes of the form 
as well. This still leaves infinitely many, but proving that is hard compared to the Euclid-style argument that there are infinitely many primes 
.
, we can always choose some 
such that 
. Denote 
. This problem then becomes equivalent to choosing a suitable , with corresponding 
, such that 
. We check 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 
, this is equivalent to ![\[ \alpha > \dfrac{-6+\sqrt{36+16(p-2)}}{8} = \dfrac{-3+\sqrt{4p+1}}{4}.\]](http://latex.artofproblemsolving.com/a/1/0/a10cfbcfd205c76a23c59a0c85e591e755a80c03.png)
, then we are done. Suppose not, then we have 
. Note that 
. By the bounds on , we get 
. Now, by assumption, 
, so if 
, we get a contradiction. Solving this inequality, this is equilavent to 
. is sufficiently large (i.e. ) and 
, we can always find some 
such that 
and .
multiple of very large. Then we know there is such that and of course we may assume that and even , simply because also divides . Now, write and observe that divides , thus divides and so is at least . This ismmediately implies if is large enough, that is is large enough. divides , obviously if 
we will done. So p=
+4. So this will lead to that any prime congruent to 1 mod 4 need also be congruent to 5 mod 8.(otherwise p-4 congruent to 5 mod 8 which cannot be ) This is clearly a contradiction by Dirichlet's Theorem!
{
}.
let us define set of positive integers 
such that 
and 
- the minimal of them.
and also 
else we can take 
instead of .
be 
. Then 
, hence 
.
. Really, if this isn’t true then there are only two variants: 
and 
since is odd. If then 
and 
- contr; if then 
and 
-contr.
, therefore 
or 
. Since 
then there are infinitely many different integers such that , so we are done.
{
} then for every 
isn’t square of integer and hence 
.
and even , simply because also divides . .
: 
has a prime divisor larger than 
which 
is prime then 
,
for positive integers 
.
must be odd for 
.
,
for positive integers 
is prime then 
must be odd for 
is of form 
for positive integers 
,
.
,
,
.
Putting this back in terms of 
which implies the bound for sufficiently large 
is pretty annoying (how do we get 
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.
.
.
since there are at most two distinct solutions to 
(you can prove this with primitive roots). So then let 
.
gives 
.
.
which is enough as 
for sufficiently large 
.
has solutions. Note that if there are solutions for 
.
then let the two smallest solutions be 
.
so 
,
.![\[p=\frac{4n+1+\sqrt{8n-15}}{2}\]](http://latex.artofproblemsolving.com/a/0/3/a038c6473ec9d00e54dbf399c2c2d8d485c43fd5.png)
Now, 
if and only if 
which is true for sufficiently large 
.
which are 
.
mod 
and 
,
.
,
.
,
.
,
,
.
,
and 
This implies the problem, since no 
In fact we show a much stronger statement: this is true for all 
and 
Then choose 
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 
.
be these solutions. Since 
,
.
.
,![\[
0 \equiv X_2^2 + 1 \equiv (2K + 1)^2 + 1 \equiv 2K(2K + 1) + 3K + 2 \equiv 3K + 2 \pmod{P}.
\]](http://latex.artofproblemsolving.com/5/5/b/55b88975aa2b877c5c773e6679b13e370d70f302.png)
Which is not possible, since 
we have 
,
.
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 
.
.
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 
,
.![\[
P \geq 2X_1 + \sqrt{P - 4} > 2X_1 + \sqrt{2X_1}.
\]](http://latex.artofproblemsolving.com/8/e/f/8ef59de561f330ffaa741fb695df079a7da3dc9e.png)
So for each 
and 
,
.
.
is not very strong, in fact, it says that 
.
then 
.
In particular, 
.
by symmetry, there exist 
such that 
,
as desired.
.
and 
which is stronger than given bound. There are two 
,
be the smaller one. Suppose that 
.
or 
.
,
which implies 
.
cannot be 
.
and this gives 
,
sufficient large, there exists a number 
such that the divisor of 
is greater than 
for very large