AoPS Academy
, the points 
and 
are taken such that the point lies between the points 
and , and 
. On segments 
and 
, points 
and 
are taken so that 
and 
. Prove that 
.
and points and on sides 
and 
, respectively, such that 
. Let 
and 
be intersection points of lines 
and 
with circumcircle 
of triangle 
, and 
and 
intersection points of lines 
and 
with side 
. If 
, prove that circle contains intersection point of angle bisector of 
with
such that 
divides 
for all positive integers 
and 
with 
.
such that 
Follows for all reals 
.
denote the set of positive integers. Find all functions 
such that for positive integers and 
![\[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\]](http://latex.artofproblemsolving.com/c/1/f/c1f8ffe04cfcc45b498f3931a4796d5c56dc04d0.png)
be an integer. Rowan and Colin play a game on an 
grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is orderly if: [list] [*]no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and [*]no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. [/list] In terms of 
, how many orderly colorings are there? is selected inside acute 
so that 
and 
. Point is chosen on ray 
so that 
. Let be the midpoint of . is tangent to the circumcircle of triangle 
.
that satisfy![\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]](http://latex.artofproblemsolving.com/f/b/b/fbb693770cd90d8a144026608074122e54da5802.png)
for all 
.
such that p is a prime, 
and 
is a divisor of 
.
,
is 
or 
.
.
)
.
be the smallest prime divisor of 
.
.
.+p$](http://latex.artofproblemsolving.com/5/f/a/5fa0acf3aa516a7ef3e24f34f5a94863ee6d9c65.png)
,
,
,
,
.
s.t. 
.
,
,
(it it does satisfy the relation, since 
)
for any prime 
and 
.
.
and 
,
.![\[p^{p-1}|(p-1)^{p}+1=p^{2}\left(p^{p-2}-{p \choose 1}p^{p-3}+
\cdots + -{p \choose p-3}p-{p \choose p-2}+1\right)\]](http://latex.artofproblemsolving.com/5/d/c/5dcb92f1390234d96e36f164d0fd147d3ad72d49.png)
.
.
is odd, so is 
)
we get 
and 
.
(from the choice of q) leads to the existence of integers 
such that 
,
.
,
must be odd. This shows that 
and 
+p$](http://latex.artofproblemsolving.com/7/7/c/77c6492e0fa69aa7e7df2a126b7acbff9f9218ff.png)
,
such that p is a prime, 
and 
is a divisor of 
.
:
.
and 
.
.
has prime divisors later) 
always holds. So 
are also solutions.
.
be the smallest primedivisor of 
.
.
.
then 
.
.
when 
and 
.
.
.
gives the solution 
.
.
.
.
.
.
when 
.
.
:
.
.
and 
.
we see that 
is increasing when 
.
we see that 
.
we get the solutions 
we find that 
and 
,
.
,
.
but not by 
,
with 
and 
,
.
.
.
we also have 
.
we get 
.
it follows that 
,
.
which is clearly true, as 
,
is strictly decreasing for 
.
are the only solutions. 
,
,
.
.
and checking 
with 
.
for some integer 
.
.
for an odd integer 
.
.
and 
.
and 
have been dealt with using, for example, LTE (giving solutions of 
and 
)
.
)
.
and 
hence, as 
is twice a factor of 
is a factor of 
so 
.
is even). Thus 
.
so 
which is a case that has already been dealt with.
because 
and 
from fermat's little theorem,so 
,
Hence, the pair 
is valid for all primes 
If 
,
But note that 
,
It follows that 
,
,
such that 
By testing these six possibilities, we find the solutions 
If 
,
such that 
Thus we obtain the solutions 
Furthermore, note that if 
is odd, and so 
,
be the least prime divisor of 
Then we have the relation ![\[(p - 1)^x \equiv -1 \pmod{q^{p - 1}} \implies (p - 1)^x \equiv -1 \pmod{q} \implies (p - 1)^{2x} \equiv 1 \pmod{q}.\]](http://latex.artofproblemsolving.com/a/3/e/a3ee50dbee88492cabd08c1c8cc2d5f26ecbf45b.png)
Hence, if we let 
,
Furthermore, since 
and ![\[(p - 1)^{q - 1} \equiv 1 \pmod{q} \implies d \mid q - 1 \implies d \le q - 1.\]](http://latex.artofproblemsolving.com/f/f/b/ffb4e81cb71a68ef88ca3f8b02670e718760ef50.png)
Now, note that if 
,
that is also a divisor of 
)
It follows that ![\[(p - 1)^2 \equiv 1 \pmod{q} \implies p(p - 2) \equiv 0 \pmod{q} \implies p - 2 \equiv 0 \pmod{q},\]](http://latex.artofproblemsolving.com/b/5/c/b5c2670ace8ab326223bce7c9c787efcc27d9f7f.png)
because we assumed that 
It follows that 
However, this contradicts 
,
,
.
,
.
.
.
where 
and 
.
.
is odd 
)
because 
and the smallest prime that divides 
and 
.
or 
.
,
so we must have 
.
.
so 
,
.
.
.
)
.
and 
or 
.
)
.
,
.
.
.
,
.
.
)
.
)
.
.
.
.
the latter implies -1
Consider the smallest prime 
.
Then 
.
.
.
implying 
.
so 
works.
.
Then as we have alread considered 
,
.
and hence 
.
.
Which holds if and only if 
,
.
we have 
.
.
.
.
.
,
and as 
then 
so 
so 
,
impossible. If 
so 
so 
which by 
.
.
for the remainder of this solution.![\[\operatorname{ord}_q(p-1) \mid \gcd(2x,q-1) = 2.\]](http://latex.artofproblemsolving.com/1/0/8/108062964a700fa669f3c37f2182b2832e033196.png)
![\[(p-1)^2 \equiv 1 \pmod{q} \iff p(p-2) \equiv 0 \pmod{q}.\]](http://latex.artofproblemsolving.com/2/1/7/2170622b92e8835580d041c94c56b7b8037606ce.png)
.![\[(p-1)^x+1 \equiv 2 \pmod{q},\]](http://latex.artofproblemsolving.com/3/f/6/3f6552a8ea8199e5af45759d7eb88c6dbfa8f484.png)
.
,![\[\nu_p((p-1)^x+1) = \nu_p(p)+\nu_p(x) \ge \nu_p(x^{p-1}) = (p-1) \nu_p(x)\]](http://latex.artofproblemsolving.com/a/8/6/a8691725ff9b8220c2e4bb38bfe4be5c8121781a.png)
![\[\implies 1+\nu_p(x) \ge (p-1) \nu_p(x) \iff (p-2) \nu_p(x) \le 1.\]](http://latex.artofproblemsolving.com/1/a/3/1a3749942850b6c469d2934eafc9875706847b71.png)
,
.
.
.![\[(p-1) \cdot v_p(x) \leq v_p((p-1)^x+1) = v_p(x) + 1 \implies (p-2) \cdot v_p(x) \leq 1.\]](http://latex.artofproblemsolving.com/0/3/0/030dfcaefac6273de25d38ae1fe6152ceb145c66.png)
If 
.
.![\[\operatorname{ord}_q(p-1) \mid 2n, p-1 \text{ but not } n \implies \operatorname{ord}_q(p-1) = 2 \implies p=q,\]](http://latex.artofproblemsolving.com/0/0/f/00f9acaa793e4b6041504507e12d8f9f55e50f48.png)
contradiction. 
,
,
divides 
.
are relatively prime and thus 
,
,
and thus 
,
.
.
,
then they have equal values, so this case has to be resolved separately.
be the minimum other prime that is a factor of 
divides 
.
,
only contains prime factors greater than 
but 
has order 
.
.
which means 
which is either 
in which case we have 
which is divisible only by 
so 
then 
.
but if 
then that means 
.
.
,
clearly there are no solutions as the division no longer holds.
,
and 
Looking at the order of 
,
.
is impossible. To see why, notice that 
,
.
is not divisible by 
.![\[ \nu_p((p-1)^x + 1) = 1 + \nu_p(x) \geq \nu_p(x^{p-1}) = (p-1)\nu_p(x). \]](http://latex.artofproblemsolving.com/5/3/f/53f97bb18238ed1b38f680bae27299620fa5411d.png)
This implies![\[ \frac{1}{p-2} \geq \nu_p(x) \geq 1, \]](http://latex.artofproblemsolving.com/2/e/1/2e12b579e0888e705f156adb2f5ad7b9b3ee7308.png)
so 
and 
and 
as it always works, ignore 
as it can be done manually), let it be 
,
.
,
we can check that 
by size, so we can eliminate this case by just showing 
,
which forces no sol, since 
,
,
,
since the left is increasing in 
and the right is decreasing , in the latter we have 
on the right is 
,
.
.
then 
and 
.
From this it's not very hard to show that 
.
then either 
or 
.
then by mod 2 we get a contradiction. If 
then by FLT 
so 
so 
is odd, so 
.
.
.
,
,
.
,
and 
.
.
.
,
so 
.
,
.
then 
.
,
then 
,
,
However, because 
,
,
.
,
and 
and 
,
,
.
and 
,
.