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.
be a prime number and 
. Prove that![\[\left| p^m - (p - 2)! \right| > p^2.\]](http://latex.artofproblemsolving.com/2/1/c/21ca8bb6e5727d48b18b7ec3b127029c4c97694f.png)
Proposed by Miloš Milićev
, 
, with 
, 
. Prove that there exists 
such that for every 
![\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\]](http://latex.artofproblemsolving.com/0/f/c/0fc36d9264eb7e103128c489aeae521a859c1fd4.png)
(For and , the notation 
represents 
. )
are non-negative real numbers and 
Prove that
and 
take turns playing the following game: There is a pile of 
stones. In his first turn, selects a divisor of and removes this number of stones from the pile. then chooses a divisor of the number of remaining stones, and removes that number of stones from the new pile, and so on. The player who has to remove the last stone loses. Show that one of the two players has a winning strategy and describe the strategy.
denote the 
th Fibonacci number, defined by the recurrence:![\[
F_0 = 0,\quad F_1 = 1,\quad \text{and} \quad F_{n+2} = F_{n+1} + F_n \quad \text{for } n \geq 0.
\]](http://latex.artofproblemsolving.com/1/b/c/1bc4aa38be3c2c64832e25376a923df611807864.png)
For a fixed modulus 
, define the set![\[
S_m = \left\{ (i, j) \in \mathbb{N}^2 : F_i \cdot F_j \equiv 1 \pmod{m} \right\}.
\]](http://latex.artofproblemsolving.com/c/0/4/c04881682fdc91ae0a7a78a807044c3658e0412e.png)
Prove that 
if and only if 
.
such that 
,![\[
f(x^2+f(y))=f(xf(x))+y
\]](http://latex.artofproblemsolving.com/d/e/d/dedd2338779fcc6723077ce586713fb4a68b9931.png)
such that![\[ f(x) + f(y) + 2xy = f(x + y) \]](http://latex.artofproblemsolving.com/e/3/d/e3dfbb279ff146cc0923115fe9d3f8d9e3488e17.png)
for all nonnegative real numbers 
and 
.
(from the set of real numbers to itself) where
for all reals 
and 
be two triangles inscribed in the same circle, centered at 
, and sharing the same orthocenter 
. The Simson lines of the points 
with respect to triangle form a non-degenerate triangle 
. lies on the circle with diameter 
.
lie in this order on the circle and form a convex, non-degenerate hexagon.
and be positive integers posessing the following property: the equation
holds for all positive integers 
. Prove that 
for some integer 
.
Now we have a few cases:
, then 
, which is obviously impossible. If 
, then 
which implies 
and 
. This yields the solutions 
and 
.
, then the entire curve of solutions 
can be checked to work.
;-; 
. Also, these edge cases are so hard for me to find without making sillies.
, 
and 
for any 
.
., then we get that 
which gives us 
. But then the discriminant must be 
, that is 
which clearly has no solution., then we get that 
. Then we get that 
. This means that the discriminant must be 
, that is 
. This gives only solution 
since the left side is strictly increasing and it exceeds 
for 
. For , from the problem statement we get that 
. This gives us that 
each of which are solutions., then clearly 
and so 
which is another solution.
, for 
.![\[ab^2 + b + 7 \mid a^2b + a + b \implies ab^2 + b + 7 \mid b(a^2b + a + b) - a(ab^2 + b + 7) = b^2 - 7a.\]](http://latex.artofproblemsolving.com/b/e/d/bedf88f0178ab546f0d6feb72de1eeecbc5e30f0.png)
We now split into two separate cases:
Then 
, so that 
, which in turn gives us 
, for some . Then we find the corresponding 
. 
First observe that 
, else 
, which is a clear contradiction for 
. Hence 
, so that we obtain ![\[ab^2 + b + 7 \leq 7a - b^2 \implies b^2(a + 1) + b + 7 \leq 7a \implies b^2 < 7.\]](http://latex.artofproblemsolving.com/4/f/f/4ff8f4a9b8de7c7005420dd68dc00d8779690198.png)
Hence we split into cases for 
:
Then ![\[\frac{7a - 1}{a + 8} \in \mathbb{Z} \implies 7 - \frac{57}{a + 8} \implies a + 8 = 1, 3, 19, 57 \implies \boxed{a = 11, 49}. \bigstar\]](http://latex.artofproblemsolving.com/2/7/0/2704b8d70627d8809f46b8ac143c2e78ec2536f9.png)
Then ![\[\frac{7a - 4}{4a + 9} \in \mathbb{Z} \implies 1 + \frac{3a - 13}{4a + 9} \implies a \leq -22, \]](http://latex.artofproblemsolving.com/e/a/8/ea834238044622767cc1849a72fa675dcf2b9e40.png)
which is impossible. Therefore no solution exists in this subcase. 
, for arbitrary , as claimed. 
, , 
for all 
.
.
, contradicting the divisibilty condition.
.
, let 
, so
.
which is always valid, hence these are always solutions.
.
we get:
This gives 
and 
.
which gives no solutions.
. Then, 
, so 
so 
.
or 
. This means that or . When , we get 
so 
so or ., we get 
so 
so 
so there are no solutions to .
. Then, 
so 
. Plugging this in means 
. Note that 
so 
. Plugging back in, we get 
, which is always true.
, , and for all positive integer .
, so 
. We can then divide into cases, if , then clearly we have no solutions by size. If , then write 
, we write 
, we can see all solutions of this form work. If , by size we still require 
, or equivalently 
. We then check 
, we require 
, equivalently 
, equivalently 
, so there are no solutions for by divisor analysis. We now check 
, we require 
, so we have 
, so we have 
. We check 
gives 
, 
, so this pair works. We then check 
, we get 
, so this pair works as well. The answers are then 
.
. Since both sides are positive: 
.
: if we set 
, then 
, We can easily check that it is a solution with being a positive integer.
: Then from earlier, 
LHS is positive so RHS should follow. Meaning
. Then we can manually check and see that 
is a solution.
. Clearly 
isn't possible due to size. If then 
After checking we find the solution 
. If then 
for some , which works.
Multiply the RHS by to get 
and then subtracting times the LHS, we get 
Thus, we have that 
we have that 
but this is absurd.
we have that 
Thus, 
If 
we have 
Clearly, the solutions are 
If 
we have that 
Multiplying by 
we have that 
which has no solutions.
Thus, we have that 
We have 
We clearly have that 
for some 
Thus, 
The LHS is very clearly relatively prime to 
so we multiply the RHS by and then apply Euclids, we have that 
If it is non-zero, this is clearly absurd. Thus, 
We get the curve 
We now proceed to solve this problem using 3 cases.
As and are positive integers,
; 
work.
For 
we have,
which is a contradiciton.
. Therefore we only need to consider the cases where and ,
Hence, 
works.,
.
and 
; 
. Now we get that 
. we want 
- which is impossible we don't have solutions in this case. 
. Now if 
we have that 
, which is impossible. If y = 1, then 
or we get the solutions (x,y) = (49,1); (11,1) If y = 2, then 
and 
and we don't have any solutions here. 
, let 
all 
work. We have to check this tho and 
which is true all solutions are 
.
and with 
and 
since i was given this with and and am too lazy to change everything
and 
for all positive integers 
We proceed as follows.
and 
so and work. Also, for all positive integers 
so 
works for all positive integers We now show there are no more solutions.
we have that 
If 
then since 
so 
a contradiction. Therefore, 
and 
then 
so 
and 
for some integers Then, 
so 
Thus, all of the solutions in this case are of this form.
then since 
we must have 
However, if 
then 
so 
or 
If 
then 
so 
Since is a positive integer, 
or 
so 
or 
yielding 
or 
then 
However, since 
we must have 
so 
which is not an integer. Thus, there are no solutions in this case.
and 
for all positive integers as desired. 
and with and since i was given this with and and am too lazy to change everything and for all positive integers We proceed as follows. and so and work. Also, for all positive integers so works for all positive integers We now show there are no more solutions. we have that If then since so a contradiction. Therefore, and then so and for some integers Then, so Thus, all of the solutions in this case are of this form. then since we must have However, if then so or If then so Since is a positive integer, or so or yielding or then However, since we must have so which is not an integer. Thus, there are no solutions in this case. and for all positive integers as desired. 
?
such that 
the equality hold for 
or 
or any
such that 
.
such that![\[f(x+y)+f(y+z)+f(z+x)\ge 3f(x+2y+3z)\]](http://latex.artofproblemsolving.com/9/8/a/98ab64d17bc7d6bf7533b0874177fc963405c1d8.png)
.
of positive integers such that 
is divisible by 
in positive integer
<0 (contradition)
(m i positive integer)
(k in positive integer)
=
then 
in positive integer![\[k\equiv 0 (mod b)\]](http://latex.artofproblemsolving.com/f/9/2/f92d5e37c71ebbca9b1cffc5dc9c2e83d2c5c9c1.png)
so 
(b >= 3and q i positive integer)
=
+
(u in positive integer)
,
,
.
,
.
,
, or
.
. But then we have
,
. And also 
,
we know 
. So we have all solutions of the form 
.
so the RHS is positive. So we have 
.
. So we check 
.
,
, giving the solutions 
. Since
,
. But since 
, we must have 
, which does not have an integer solution.
for positive integers 
then 
from here
so 
and 
so
then n divides 
, too.
.
then 
from here we will have
.
then 
also 
then we have to have 
here 
or 
otherwise inequality will lead to contradiction
then 
and 
we will have 
and 
is divisible by 
so 
is an integer, too.
those are 
)
by simliar of previous we will show that there is no solution for 
then it is easy to see there are no solutions.
,
.
is divisible by 
,
,
and we have the solution 
or 
.
can't be more than 
,
to 
and 
works, this gives the solutions 
and 
.
then 
and we have no solution.
,
,
since if 
then the left side is less than the right side (we would have 
,
,
)
.
since 
.
,
.
and 
work, giving the solutions 
.
are positive, then 
.
or 
.
so we must have 
and we must have 
since otherwise we have:
.
or 
.
and so we write:
.
must be positive. We rewrite this as:
and since 
is positive, the denominator must be positive and so 
.
into this expression in turn shows that only 
and 
give integer values for 
and 
.
.
.
,
does not give integer 
.
.
which satisfies the problem must also satisfy:
.
.
which is an integer.
?
?
,
?
.
?
is not actually used for the algebraic manipulations, but to set aside a case that forces 
,
,
.
.
(division by 
on the RHS of 
to be an integer?
must divide 
.
and 
and 
for all 
because 
that is impossible for 
therefore 
if we prove all the divisors we have that 
no solutions here
![\[y(x^2y+x+y)-x(xy^2+y+7) = y^2-7x.\]](http://latex.artofproblemsolving.com/8/2/c/82c9525fecfa2fa1730f2246886d7780f282fa74.png)
.![\[|xy^2+y+7| > |y^2-7x|,\]](http://latex.artofproblemsolving.com/c/3/9/c39303ba228bd6baedb223bc571b80db96f57d24.png)
implying there are no solutions, unless 
where we get the pairs 
.
![\[xy^2+y+7 \mid y(x^2y+x+y)-x(xy^2+y+7)=y^2-7x \]](http://latex.artofproblemsolving.com/3/e/7/3e7266ae7abb0ab975b5796a73c8509c5929457c.png)
which implies that 
which gives 
or 
or 
which Convsersely always works. The edge cases give 
and 
as solutions.