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 positive reals, s.t 
, prove that :
of positive integers for which ![$a + [a, b] = b + [b, c] = c + [c, a]$](http://latex.artofproblemsolving.com/a/1/0/a10eb969d7e061a14a11cfeccdc043f4fbbfbc1b.png)
.![$[a, b]$](http://latex.artofproblemsolving.com/d/a/2/da2e551d2ca2155b8d8f4935d2e9757722c9bab6.png)
denotes the least common multiple of integers 
.
, determine the number of residue classes modulo which can
modulo , where 
and 
are arbitrary integers.
and 
Show that
Where 
and 
Show that
Where
, 
.
.
are fixed positive integers.
are fixed positive real numbers.
and 
. But right answer is only . Should it be reported somewhere in Aops?
chess board. The allowed move is to add an integer 
to each of two adjacent numbers in such a way that non-negative numbers are obtained. (Two squares are adjacent if they have a common side.) Find a necessary and sufficient condition for it to be possible for all the numbers to be zero after finitely many operations.
be positive integers such that ![$\sqrt[n]{a}+\sqrt[n]{b}$](http://latex.artofproblemsolving.com/d/a/a/daaf95cd3381060559529aa4a318e97b727eac31.png)
is an integer. Prove that 
are both the 
th power of 
positive integers.
+1 w
be positive integers such that 
is not squarefree. Prove that there exist a positive real 
, such that 
and 
are coprime for any positive integer .
implies f is injective and surjective on 
where 
is constant.
implies that 
. Now 
gives
thus 
(1) by the injectivity of 
.
rewrites as 
and since 
is surjective on 
we have that 
for all large x and 
.
(x large) in (1) we obtain that 
for all x large and 
.
. Now we compare 
with 
to get that 
for all large 
. Thus 
for all large x.
in (1) we conclude that c=0 implying that 
for all large x. where is large implies that 
for all 
.
denote the assertion that
gives us:
gives us:
we get:
gives us:
which means that .
2. 
(from 1)
(from 2)
(from1)
for all positive reals
is injective and surjective on 
. Also 
is a fixed point, and so 
. is surjective for big enough values, take 
so that 
, and take a such that 
.
, such that 
for all positive reals .
.
be our assertion then, is injective.
let 
, then 
is injective as desired.
. Hence, has a fixed point. Call it 
. Consider an arbitrary positive real number 
. Then 
gives us 
Furthermore, 
gives us 
Finally, 
gives us 
However, 
Hence, 
, which means 
. Since was arbitrary, the indentity function is the only candidate for the solution, and it's easy to check that it is a solution.
FE, nevertheless here is my solution. denote the above assertion. that 
Then by , we have that 
which by and 
gives us that 
meaning that 
for all 
. 
, which clearly works. be the given assertion. is injective.
, compare 
and 
to obtain 
. 
of real numbers good if
As takes only positive values, so it isn't hard to see that 
must have the same sign. Also, as is injective, so either of being equal to 
implies the other to equal too, i.e. if 
then we must have 
.
, for
the pair is good.
satisfying
Compare 
and 
. Using injectivity of we obtain
It isn't hard to see that our Claim follows. is good, then 
is also good.
So we are done by Claim 2. has arbitrarily large fixed points. gives 
As 
, so as becomes large, 
also becomes large, as desired. is good, then we must have 
. is good iff 
is good. We may assume are positive now (case is direct). We want to show 
. Pick a large for which 
. Pick largest 
for which 
. Then,
As can become arbitrarily large, so 
can become arbitrarily large, which implies our Claim. is good, we must have 
.
. By Claim 3 we know is also good. Using Claim 5 we obtain
This forces , as desired. ,
This gives that
for some constant 
. We must have 
, say because:
in . as a fixed point by Claim 4.
( 
) ) and we get 
( 
) ) and we get 
( 
)
. ( 
) ) and ( ) we get our function is 
, which clearly works. We now show that it is the only one; let denote the assertion. has arbitrarily large fixed points. we have that 
, which implies has a fixed point. Suppose is a fixed point of . Then, 
gives 
, which means 
is also a fixed point of . By induction, 
is a fixed point of for any positive integer , which proves our claim. is surjective on some interval 
, where 
.. Then, note 
can take on any positive real value greater than 
simply by varying ; hence proved. be any positive real. From our two claims, we know there exists some positive real such that 
is a fixed point of . Fix as such a value; then, we must have 
, or 
, which implies 
, for some nonnegative real 
. Plugging this into the given equation yields 
, so , as desired. 
to mean 
for all .
gives 
, and so 
is a fixed point. is a fixed point, then 
gives is also a fixed point. So 
too.
into the original equation means 
.
.
. It is easy to check that this works. Now we will show that it is the only solution. is injective.
as desired. is kinda surjective, i.e., hits everything above for some constant .
. Let be a number greater than . Set 
.
yields 
.
, there exists a number 
so that 
as desired.
for all .
in the original equation.
as the "inverse" of , i.e. the function such that 
. By Claim 1 and Claim 2, this function exists and is injective.
satisfies Cauchy's additive functional equation.
By Claim 1, this implies
However, by Claim 3, we have
as desired.
for some constant . It is easy to check that 
, so we have 
for all 
.
. For an appropriately large , we have
. Prove that:
,
which works.
(1)
gives: 
or
Let this be 
gives: 
or
Let this be 
gives: 
(1)
gives
.
o
or 
and 
and 
by induction on 
and finish.
.
for any x positive real.
.
.
.
.
,
![\[P(x,x) \implies f(x+2f(x))=x+2f(x)\]](http://latex.artofproblemsolving.com/8/a/2/8a2c132bc8b4cfff2afcc8368300116a9b2a2769.png)
![\[P(x+2f(x),x)\implies f(2x+5f(x))=2x+5f(x)\]](http://latex.artofproblemsolving.com/e/e/d/eed64f48e5707afd01fad5a7250f6d3c78e39dfa.png)
![\[P(x,2x+5f(x))\implies f(3x+6f(x))=2x+7f(x)\]](http://latex.artofproblemsolving.com/7/8/e/78e134243e1cb521917fe9ef8d2b11d2e1ecb619.png)
![\[P(x+f(x),x+f(x))\implies f(3x+6f(x))=3x+6f(x)\]](http://latex.artofproblemsolving.com/1/5/b/15bd94db3d6cd100527fcf1bf3e445eea96f299e.png)
so ![\[2x+7f(x)=3x+6f(x) \implies f(x)=x. \ \ \blacksquare\]](http://latex.artofproblemsolving.com/b/2/c/b2ce189543707bc0daa6112588fa27fbc067b413.png)
Combining 
clearly works.
is constant, so 
to get:
and we have:
to get:
Therefore we have 
gives:
gives:
gives:
This means that 
so we are done.
gives 
gives 
is the only solution, and this function obviously works.