[/quote]
,
is sabroso, since 
.
and 
Prove that
satisfy the following two conditions:
exist such that 
.
,
.
then prove 
Equality holds iff 
or 
and 
.
satisfies the following three conditions:
All elements of 
.
.
is the foot of the altitude from 
of triangle 
.
and 
points 
and 
are marked such that the circumcircles of triangles 
and 
are tangent, call this circles 
and 
respectively. Tangent lines to circles 
.
.
be non-negative real numbers such that 
Prove that
satisfies the equation:![$\sum_{k=1}^{n}\frac{1}{2}(1-(-1)^{[\frac{n}{k}]})a_{k}=1$](http://latex.artofproblemsolving.com/3/2/1/3216129b3e42dfb49eb8bb764fafa1b9fbf28882.png)
,
.
and 
of the set 
,
for all 
.
.
be a point of 
such that 
is tangent to the circumcircle of 
and 
be the circumcenters of triangles 
and 
,
be the midpoints 
.
through 
such that for all 
,
.
be a function over reals sich that 
is surjective.
for all reals 
,
Y by nguyendangkhoa17112003, anantmudgal09, abdelkrim, Hexagrammum16, Plasma_Vortex, Seicchi28, The_Maitreyo1, Carpemath, sriraamster, SHREYAS333, TheMathLife, samuel, Vietjung, lifamily, AwesomeDude86, Toinfinity, OlympusHero, A-Thought-Of-God, L567, RudraRockstar, samrocksnature, Palindrome_, centslordm, megarnie, Math_Is_Fun_101, Lamboreghini, veirab, EpicBird08, Adventure10, deplasmanyollari, lian_the_noob12, ItsBesi, buddyram, MathIQ., Rounak_iitr, Akacool, cubres, NicoN9
Let 
be the set of integers. Determine all functions 
such that, for all integers 
and 
, 
Proposed by Liam Baker, South Africa
be the set of integers. Determine all functions 
such that, for all integers 
and 
, 
Proposed by Liam Baker, South AfricaThis post has been edited 2 times. Last edited by djmathman, Jul 18, 2019, 4:41 AM
and 
for some 
constant.
,
for all 
.
for this one, we quickly obtain that 
,
.
by symmetry and so the condition becomes 
It is equivalent to 
This means 
is linear on even numbers. Notice that 
,
,
,
for all 
for all 
(divide the case when 
is negative and when 
or 
for some constant 
gives 
, or for every 
in the image of 
.
A simple variable-switching and some translation gives the well-known Cauchy equation. Hence 
after we reach the last step in @above's solution. Cauchy ends it.
=>
(
(
and 
in (1) and use (2), we have 
,
an
,
.
.
.
to get 
,
.
this is just 
over integers implying 
and 
.
or 
and 
or 
for an arbitrary integer 
.
and 
denote the assertion. 
gives us 
.
gives 
and 
gives 
,
.
to both sides of the original equation, we get ![\[f(2a)+2f(b)+f(0)=2f(a)+2f(b)=f(f(a+b))+f(0)=3f\left(\frac{2}{3}(a+b)\right) + f(0).\]](http://latex.artofproblemsolving.com/5/3/f/53f518564aa56672616c1c7f7c0b87412b108d9c.png)
Now, dividing both sides by 
and subtracting 
from both sides gives us ![\[g(a)+g(b)=\frac{3}{2}g\left(\frac{2}{3}(a+b)\right),\]](http://latex.artofproblemsolving.com/a/e/f/aefb65fc83b22262709d7b1629f7657c6fed7e45.png)
where 
,
,
,
by induction. From this, we get ![\[g(a)+g(b)=3g\left(\frac{a+b}{3}\right),\]](http://latex.artofproblemsolving.com/3/1/6/31682a2b5cf4f6944826fc627a95071eacda45db.png)
and by plugging in 
,
,
,![\[g(a)+g(b)=g(a+b).\]](http://latex.artofproblemsolving.com/7/e/6/7e6c53c7ae8ba1cece6508d0109c5158f6186b5f.png)
Now, since 
satisfies Cauchy and is over 
,
.![\[2k(a+b)+3c=k^2(a+b)+c(k+1),\]](http://latex.artofproblemsolving.com/4/0/5/405678b68576446e1125bf77ed643ac7c22b2e71.png)
so either 
or 
and either 
and 
,
being zero for all Integer 
for some constant 
:
,
which is constant.
,
and 
for any constant 
.
.
.
.
.
.
.
and we get 
.
and we can transform it to 
.
,
.
.
,
- impossible. Now checking the values for k we get that k = 2 works and if
or 
are the only solutions. Checking these work 
for some integer constant 
and 
.![\[f(2a)+2f(b) = f(0) + 2f(a+b).\]](http://latex.artofproblemsolving.com/0/8/e/08e7f6e67090f29a533eafc1e0d7c4662cfa65df.png)
Setting 
in the equation above gives us that ![\[f(a)+f(b) = f(0) + f(a+b),\]](http://latex.artofproblemsolving.com/b/3/c/b3c727963078d956e469533f33d0d8f4ed6ba4fb.png)
which means that 
satisfies Cauchy's equation – therefore, 
be the assertion, then we take 
and subtract to get 
,
gives 
.
for 
,
. Thus 
,
,
.
denote assertion into the original equality.
we get that 
,
.
here, we get 
.
or 
.
(using the evaluations 
)
for all 
,
for any constant 
and 
,
for all 
.
.
yields 
,
yields 
,
for all 
are the only linear solutions. Compare 
to get that 
is constant and obviosly 
.
and 
.
to get 
.
to show that it is indeed linear which finishes.
while 
Therefore the assertion becomes 
so the function 
thus 
for integer constants 
Plugging this back into the assertion yields either 
for some constant 
These are the only solutions, and it is not hard to see that these work.
for any constant 
,
,![\[2 f(a) + 2 f(b) = f(f(a + b)) + f(0). \]](http://latex.artofproblemsolving.com/1/0/c/10cf4ab93a27cecdc6e0936399afb5c3ab8e1e40.png)
Now asserting 
gives us 
,
yields![\[ 2 f(x + 1) + 2f(-x) = 2 f(x + 1) - 2 f(x) + f(f(0)) + f(0) = f(f(1)) + f(0) \]](http://latex.artofproblemsolving.com/e/9/a/e9a0478988c22d6064309a9fd26072f3f9b47658.png)
![\[\implies f(x + 1) - f(x) = \dfrac{f(f(1)) - f(f(0))}{2}. \]](http://latex.artofproblemsolving.com/c/6/4/c64114b46552fc4ac82623f07954f3aef1b43812.png)
Since the RHS is constant, 
as our claimed solution set. We are done.
for any integer 
term. This motivates us to set 
.
.
we find that 
so 
.
.
we notice is always a constant, so this implies 
.
.
,
.
,
.
whose solution is only 
.
,
.
and 
gives 
.
gives 
gives 
Thus, 
implies the only solutions are 
where 
