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.
is a perfect square number.
consecutive even numbers, such that the square of their product is divisible by the sum of their squares.
consecutive even numbers, such that the square of their product divides the sum of their squares.
be a function that satisfies the following equation for all real numbers 
:![\[
f(x^2 + x + 3) + 2f(x^2 - 3x + 5) = 6x^2 - 10x + 17.
\]](http://latex.artofproblemsolving.com/f/6/2/f625b922656061221157fdb5a3b958cfa7d57108.png)
Find the value of 
.
prime and 
a positive integer. Determine all pairs 
satisfying the equation: 
be a positive integer and let 
and 
be sets of integers satisfying the following conditions:
, 
and is a subset of 
, 
iff 
.
be a prime such that: 
. Simplify:
be a given integer. The Mint issues coins of 
different values 
, where each 
is a positive integer (the number of coins of each value is unlimited). A set of values 
is called lucky, if the sum 
can be collected in a unique way (namely, by taking one coin of each value).
with 
(b) Prove that every lucky set of values satisfies 
is a given prime number.
is a set which 
.
non-empty subsets of such that 
can be written in the form
where 
are non negative integers, such that
for every 
.
for 
. It is easy to check that these work.
denote the given assertion. 
gives 
and 
gives 
. From 
we get 
. Thus becomes 
. It follows that![\[
f(x+y)=f(x)+f(y)\ \ \ \ \ \ \ \forall(x,y)\in\mathbb{R}^+\times\mathbb{R}^-.\tag{*}
\]](http://latex.artofproblemsolving.com/9/b/3/9b382f12b114dc17049ff9b665afadca3ebf9ae2.png)
for all 
.
and 
are both positive, 
so we are done. If exactly one of and are positive, we are done by 
. If and are both negative,![\[
f(x+y)=-f(-x-y)=-f(-x)-f(-y)=f(x)+f(y)
\]](http://latex.artofproblemsolving.com/1/b/6/1b6816748309b461ab5a46c76e6e997ac0612a47.png)
by case 
so we are done. 
so 
, as desired. 
for any constant 
, and it's easy to see that this satisfies the equation.
, we get , and plugging in 
, we get 
. From this, we get 
, giving 
. and then , we get 
, which further becomes since . This implies that if 
or 
, we have 
.
, and setting 
gives us that also holds true when both 
. Finally, by negating everything and using , this also holds true when both 
. Hence, 
satisfies Cauchy.
into and use the fact that satisfies Cauchy to get ![\[f((x+1)^2)=f(x^2)+f(x)+f(x+1)=(x+1)f(x+1)=xf(x)+xf(1)+f(x)+f(1),\]](http://latex.artofproblemsolving.com/4/0/2/4027f798cfe379d43cf7f585a710d2d8933d86c6.png)
and by using , we can cancel things to get 
, giving where 
.
. These work since 
. Now we need to show these are the only solutions. Let x = 0, we get 
. Let y = 0, we get 
. By these two we get that f is odd and also that 
. So now using what we got we can write the starting equation as 
. Using this and the fact that f is odd we get that f is additive. Now plugging in x = x + 1 in and using that f is additive we get 
which after clearing things up gives us that 
there are no other solutions and we are ready.
for some constant . It is easy to verify that this satisfies the given condition. Now we will prove that it is the only solution.
.
..
for positive . Since is odd, this is also true for negative .
, and is odd, we can directly use Cauchy to finish 
and 
. It is easy to verify these both work. Now, we will show that these are the only functions satisfying the given conditions. Plugging in 
,![\[f(x^2) = xf(x)\]](http://latex.artofproblemsolving.com/d/7/8/d784a31622ac92abd632d594076ec4c054df76be.png)
Plugging in 
,![\[f(-y^2) = -yf(y) = -f(y^2)\]](http://latex.artofproblemsolving.com/8/f/5/8f502d7303fdc1659acd551919d8cfe6e2cddc62.png)
So, 
for all 
, thus it is odd. Now, we prove the following claim: is additive.![\[f(x^2 - y^2) = f(x^2) - f(y^2) \implies f(a - b) = f(a) - f(b)\]](http://latex.artofproblemsolving.com/6/3/b/63b96640347c304a638d22b2829295552979b029.png)
for all nonnegative reals 
. Now, set 
, for nonnegative 
. This gives![\[f(y) = f(x+y) - f(x) \implies f(x) + f(y) = f(x + y)\]](http://latex.artofproblemsolving.com/b/c/6/bc6e20512b330e88a99c1b414fb6b1b117c2f51e.png)
for all 
. Since is odd, we have![\[-f(-x-y) = f(x + y) = f(x) + f(y) = -f(-x) - f(-y) \implies f(-x-y) = f(-x) + f(-y)\]](http://latex.artofproblemsolving.com/b/6/9/b69fa62cd1336afe1422f32c8a9fea2ef40b6dee.png)
so is also additive over all the nonpositive reals. Now, rewrite![\[f(x) - f(-y) = f(x + y) \implies f(x) = f(x + y) + f(-y)\]](http://latex.artofproblemsolving.com/f/3/1/f3112d0d7d22c07376e163921b03e3074db3a0f2.png)
We can set 
for any 
, therefore the above equation rewrites as![\[f(a -y) = f(a) + f(-y)\]](http://latex.artofproblemsolving.com/7/d/f/7df76e980385a4b9984df6a66eb89d2ba634068d.png)
So, 
is also true for all pairs 
with 
. is additive over all the real numbers. Thus, we have![\[f((x + 1)^2) = f(x^2) + f(2x) + f(1) = xf(x) + 2f(x) + f(1)\]](http://latex.artofproblemsolving.com/f/3/3/f33b06811523575e439f27b19d791f52441d6ac3.png)
![\[f((x + 1)^2) = (x + 1)f(x + 1) = (x + 1)f(x) + xf(1) + f(1)\]](http://latex.artofproblemsolving.com/7/d/4/7d4eb1e6de5ca74c8303a4d1663dd507105790e3.png)
Setting the two equations equal, we get![\[(x + 2)f(x) = (x + 1)f(x) + xf(1) \implies f(x) = xf(1) \implies f(x) = cx\]](http://latex.artofproblemsolving.com/9/0/7/907af73bb2cfd7c2138784a7262ae84d6bc1f711.png)
for some 
, so we are done.
be the assertion that the functional equation holds for 
. Plugging 
gives and plugging 
gives is odd. Plugging gives for all real . So for nonnegative real 
gives 
, and we can extend this to all negatives because is odd. So for all real . Plugging this into the original equation gives 
, so is the only solution.
denote the assertion. Then 
Therefore 
for all nonnegative 
Therefore, for positive we have 
As 
is odd, it follows that is additive. Hence 
This solution clearly works.
for any constant which clearly works. Now we prove they are the only ones.
and 
for nonzero implies is odd. Now gives 
so in fact . Imposing 
, replacing 
yields for all 
.
for 
, we have![\begin{align*}
2 f(x) + f(1) &= f(2x + 1) \\
&= (x + 1) f(x + 1) - x f(x) \\
&= (x + 1) [f(x) + f(1)] - x f(x) \\
&= f(x) + x f(1) + f(1) \\
\end{align*}](http://latex.artofproblemsolving.com/6/7/b/67b8eee6689b4cf05c8d019adca40964625e3f8b.png)
so 
for . But is odd, so for some constant as needed.
.
as seperate cases, we see that 
Thus, we also see that 
or 
For 
we see that the domain is all positive, and thus, 
Now we substitute 
to get that 
Thus, we are done.
and 
.![$\sqrt[3]{ab-1}+\sqrt[3]{bc-1}+\sqrt[3]{ca-1} \le \sqrt[3]{(a+b+c)^2}$](http://latex.artofproblemsolving.com/9/6/3/9633b5849794285e44519c745c1f616eceb77711.png)
.
Let 
.
and 
.
be the set of real numbers. Determine all functions 
such that ![\[ f(x^2 - y^2) = x f(x) - y f(y) \]](http://latex.artofproblemsolving.com/6/8/f/68f78e0df627f6ac04052433c01627bf16f0af8e.png)
for all pairs of real numbers 
.
(1)
(2).
be positive reals. Then
,
.
(3).
(4)
(5).
can be positive or negative.
,
to get 
for 
.
,
to get 
.
,
to get 
.
is constant, let 
.
for all real 
.
for all 
for every real x.
.
such that 
and 
we find 
,
.
(see Paladin8's solution) we have 
,
.
,
,
.
and using 
you get the result...
a
,
are linear functionals, all such functions f would satisfy the functional equation. I do have a question, though: do there exist linear functionals from R to R that are not linear functions?
.
,
and similarly, 
.
.
and 
,
.
is a continuous function. Hence, we can take the derivative of it. Letting 
,
is an infinitely small number, we have that 
.
is constant, so 
is constant. It follows that 
is linear for positive 
.
,
.
,
.
.
is negative, then 
.
.
"
and 
are collinear since any two points are collinear. We prove through induction that 
is on this line. By the inductive hypothesis, 
is on the line. Now, 
,
,
given a rational and 
and 
to our conjecture from Cauchy's functional equation that 
.
.
.
for one of 
.
gives us 
,
.
when at least one of 
and thus 
after performing the same manipulation on the 
term. Therefore, as both terms can be negative, this finishes off the second case.
which is also equal to 
.
so the solutions are 
over all real constants 