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and 
be positive real numbers such that 
. Prove that
When equality holds?
satisfy the equality, 
prove that the triangle must have a right angle.
is a quadratic trinomial such that there exists 
real numbers that satisfies 
and 
. Prove that 
are the only numbers satisfying 
and 
![\[
a_7 \cdot 3^7 + a_6 \cdot 3^6 + a_5 \cdot 3^5 + a_4 \cdot 3^4 + a_3 \cdot 3^3 + a_2 \cdot 3^2 + a_1 \cdot 3^1 + a_0 \cdot 3^0
\]](http://latex.artofproblemsolving.com/0/c/e/0ce2c259010b678238f11d35fc1328ea76f945f6.png)
where 
for 
?
sides and 
sides, respectively. The total number of sides is 33, and the total number of diagonals is 243. What are the values of and ?
be a polynomial with integer coefficients such that 
and 
for some integer 
. Prove that there exists no integer 
such that 
.
for some real number 
and ![\[ a_n=na_{n-1}+(n-1)(n!(n-1)!-1) \]](http://latex.artofproblemsolving.com/4/1/f/41f0cc7b041a8bba056882b5c794503195c374fd.png)
for integers 
. Given that 
, and given that 
, where 
and 
are positive integers whose greatest common divisor is 
, compute 
and 
are in the coordinate plane such that 
. Let 
denote the locus of all points 
in the coordinate plane satisfying 
, and let 
be the midpoint of 
. Points 
and 
are on such that 
and 
. The value of 
can be expressed in the form 
, where 
and 
are relatively prime positive integers. Find 
.
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 
Then 
in terms of 
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).
,
.
(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.
.
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 