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.
for which 
is the square of an integer
of natural numbers with elements with odd number. A nonempty subset 
of is called 
if the product of the elements of divisible by the sum of the elements of , but not by its square. It is known that the set itself is good. Determine the maximum number of subsets (including itself). make the equation ![\[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\]](http://latex.artofproblemsolving.com/e/7/f/e7fb8c20a43535b5aaed5ab254f1ef043263c62b.png)
true?
be an integer greater than zero. Then, the value of the sum of the reciprocals of the cubes of the roots of the equation![\[
mx^4 + 8x^3 - 139x^2 - 18x + 9 = 0
\]](http://latex.artofproblemsolving.com/a/3/a/a3a9bca412e927632e4921c51f2a6593409031d9.png)
is equal to:
for which there exists an even natural number 
such that the number![\[
(a - 1)(a^2 - 1)\cdots(a^n - 1)
\]](http://latex.artofproblemsolving.com/9/c/a/9caf4eeb82ff46b5ba55ab4b6bc28f0cace586ec.png)
is a perfect square.
and 
be on segment 
of an acute triangle 
such that 
and 
. Let and 
be the points on 
and 
, respectively, such that is the midpoint of 
and is the midpoint of 
. Prove that the intersection of 
and 
is on the circumference of triangle .
, then prove that the following inequality holds:![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
be triangle, inscribed in parabola. Tangents in points 
forms triangle . Prove that 
. (
is area of triangle 
).
such that the equality 
holds for all pairs of real numbers 
.
, in which 
: 
, but the proof is important.
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
be an acute triangle with circumcircle 
. Let 
be the midpoint of 
and let 
be the midpoint of 
. Let 
be the foot of the altitude from 
and let 
be the centroid of the triangle . Let 
be a circle through and that is tangent to the circle at a point 
. Prove that the points 
and 
are collinear. with integer coefficients such that ![$P\left(\sqrt[3]5+\sqrt[3]{25}\right)=5+\sqrt[3]5$](http://latex.artofproblemsolving.com/5/8/e/58ec2a921cd988bd5ae017a8653544ddd758225f.png)
.
and 
. Let 
be the assertion. We see that is non-constant. is injective at 
.
gives 
or 
thus, 
and 
. Now suppose that 
where 
. 
yields 
and this gives 
which does not hold.
implies 
.![\[f(-xy)+2xy-1+f(x)f(y)=f(xy-1)+f(x)f(y)=2xy-1\]](http://latex.artofproblemsolving.com/1/3/7/137ec102980fa71caf1e92d66172a51b90ff86ad.png)
Hence 
. Pick 
in order to get 
so 
or 
., then is odd. Comparing with 
implies![\[-f(xy+1)-f(x)f(y)=f(-xy-1)-f(x)f(y)=-2xy-1=-f(xy-1)-f(x)f(y)-2\]](http://latex.artofproblemsolving.com/8/4/e/84ed0b0b42ca0bd50ed1103bdcd1c98a23f5fd44.png)
This gives 
. Compare and 
which implies![\[f(xy+2y-1)+f(x)f(y)+2f(y)=2xy+4y-1=f(xy-1)+f(x)f(y)+4y\]](http://latex.artofproblemsolving.com/0/d/4/0d41fcc96dab6159121a32f449e2956b63364d37.png)
Thus, 
or 
or 
. Pick 
to get 
so 
which means is additive. Since is both additive and multiplicative, we get which holds., pick 
to see 
.![\[2xy-1=f(xy-1)+f(x)f(y)=f(xy)+2xy-1+f(x)f(y)\]](http://latex.artofproblemsolving.com/e/4/5/e45331a0fac44124cc961b0f987a41e07198084f.png)
Hence 
. Set 
. We have 
and 
. We get 
for 
. Plugging 
yields 
. Compare 
with 
to get![\[g(xy+ny)+g(x)g(y)=(x+n)^2g(y)+y^2g(x)\]](http://latex.artofproblemsolving.com/c/a/9/ca939130777b9c0a8593428c47ab5e7e201fa82a.png)
Thus, 
.![\[g(xy+ny)-g(xy)=(n^2+2nx)g(y)\]](http://latex.artofproblemsolving.com/8/f/4/8f468af6475dfa74189174ab786831eafc4d2a98.png)
Now 
gives 
and pick 
.![\[g(x+y)-g(x)=(n^2+2n^2.\frac{x}{y})g(\frac{y}{n})=n^2(1+\frac{2x}{y})g(\frac{y}{n})=\frac{2x+y}{y}.g(y)\]](http://latex.artofproblemsolving.com/4/e/a/4ea78149ee3562e78ffdf187ff581c7bd7e788bc.png)
and swap to observe 
or 
. Let 
which gives 
thus, as desired.
be the assertion.
: . Since the constant function is not a solution, we must have and .
: 
.
: 
.
Replacing 
with 
and using 
gives:
Let 
, 
, and note that 
The LHS must be symmetric:
Setting 
:
Solving with 
gives , and 
gives . Verification is straightforward.
be the set of real numbers. Find all functions 
such that, for all real numbers 
and 
, the following equation holds:
be the assertion 
.
then is constant, but none of the constant solutions work, so and .
Set 
, then 
for all 
(we can see that it's also true for )., then 
so 
. But since 
we must have 
which fits.
Set , then for , but we can see that it's also true for , so 
for all 
which fits.
is an obvious solution. Now, let 
or 
.
.
