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.
denote the set of all real numbers. Find all functions 
such that ![\[ f\left( x^{2}+f(y)\right) =y+\left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}. \]](http://latex.artofproblemsolving.com/8/5/c/85c37ee472f918eace12c8cd7bef6766e617d2aa.png)
(where 
are the digits and 
) we define the \textit{permutation sum} as the value 
For example the permutation sum of 20253 is 
Let 
and 
be two fivedigit integers with the same permutation sum.
.
pairwise distinct positive integer numbers 
, find the maximum possible number of equal numbers among the fractions of the form![\[
\frac{a_i^2 + a_j^2}{a_i + a_j}
\]](http://latex.artofproblemsolving.com/3/5/3/3537ddec6d793d3831ba82a1e13b98d43b1c29bc.png)
be the incircle of triangle 
. Line 
is parallel to side 
and tangent to . Line 
is parallel to side 
and tangent to . It turned out that the intersection point of and lies on circumcircle of 
, we have:
satisfy all conditions below:
for all 
for all 
for all
that are solutions of the equation 
.
![\[
2^a + 3^b + 1 = 6^c
\]](http://latex.artofproblemsolving.com/1/9/e/19eb5efa2ccdf54cd277339f8d3b99f75168779d.png)
are such that 
, find the minimum value of expression 
of prime positive integers 
such that number 
equals to a square of an integer.
be real number greater than 
. Let![\[S=\log_a {bc}+\log_b {ca}+\log_c {ab}\]](http://latex.artofproblemsolving.com/5/9/e/59e1a33017116963165f744b49b8515f126c64ef.png)
.
for which:
for all 
.
be a cyclic pentagon inscribed in a circle of centre 
which has angles 
.
and 
meet at a point belonging to the diameter 
.
.
and 
and let 
be the point on the circumcircle such that 
is a diameter. We'll perform some angle chasing before we continue. Let ![\[a=\overarc{AB},b=\overarc{BC},c=\overarc{CD},d=\overarc{DE},e=\overarc{EA},\]](http://latex.artofproblemsolving.com/e/9/4/e9470b7a6c9283a2313703364d7de9904c1ed0c3.png)
. By pentagon properties 
.
Solving we get 
.
.
and 
so 
.
Now we'll consider 
.
.
and 
.
.
so 
.
so:
In order for 
so 
.
.
Which is true. So 