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 natural number. We write out the fractions 
, 
, 
, 
such that they are all in their simplest form. Let the sum of the numerators be 
. For what 
is one of and 
odd, but the other is even?
be a natural number. We write out the fractions , , , such that they are all in their simplest form. Let the sum of the numerators be . For what is one of and odd, but the other is even?
be the sum of the positive integers less than or equal to and co-prime to . Since 
, by pairing, 
. Then note that, 
and for any 
, 
. Therefore, 
. Since 
and 
are multiplicative, so is 
.
All numbers below are supposed to be nonnegative integers.
has the same parity for 
thus it's suffice to prove, that the sum 
is odd. Note that 
Hence there are exactly 
suitable values of 
which is odd. The conclusion follows.
will have an even numerator if and only if its 2-adic valuation is strictly greater than the 2-adic valuation of n.
.
an
with 
odd.
.
a
will always have different parities. Therefore all n are solutions.
if 
, where 
is even and 
is odd.