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such that 
for all 
is called 
if there exists a permutation 
of the numbers 
, such that:
and 
have different parities for every 
;
the sum 
is a quadratic residue modulo 
for every 
.
, let 
. Point 
lies on segment 
such that 
. Let 
and 
be the circumcenter and incenter of triangle 
, respectively. Let 
and 
be the circumcenter and incenter of triangle 
, respectively. Suppose that lines 
and 
intersect at point 
. Find the length of 
.
fair coins are flipped. If 
of them land heads, find the expected value of 
.
is such that 
for any positive integers 
. Assume there exists a positive integer such that 
for all positive integers 
. Determine all possible values of 
.
is such that for any positive integers . Assume there exists a positive integer such that for all positive integers . Determine all possible values of .
and 
Prove that
and 
.
Let 
.
satisfies the condition and 
.
.
and 
,
for all 
.
,
for all non-negative integers 
and 
,![\[
g(a, b) + g(c, b) = g(ac, b^2).
\]](http://latex.artofproblemsolving.com/9/1/d/91dd66cfb45616ccfaa1691e942e816060decf3d.png)
denote the assumption on 
in this equation. For 
,
depends only on the value of 
for all prime numbers 
.![\[
h(x) = \sum_{p \text{ prime}} h(p)\nu_p(x). \tag{2}
\]](http://latex.artofproblemsolving.com/6/8/b/68b83a00528564c6a54d5e2ac13d438b4266babf.png)
or 
.![\[
h(45) = g(45, 1) = 2g(3, 1) + g(5, 1) = 0.
\]](http://latex.artofproblemsolving.com/8/e/0/8e02c396ee210441ab72d69160d7f57de8ab2805.png)
and 
follows that 
.
we get that 
.
from 
,
,
,
.
,![\[
g(n, x) = g(n, x^n) + (x^2 - 1)h(x) + (x^4 - 1)h(x) + \ldots.
\]](http://latex.artofproblemsolving.com/4/4/b/44b899e553765ac801138cdc87c7900ea3c368da.png)
for some 
,![\[
g(n, x) + (x^2 - 1)h(x) = g(n, x^2) \le g(n, (x + 1)^2) = g(n, x + 1) + (x^2 - 1)h(x + 1).
\]](http://latex.artofproblemsolving.com/9/d/0/9d07a67f85a3666f1bb39273106e7079e9f29a88.png)
,
are constants. Therefore 
for all 
from the beginning of the problem to case 
.
or 
,
for every prime 
because 
is monotonically increasing. Let 
.
,
(the case 
is analogous). Then we have![\[
h(a^k) = kh(5^k)(7) = kh(5)(7) = h(b^k).
\]](http://latex.artofproblemsolving.com/0/1/7/017694b47f0bd9755bf3e2458aaea70e3f458104.png)
.
,![\[
q + 1 = \frac{a^{q + 1} - 1}{a - 1} > \frac{b^{q + 1} - 1}{b - 1} > b^q + 1.
\]](http://latex.artofproblemsolving.com/f/5/f/f5fdfd3799129cc139d96f18b8d683e47f815745.png)
.
,
and 
.