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invertible, for an infinite number of 
:
such that 
is a scalar
![$\{x_n\} \subseteq [0,1]$](http://latex.artofproblemsolving.com/8/e/4/8e4206325387485d73765c9d7070ecb1ce18ff60.png)
, there exists some constant 
, for any 
, among the sequence 
and 
could be chosen to satisfy 
and 
.
and 
be positive integers. Prove that![\[
a^2 + \left\lceil \frac{4a^2}{b} \right\rceil
\]](http://latex.artofproblemsolving.com/5/6/2/562d003cf06ca46031cec9e4e968f8b67784225f.png)
is not a perfect square.
matrices
matrices matrices
, if f ' (0)=p and f ' (5)=q, then prove f ' (-5) = q
is never a square of an integer.
are complex numbers with absolute value 
. For 
define 
. Given that the sequence 
converges. Show that,
.
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 .
Then 
is
![\[
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
\frac{1 - e^{-2} \cos\left(2\left(u + \tan u\right)\right)}
{1 - 2e^{-2} \cos\left(2\left(u + \tan u\right)\right) + e^{-4}}
\, \mathrm{d}u
\]](http://latex.artofproblemsolving.com/9/1/1/91165e6b531233c56b799ed32ded982d57b6c87a.png)
satisfies the condition and 
.
.
and 
,
for all 
.
,
for all non-negative integers ![\[
g(a, b) + g(c, b) = g(ac, b^2).
\]](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 
.