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.
, such that:
.
, 
, with 
, 
. Prove that there exists 
such that for every 
![\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\]](http://latex.artofproblemsolving.com/0/f/c/0fc36d9264eb7e103128c489aeae521a859c1fd4.png)
(For and , the notation 
represents 
. )
are nonzero real numbers such that![\[
\frac{1}{abc} + \frac{1}{a} + \frac{1}{c} = \frac{1}{b},
\]](http://latex.artofproblemsolving.com/6/d/9/6d9aabd8130b99d7128de9a3f063eb4c7124febf.png)
let 
be the maximum value of the expression![\[
\frac{4}{a^2 + 1} + \frac{4}{b^2 + 1} + \frac{7}{c^2 + 1}.
\]](http://latex.artofproblemsolving.com/a/b/2/ab28769126365574d4c0108500c370826082de6b.png)
Determine the sum of the numerator and denominator of the simplified fraction representing .
be a circle and 
a line in its plane. Let 
be a point on , located outside of . Let 
and 
be the tangents from to , where 
and 
are distinct points on . Let 
and 
be two points on . Lines 
and 
intersect line in two points 
and respectively 
. Lines 
and 
intersect the second time circle in points 
and 
. Prove that the tangents from and to are concurrent on line .
and 
be the roots of the equation 
where 
is an odd integer. Let 
is an integer
) = 1 .
denote the sum of all integers 
such that 
and exactly one of 
and 
is divisible by 
. Compute the remainder when is divided by 
.
is given in the plane. The incircle of triangle 
has center 
and is tangent to diagonal 
at 
. Let 
be the center of parallelogram . Show that 
and 
are parallel.
is valid while 
is not)
of positive integers such that 
is an integer.
are positive integers such that for every natural number the set of prime factors of 
are same. prove that 
.
square is dissected in to 9 rectangles by lines parallel to its sides such that all these rectangles have integer sides. Prove that there are always two congruent rectangles.
Prove that
Where 
such that 
?
there is such an 
prime factors.
works for every 
.
s.t. for 
has a prime factor 
larger than 
.
.
,
has a prime factor 
.
,
(I'm using the well known and easy to prove fact that the only positive integer solution 
to 
is 
)
.
has a prime factor 
which is coprime to 
,
,
will have exactly 
.
?
.
.
(mod 
(mod 
.
,
(mod 
,
(mod 
(mod 
(mod 
(mod 
(mod 
is not divisible by 
positive integers, such that 
and 
is not a 2-power.
such that
is the canonical factoring of 
works for the base case. Now, we will show that we can repeat an algorithm so that we can find a prime factor, 
,
,
and 
has one more prime divisor than before. We use induction to find this algorithm. For the base case 
is a factor of 
,
,
.
.
.
of both factored terms by computing the remainder of the second when divided by the first. Plug 
in, so we get that the remainder is 
and 
.
.
.
divides 
.
,
.
,
once. Yet, 
for 
,
prime divisiors, whereas 
has only one 
)
(vide grobber's solution, which I do not completely understand yet, but it seems very nice and simple)
but 
for 
unless 
.
so set 
for 
.
.
for which we arrive at 
.
for 
.
with 
for 
and 
we can find 
with 
By Zsigmondy's there exists a prime 
such that ![\[q\mid 2^{9\cdot p_1\cdot p_2\cdots p_k}+1, q\nmid 2^{9\cdot p_1\cdot p_2\cdots p_{k-1}}+1\]](http://latex.artofproblemsolving.com/f/d/2/fd2b4f0263ecc5c947fc7d1125ba6e0341fb90cd.png)
Set 
which by Zsigmondy's doesn't equal 
for 
.
using the inductive hypothesis and Chinese Remainder Theorem. We are done.
.
is Korean. Now, the problem is solved upon the following claim:
is Korean, there exists a prime 
is Korean too.
,
.
it follows then that 
.
since 
;
by Chinese Theorem, since 
.
and wlog let 
.
and so by the minimality of 
we get that 
or 
.
which is absurd while in the second case we get 
.
and so by the choice of 
.
(
.
(
all satisfay the condition.Now,it is enough to prove that numbers of the form 
have infinitely many primes dividing them,but this is easy to prove,since we have for 
divides 
so suppose opposite.Now,we just need to prove that 
and 
can't have identical sets of primes for 
,
is at most 
for 
satisfies the condition. We aim to find a prime 
satisfies the condition and 
,![\[2^{pn} \equiv -1 \pmod{p} \implies 2^{2n} \equiv 1, 2^n \not\equiv 1 \pmod{p}.\]](http://latex.artofproblemsolving.com/b/f/5/bf534b924dcbfe7538a8f8bd6d95020c358ce7dc.png)
with the desired 2000 prime divisors. 
with exactly ![\[n=p_1^{e_1}p_2^{e_2}\dots p_k^{e_k},\]](http://latex.artofproblemsolving.com/b/c/6/bc665170dca5ca340fa4429a2d3c42f084fdf0d3.png)
such that 
,
such that for all positive integers 
,
satisfies the problem condition, where we can make 
,
also satisfies problem conditions. This will then prove the part where we can make ![\[n\mid 2^n+1 \iff \nu_q(n)\leq \nu_q(2^n+1),\]](http://latex.artofproblemsolving.com/f/1/c/f1c4e7b144fe2b6ee39f4c93d8a45b3e99af3763.png)
and by LTE (which we can use, since 
)![\[\nu_q(nq)=1+\nu_q(n)\leq 1+\nu_q(2^n+1)=\nu_q(2^{nq}+1),\]](http://latex.artofproblemsolving.com/e/6/3/e63dddcbdaa94ecd251ff4d264da6806a375e6d0.png)
which means that 
also satisfies problem conditions.
such that for all positive integers 
and 
as ![\[r\mid 2^n+1 \iff ord_r(2)\mid 2n,\]](http://latex.artofproblemsolving.com/8/4/d/84ddbdb1d93aa00f7c27c9e2a3547ed6d4058e20.png)
and since 
,
.
for some 
we can choose such that 
.
also satisfies problem conditions. This, combined with 
satisfies problem conditions for any 
we chose. We can do this by making another ![\[\nu_r(rn)=1\leq \nu_r(2^{rn}+1)=\nu_r(2^n+1)+1,\]](http://latex.artofproblemsolving.com/6/b/f/6bffd0c3287661e0940e064fb4b3a78b881500f6.png)
which we can do since we know that 
and 
,
.
,
.
distinct prime factors, then there also exists such an 
distinct prime factors. By repeating this logic we can find an 
distinct prime factors.
.
,
that does not divide any 
for smaller 
has order 
modulo 
and thus cannot divide 
and consequently 
.
has 
more distinct prime factor than 
;
that is not a factor of 
,
.
.
,
for some 
satisfying 
.
is a factor of 
.![\[ \nu_{p_j}(2^{n_{i+1}} + 1) = \nu_{p_j}(2^{n_j} + 1) + \nu_{p_j}\left(\frac{n_{i+1}}{n_j}\right) = \nu_{p_j}(n_{i + 1}) + \nu_{p_j}(2^{n_j}+1) - \nu_{p_j}(n_j). \]](http://latex.artofproblemsolving.com/9/7/2/9722c24beea4031b9eeaf19c7b8a906ca07e4fff.png)
However, the strong inductive hypothesis implies 
,
.
.
,
.
.![\[ \nu_{p_j}(2^{n_i} + 1) = \nu_{p_j}(2^{n_j} + 1) + \nu_{p_j}(n_i) - \nu_{p_j}(n_j) = \nu_{p_j}(2^{n_j} + 1) + \nu_{p_j}(n_i). \]](http://latex.artofproblemsolving.com/5/2/4/52413dd03b3c0ddc60176c54fc5fc0e314b6a436.png)
We can raise ![\[ {p_j}^{\nu_{p_j}(2^{n_i} + 1)} = {p_j}^{\nu_{p_j}(2^{n_j} + 1)} {p_j}^{\nu_{p_j}(n_i)}. \]](http://latex.artofproblemsolving.com/5/5/3/553eb07b4731dbf21b7a9b1f8083dc10187b6545.png)
Doing this for every prime factor ![\[ 2^{n_i} + 1 = n_i \prod_p p^{\nu_p(2^{n_{j(p)}} + 1)}. \]](http://latex.artofproblemsolving.com/3/8/9/389528853cc0fe16384359c3ec65acebca9f3ec1.png)
For all 
.![\[ 2^{n_i} + 1 \leq n_i \prod_p (2^{n_{j(p)}} + 1). \]](http://latex.artofproblemsolving.com/c/7/8/c7898a2cbf855263c6ce748e039891c1663f710e.png)
Since 
is unique and in the set 
.![\[ 2^{n_i} + 1 \leq n_i \prod_{j=0}^{i-1} (2^{n_j} + 1). \]](http://latex.artofproblemsolving.com/e/f/7/ef7ae22bff327250bde2ca088a115ceaf97aa7a1.png)
Taking the log base ![\[ n_i < \log _2 (2^{n_i} + 1) \leq \log _2 n_i + \sum _{j=0}^{i-1} \log_2(2^{n_j} + 1) < \log_2 n_i + \sum _{j=0}^{i-1} (n_j + 1) = \log_2 n_i + i + \sum_{j=0}^{i-1}n_j. \]](http://latex.artofproblemsolving.com/a/0/8/a087e20f7612b331731fd664dfd9d35d9f933d00.png)
Now, in order to achieve a bound on 
for all 
.
for all 
.![\[ n_i < \log_2 n_i + i + \frac12 n_i \leq \log_2 n_i + \log_3 n_i + \frac12 n_i \iff n_i < 2(\log _2 n_i + \log _3 n_i). \]](http://latex.artofproblemsolving.com/0/8/5/085143f0233a99c4b56faba183d4f9cd66636bdf.png)
This inequality obviously cannot be satisfied as 
?
where 
are distinct primes such that 
.
,
.
,
,
.
,
For 
,
.
,
so we are done.
and 
,
,
,
and 
and 
are coprime, then 
.
are odd.
By LTE, 
.
and all three numbers are pairwise coprime, 
.
in the following manner. Let 
(so that 
)
,
,
all 
are distinct,
and 
for each 
and the greatest common divisor of the terms in the parenthesis is 
has all its prime divisors distinct from 
and 
,
,
as well.
for each 
.
.
.
for each 
.
,
if 
iff 
.
.
for which 
and 
.
.
,
.
.
and 
by Zsigmondy.
times on 
.
,
has a prime divisor greater than 
.
,
,
by Mihailescu's theorem. Since 
dividing 
,
and 
.
has a prime divisor 
are coprime, 
.
,
,
times finishes. 
there exists some positive integer 
.
for which 
.![\[2^{3n_r}+1=2^{3^{r+1}\cdot p_1p_2\dots p_{r}} +1\]](http://latex.artofproblemsolving.com/0/d/a/0da22d119e6c029f8126edc8641714309b294897.png)
Now, by Zsigmondy's Theorem there exists a prime 
but 
.
.![\[3n_r \mid 2^{n_r}+1 \mid 2^{3n_r\cdot p_{r+1}}+1\]](http://latex.artofproblemsolving.com/d/b/7/db771902368843b8965316e6e935469ac5c7d695.png)
since by Lifting the Exponent Lemma, 
.![\[2^{3n_r\cdot p_{r+1}}+1 \equiv 2^{3n_r}+1 \equiv 0 \pmod{p_{r+1}}\]](http://latex.artofproblemsolving.com/0/a/f/0afa2cd6f6f7bbd562f07253e3128aca42b85d24.png)
by construction. Thus, 
so we can let 
which completes the induction and proves the result.
so that 
divides 
.
.
is also divisible by 
be a construction for 
is a construction for 
has a valid construction. Let 
be a prime dividing 
such that 
I will prove the existence of such a 
and 
Thus, since 
there must exist a non-3 value greater than 1 that divides 
By picking a 
is a valid construction for 
We have that 
and 
instead