[/quote]
,
of natural numbers such that for every pair of natural numbers 
and 
,![\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]](http://latex.artofproblemsolving.com/1/6/7/167679c1707b957d87311298ea5b72347a9bdc45.png)
such that 
for all 
such that for all real numbers 
and 
,![\[f(xf(y))+f(y)=f(x+y)+f(xy).\]](http://latex.artofproblemsolving.com/d/2/f/d2ff6ed448cf39e7ec9bce6b944965d5e89b9878.png)
table filled with natural numbers, we say it is a divisor table if:
-
,
-
,
for every 
.
is given. Determine the smallest natural number 
,
to tangent 
disjoint from 
on 
and 
.
is tangent to two fixed circles.
be a triangle and let 
be its incircle. Denote by 
and 
the points where 
and 
,
and 
the points on sides 
and 
,
the point of intersection of segments 
and 
.
is denoted by 
.
.
![\begin{align*}
\frac{a}{c} &= b-\sqrt{b}+c \\
\sqrt{\frac{a}{c}} &= \sqrt{b}+1 \\
\sqrt[4]{\frac{a}{c}} &=\sqrt[3]{b}-1
\end{align*}](http://latex.artofproblemsolving.com/0/c/d/0cd22d36acedbcb10d76d1b884aa35418273ed52.png)
be positive integers. Prove that the expression![\[\frac{\gcd(a+b,ab)}{\gcd(a,b)}\]](http://latex.artofproblemsolving.com/1/b/7/1b78a2027c3b7156fbb566901d2cfd088cc09f00.png)
is always a positive integer, and determine all possible values it can take.
be a convex pentagon such that 
,
,
.
to 
are concurrent.
that satisfies: for all different 
,
?
and internal bisector 
.
again at 
and 
cuts 
.
lie on 
.
at 
respectively. Let circumcircle of 
cuts 
.
.
satisfying
Prove
the following equation :
be prime numbers. Prove that if 
is a perfect square, then 
is also a perfect square.
Y by mmkkll, feranjos, CrazyFok, Littlelame, emmelin, KhaiMathAddict, cubres, ItsBesi
Determine all composite integers 
that satisfy the following property: if 
, 
, 
, 
are all the positive divisors of 
with 
, then 
divides 
for every 
.
that satisfy the following property: if 
, 
, 
, 
are all the positive divisors of 
with 
, then 
divides 
for every 
.This post has been edited 7 times. Last edited by v_Enhance, Sep 18, 2023, 12:33 AM
Reason: minor latex pet peeve
Reason: minor latex pet peeve
be the smallest two prime divisors of 
.
for some 
,
and 
for some nonnegative integer 
.
,
and 
which is a contradiction.
.
,
and hence 
and hence 
and hence 
and hence 
and hence 
and hence 
etc.
and hence 
is a prime power which indeed works.
one can get that 
so 
,
in the same way, which leads to 
and 
for prime 
so 
and this is contradiction
for distinct prime divisors 
for some prime divisor ![\[\frac{\frac{n}{p} + n}{\frac{n}{q}} = \frac{q(p+1)}{p},\]](http://latex.artofproblemsolving.com/0/5/c/05c967bf45ae5d33ffba891e1d9c652bc72acc03.png)
obviously not an integer.
and 
or 
;![\[ \frac{\frac{n}{p} + \frac{n}{p^2}}{\frac{n}{q}} = \frac{q(p+1)}{p^2}, \]](http://latex.artofproblemsolving.com/5/0/5/5059120c1c8d485fe7447ba05c92c0a5593df531.png)
obviously not an integer either.
.
Then, for some 
(specifically, the smallest 
)![\[\frac{n}{q}\large \mid\frac{n}{p^k}+\frac{n}{p^{k-1}}=\frac{n}{p^k}(k+1).\]](http://latex.artofproblemsolving.com/a/d/5/ad52ef76dad656f90976900413d51f04a0a18c4a.png)
This means that for some integer 
,![\[\frac{n\cdot l}{q}=\frac{n}{p^k}(p+1) \iff l\cdot p^k=pq+q\]](http://latex.artofproblemsolving.com/6/8/1/68157585856dc13142624a2d5fc58ee9edff7020.png)
so 
.
distinct prime factors as desired.
for all primes 
,
.![\[
\frac{n}{q} \mid \frac{n}{p^{k-i}} + \frac{n}{p^{k-i-1}},
\]](http://latex.artofproblemsolving.com/4/0/a/40aaa0ef28c0aca6ab6f5bb96699eb71b6d2046a.png)
but this is obviously false. We are done.
for distinct prime divisors 
for some prime divisor 
or 
;
.
is divisible by 
,
divides 
.
.
.
,
for some 
,
.
so 
are both divisible by 
is not.
for primes 
.
which is not an integer.
which are not integers. Thus only 
for some primes 
and some 
.
and hence
Which is impossible unless 
,
to be an integer, which is impossible.
,
and 
be the first two primes of 
be the largest number, such that 
![\[\frac{n}{p_2},\frac{n}{p_1^s},\frac{n}{p_1^{s-1}}\]](http://latex.artofproblemsolving.com/7/a/2/7a2e24917485ff73d99d64ee636ca0a168b0b005.png)
are consecutive divisors of 
Which is a contradiction, hence 
,
,
on both sides, the left side has 
,
,
This does not hold true
for some prime 
.
, 
.![\[d_i=p_{i-1}\mid p_i + p_{i+1}=d_{i+1}+d_{i+2}\]](http://latex.artofproblemsolving.com/d/7/e/d7ec55f74aa6c1f73d60b87f3a29dbf4713c30d3.png)
quite clearly.
for all 
for some 
and 
as the divisors are arranged in increasing order. Now, if 
which is a very clear contradiction since 
are both primes. Thus, there are no such 
where 
work evidently. We now show these are the only ones.
.
![$$\implies \boxed{d_{i}d_{i+1} \mid d_{i+2}(d_{i}+d_{i+1})} [\because i \mapsto m-i-1]$$](http://latex.artofproblemsolving.com/c/1/6/c16495b4b20ccabcb4b9e38e9198166de8adc88a.png)
Now 
gives, 
since 
.
is a prime number (say 
has any prime factor 
which is a blatant contradiction. Thus 
as it can not be 
otherwise 
again, a contradiction. Now by an easy strong induction and similar argument we can conclude that 
.
where 
)
.
-
:
,
.
,
and 
and 
,
.
and 
,
.
,
.
so we must have 
.
.
or 
.
,
,
for a prime 
,
,
is just 
,
always divides 
.
and 
.
.
,
.
to be an integer, but 
is clearly not an integer. There are no cases here.
for 
.
must be an integer, but this is clearly impossible.
be the smallest two prime divisors of 
Let 
be the two consecutive divisors. Hence, we get 
so 
or 
All perfect powers work.
so 
.
,
.
(easy to see why)
,
with 
are the only solutions (easy to check why these work).
and 
(assume 
as the case 
is trivial). One may notice that 
,
and 
.
From the above we say 
,
Hence applying this argument in general yields 
So 
So if some prime 
where 
So there is no other prime 
dividing 
It is easy to see that this works.
