be an isosceles trapezoid with 
.
be the midpoint of 
.
and 
the circumcircles of the triangles 
and 
,
be the crossing point of the tangent to 
with the tangent to 
.
is tangent to 
is a positive integer. Call all positive divisors of 
and 
beautiful divisors and difference of any 
we have 
.
.
,
,
,
,
meet the sides 
,
,
at points 
,
,
,
be the incenter of triangle 
,
be the foot of the perpendicular from 
.
is cyclic, then the center of its circumcircle lies on segment 
.
Y by Amir Hossein, blackbluecar, jasperE3, megarnie, puntre, Adventure10, Mango247, deplasmanyollari, ohiorizzler1434
Find the smallest positive integer 
or show no such exists, with the following property: there are infinitely many distinct -tuples of positive rational numbers 
such that both
are integers.
or show no such exists, with the following property: there are infinitely many distinct -tuples of positive rational numbers 
such that both
are integers.This post has been edited 1 time. Last edited by Amir Hossein, Jul 10, 2018, 4:20 PM
Reason: Fixed.
Reason: Fixed.
fails is not too difficult, and proving 
can be shown by simple vieta jumping.
fails. Let 
where 
.
therefore 
.
hence 
.
,
therefore 
.
has infinitely solution 
as 
is one of them and Vieta tells us that if 
is solution then so is 
.
It's easy to check that 
and 
which are both integer. Hence we are done.
isn't very hard so I'll omit soz)![\[ab+\frac{a}{a+b}+\frac{b}{a+b}\in \mathbb{Z}\]](http://latex.artofproblemsolving.com/a/1/e/a1e88a6bf32303d28ce0866a59197eadb5db0eda.png)
![\[\frac{1}{ab}+\frac{a+b}{a}+\frac{a+b}{b}\in \mathbb{Z}.\]](http://latex.artofproblemsolving.com/2/e/1/2e1c4a5b67d02515937f439c067be1e49e6abf13.png)
The first one is trivially true; the second one would be true if 
.
coprime. Then you would need 
and 
.
is a solution to this. Now for a certain 
,
is also a solution, and so ![\[\frac{(\frac{a^2+1}{b})^2+1}{a},\frac{a^2+1}{b}\]](http://latex.artofproblemsolving.com/5/9/e/59e0efed13094f4f6a67a77953aca997db8754ee.png)
is also a solution. All that remains is to show that if 
,
clearly fails.
and 
for integers 
and 
.
with 
;![\[ q = \frac{1}{a} + \frac{1}{p-a} = \frac{p}{a(p-a)} = \frac{py^2}{x(py-x)}. \]](http://latex.artofproblemsolving.com/5/d/d/5dde780764a848692f2479f9c4a322577fbdfb2e.png)
Then 
,
and obtain ![\[ q = \frac{xk \cdot y^2}{x^2(ky-1)} = \frac{ky^2}{x(ky-1)}. \]](http://latex.artofproblemsolving.com/c/8/6/c86950147bcf01456884a52f8f58cf3d20cb6151.png)
As 
,
,
.
then 
,
,
or 
.
,
and 
are both half-integers, and so we conclude 
.![\[ \left( \frac{1}{1+x+y}, \frac{x}{1+x+y}, \frac{y}{1+x+y} \right) \]](http://latex.artofproblemsolving.com/9/d/8/9d83ab856ebbdc2efa30659177d787c4889bd9fb.png)
where 
,
,
are positive integers (in fact if we pick 
this is WLOG). Then it suffices that ![\[ \frac{1+y}{x} + \frac{1+x}{y} \in {\mathbb Z} \]](http://latex.artofproblemsolving.com/4/b/3/4b3e8bcbcebfddaa4b45f0f3edad8cb0dd9c9e1f.png)
which is a famous MOP 2007 problem solved by Vieta jumping (there are in fact infinitely many with 
)
.
.
clearly only 
works. For 
let 
and 
with 
.
divides 
and 
divides 
.
divides 
and because 
it follows that 
.
,
and 
,
,
and 
are both integers giving the pairs 
and 
.
with 
and 
an integer (this actually follows if the original problem is true, so it can't make the problem false and it adds a nice bit of structure). Let 
and the analogous where 
are positive integers. Then
to be an integer. We now impose 
,
to be an integer. Assume that for some positive integers 
this ratio is equal to 
,
.
,
then this other root is an integer greater than 
that gives the ratio 
.
satisfies the desired equality with 
,
Why not say he was a co proposer?
It is easy to see that 
where 
.![\[a_1+a_2+a_3=xy+\frac{2xy+x^2+y^2}{xy-1}=xy+2+\frac{x^2+y^2+2}{xy-1}.\]](http://latex.artofproblemsolving.com/2/3/7/23790948a6c379b7186720dd7b48bf3a3e93f5d3.png)
Hence we want infinitely many 
such that 
Define the sequences 
and 
such that
It is easy to check by induction that all the pairs 
work. Also, the value 
is strictly increasing. Thus, we get infinitely many pairs 
And we are done. 
which easily we can choose 
which such in this.
integer.
doesn't ensure 
be integer and 
.
must be an integer. You only take such that 
is an integer
be integer then should 
also be integer which means
.
fail.
to be a solution, and we achieve this by reducing special cases (for example, I choose 
as my special case) into a generalized pell equation on which I impose a few parity restrictions to get a standard pell equation so that I can write solutions in terms of the fundamental solution (specifically for this one it was 
)
and 
must be integers, so 
is the only solution.
and 
with 
and 
to be an integer. From this we obtain 
,
and similarly 
being an integer, we also need 
are integers, which only has finitely many solutions.
such that:
and
are integers.
.
and 
(note that 
if 
then consider the good pair 
so 
then 
works.
We have 
and 
so we can divide both 
by 
and 
where 
is an integer. We have 
must be an integer, so 
.
We know that 
becomes 
becomes 
Now, let 
so 
is an integer. Thus, we want 
integer. Let 
be a valid pair then 
so 
has integer solutions. We have 
is a valid solution, and bounding confirms: if 
then 
as long as 
This, it comes down to finding a valid 
so we're done.
and 
where 
.![\[ \frac{p}{q}+\frac{r}{s} = \frac{ps+rq}{pq} \in \mathbb{N} \text{ and } \frac{q}{p}+\frac{s}{r} = \frac{ps+rq}{rs} \in \mathbb{N}\]](http://latex.artofproblemsolving.com/9/0/a/90ab30247bc820303993ca39e53130bd9bcdb8f4.png)
So, 
,
all divide 
.![\[ p \vert ps+rq \implies p \vert rq \implies p \vert r \text{ but } r \vert ps+rq \implies p \vert ps \implies r \vert p\]](http://latex.artofproblemsolving.com/f/1/0/f1034ce1999a62022a1755abdca3f24b3851fd14.png)
So, 
.
.![\[ \frac{1}{a_1}+\frac{1}{a_2} = \frac{2}{a_1} \in \mathbb{N} \implies a_1=\frac{2}{k} \]](http://latex.artofproblemsolving.com/1/3/a/13af42dd08f3118228ccd12a0d0c36a64773bfb4.png)
for some 
.![\[ a_1+a_2 = 2a_1 = \frac{4}{k} \in \mathbb{N}\]](http://latex.artofproblemsolving.com/2/9/7/2975a774c277763fb517c02941ec59a081b1af3f.png)
So, there is only finitely many choices of ![\[ (a_1,a_2,a_3) = \left (\frac{1}{m+n+1}, \frac{m}{m+n+1}, \frac{n}{m+n+1} \right ) \]](http://latex.artofproblemsolving.com/7/f/2/7f26078d5a0bbd592e474ef165e69e198284be28.png)
This boils down to proving ![\[ \frac{m+1}{n} + \frac{n+1}{m} \in \mathbb{N} \]](http://latex.artofproblemsolving.com/f/9/a/f9ab3659959c64771dddc49063106563c8978cb3.png)
for infinitely many pairs 
.![\[ \frac{m+1}{n}+\frac{n+1}{m} =4 \]](http://latex.artofproblemsolving.com/4/9/b/49b7a875a63ad486bc338eac63474889b3f4dee9.png)
infinitely often. Indeed, notice that 
works. Now assume that there exists some pair 
is maximized. Assume wlog 
![\[ \frac{m+1}{n}+\frac{n+1}{m} =4 \implies m^2-(4n-1)m+n^2+n=0 \]](http://latex.artofproblemsolving.com/b/1/1/b11395e681b906744711fe1159e5d1fffb28de3a.png)
which is a quadratic in terms of 
.
be the other root. First, notice that 
.
.![\[ n^2+n = mm_1 \leq m^2 \implies n<m \]](http://latex.artofproblemsolving.com/2/1/8/2180febb28f8cfa3745db8c8a98eef1ff61a3bf6.png)
Contradiction.
.
and 
in reduced form. Then 
and 
.
,
.
with 
.
,
.
a positive integer.
,
.
is an integer, but as 
,
(as in the 
which obviously has finite solutions. We claim the answer is n=3, constructed by 
,
by Vieta jumping 
.
for positive integers 
.
for 
.
for 
.
,
,
.
.
is an integer, but this is trivial by Vieta Jumping.
.
such that 
and 
are both integers. Let 
and 
.
and 
are integers. Thus there are finitely many pairs of positive rationals 
and 
are both integers. Take 
.
and we only have to prove that 
is an integer. Thus we only need to consider the following claim:
is an integer.
.
and assume 
.
and let 
other than 
and 
,
, so if 
is a solution, then 
is also solution and 
implies we can generate infinitely many solutions from 
and we can generate infinitely many solutions from 
.
.
,
.
that 
.
and 
.
,
.
,
)
.
.
:
.
.
.
are roots, and 
,
.![\[b_0 = \frac{c^2 + c}{b} > \frac{c^2 + c}{c} > c.\]](http://latex.artofproblemsolving.com/2/3/7/2379a750a4bcae85e8f37fb426004fe16aa10ed6.png)
Then 
,
is not possible is easy
integer we just let 
.
to make that integer [also making 
distinct]
, NOW this is an old HMMT problem which was solved by infinite descent.
and 
then 
.
,
so we want 
.
are coprime we want 
and similarly, we want 
.
and 
then 
,
which means 
and hence 
.
forced and hence 
,
,
and 
where 
is integer, and we show that 
is integer.
.
is integer, and that integer has to be 3.
,
.
is a solution and so is 
.![\[(1,1)\longrightarrow(1,2)\longrightarrow(5,2)\longrightarrow(5,13)\longrightarrow(34,13)\longrightarrow\cdots\]](http://latex.artofproblemsolving.com/9/1/4/914c3130bc96ed4ed707b35424a80f00b7eb9cd8.png)
So, 
is an integer, but taking 
with a prime dividing 
or 
implies noninteger, so 
and we can see 
which is finite.
.![\[F_k+F_{k+4}=F_k+F_{k+2}+F_{k+3}=F_k+F_{k+1}+2F_{k+2}=3F_{k+2}\]](http://latex.artofproblemsolving.com/a/1/4/a14ae197af5ec2c9a417613f3213323bd2199e7f.png)
.
for odd 
works since 
.![\[-1=F_{k+2}^2-F_kF_{k+4}=F_{k+2}^2-(3F_{k+2}-F_{k+4})F_{k+4}=F_{k+4}^2-3F_{k+2}F_{k+4}+F_{k+2}^2=F_{k+4}^2-(3F_{k+4}-F_{k+2})F_{k+2}=F_{k+4}^2-F_{k+2}F_{k+6},\]](http://latex.artofproblemsolving.com/8/4/4/84478d647f025399aaadf498204117e3baa06e50.png)
completing the inductive step.![\[(a_1,a_2,a_3)=\left((F_k+F_{k+2})(F_{k+2}+F_{k+4}),\frac{F_k}{F_{k+2}}+1,\frac{F_{k+4}}{F_{k+2}}+1\right)\]](http://latex.artofproblemsolving.com/1/7/2/172b75b0a7d8822c00ef5595f70e9d6a78fe133f.png)
works, which finishes.
