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and 
, there cannot exist infinitely many 
such that![\[
\sqrt{n + a} + \sqrt{n + b} \in \mathbb{Z}.
\]](http://latex.artofproblemsolving.com/5/4/d/54d20546de5736770a9929ac86163e3b06553b7d.png)
![\[
x = \sqrt{n + a} + \sqrt{n + b} \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/6/6/8/668d53723d307556536736fda0e4a19be848834f.png)
![\[
x^2 = (\sqrt{n + a} + \sqrt{n + b})^2 = (n + a) + (n + b) + 2\sqrt{(n + a)(n + b)}.
\]](http://latex.artofproblemsolving.com/1/9/f/19f2627a432354a757d3c9e8ba42e96f83f46726.png)
So:![\[
x^2 = 2n + a + b + 2\sqrt{(n + a)(n + b)}.
\]](http://latex.artofproblemsolving.com/0/7/3/073602693604bbf3cb1f5c1c2e5051e8994a5e78.png)
![\[
\sqrt{(n + a)(n + b)} \in \mathbb{N}.
\]](http://latex.artofproblemsolving.com/1/1/4/11441fab3ba768fee7a0eef4dbf68124d6637aec.png)
![\[
(n + a)(n + b) = k^2.
\]](http://latex.artofproblemsolving.com/0/e/c/0ec5e7bac2aebe9ec507727ed21f7ca028655222.png)
Assume 
. Then we have:![\[
n + a \mid k \quad \text{and} \quad k \mid n + b,
\]](http://latex.artofproblemsolving.com/7/5/f/75fc52b6662df3a4c0cb2f40f49c4b3351c90a4d.png)
or it could also be that 
.![\[
(n + a)k_1 = k \quad \text{and} \quad kk_2 = n + b.
\]](http://latex.artofproblemsolving.com/1/f/5/1f5936241f7bebcf15c68fed5267bbfad2d40c2b.png)
![\[
k_1 k_2 = \frac{n + b}{n + a}.
\]](http://latex.artofproblemsolving.com/d/4/c/d4c1958e3554b58c3def282070f849842cc10912.png)
, we have:![\[
k_1 k_2 = 1 + \frac{b - a}{n + a}.
\]](http://latex.artofproblemsolving.com/3/0/6/306a6a00c44d4e4db1b36c7e588f684f28fdbf44.png)
, 
must be an integer, which is not possible..
of the integers with the following property: for any integer 
there is exactly one solution of 
with 
.
be an acute triangle with incentre 
and 
. Let lines 
and 
intersect the circumcircle of at 
and 
, respectively. Consider points 
and 
such that 
and 
are parallelograms (with 
, and 
). Let 
be the point of intersection of lines 
and 
. Prove that points 
, and are concyclic.
, find all functions 
such that![\[f(x+y+f(y))=f(x)+f(ay)\]](http://latex.artofproblemsolving.com/7/5/d/75d1ac71fa0c2e53fdfc8e198b109d18e935fe74.png)
holds for all 
.
and 
, in 
is isosceles and 
but due to cyclic quadrilateral 
so, 
Now, note that 
(ASA - congruency axiom). ![\(\Rightarrow CG = PD \Rightarrow \Delta BFC \cong \Delta APC \Rightarrow \angle BCF = \angle ACP \Rightarrow \measuredangle {[BC,PC]} = \measuredangle {B/2} \quad \square\)](http://latex.artofproblemsolving.com/4/3/c/43c7d8dd365e97eb07248ae928fc9a1bd6d948e0.png)
we have
So, 
With
we get
Hence, 
which is enough to show that 
and , in is isosceles and but due to cyclic quadrilateral so, Now, note that (ASA - congruency axiom). 
and 
? It seems like a really arbitrary point one would consider?
is midpoint of arc 
.
=>
are colinear.
=> triangle 
and 
are similar =>
=>
is parallelogram =>
be the feet of the internal, external angle-bisectors of 
onto 
Let be the incenter of 
and 
be the midpoint of 
It's well-known that 
is an isosceles triangle.
is a parallelogram.
be the point so that 
is a parallelogram. Then we have 
, and so 
is an isosceles triangle. Hence, we have that 
However, 
means that we should have 
if the problem is to be believed, which would uniquely determine as the incenter 
being a parallelogram, which we'll now show.
such that 
Let 
We have that 
where 
denotes the point at infinity along line 
, and so the lemma is proven.
Similarity that 
, and so in particular 
With the lemma, we now conclude that is a parallelogram, as desired.
,
meet sides 
respectively. The tangent at 
to the circumcircle of 
extended at 
.
.
.
. 
.