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.
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..
+1 w
be reals such that 
and 
Prove that
Let be reals such that 
and Prove that![$$ a^2+b^2+c^2\geq \frac{3}{ \sqrt[3]{2}}$$](http://latex.artofproblemsolving.com/6/2/2/62211222c148c86d65526d86dde603cb6598925f.png)
![$$ a^2+2b^2+c^2\geq 2\sqrt[3]{4} $$](http://latex.artofproblemsolving.com/3/8/3/383d38cc87c1b1ea1e17e10e4bb95bed79329786.png)
, points 
are chosen such that when going around the triangle, the points occur in the order 
. It is given that![\[
AD_1 = AD_2 = BE_1 = BE_2 = CF_1 = CF_2.
\]](http://latex.artofproblemsolving.com/4/c/f/4cf9e7bc009736836bcd10c1d8ba46bd3a91cd7b.png)
Prove that the perimeters of the triangles formed by the triplets 
and 
are equal.
is call 
if it satisfies the following properties:
and 
for all indices 
.
.
for which: Every sequence, there always exist two terms whose diffence is not less than 
. (where is given positive integer)
into 
or 
can sometimes be quite complex. That's why I’ve written a program to assist with this process.
+1 w
. The tournament adopts an elimination system, and the winner advances to the next round. What is the probability that the Argentine team will meet the Brazilian team in the entire tournament?
. 
. 
. 
. 
. 
Prove that
Let 
Prove that
Let 
Prove that
distinct points in a plane such the circumradius of any 
of them is the same.
, satisfying
are consecutive integers. is an integer.
, satisfying are consecutive integers. is an integer.
, with 
. By Heron's formula, the area of the triangle is 
. In order for this to be an integer, we must have 
for some integer . Using 
, we get that both 
and are even (If is odd, then 
, which cannot result in a quadratic residue of ). Thus, 
is divisible by , so if exists, then is an integer.
since 
. Thus, let 
and 
. Substituting these values in gives us the Pell equation 
. This has fundamental solution 
. Thus, using the recursive formula 
, we get the following solutions: 
. These give the side lengths triplets of 
.
and 
.
Prove that
Where 
