AoPS Academy
+4 w
[/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.
+1 w
and 
, let us consider the equation
[list=a]
and 
, find the least positive integer 
satisfying this equation. and , there exist infinitely many positive integers satisfying this equation.
denotes the greatest common divisor of positive integers 
and 
.)
, and 
, a triangle can be formed. Could it happen that from segments of lengths 
a right-angled triangle can be formed? for which it is possible to arrange all divisors of that are greater than 1 in a circle so that no two adjacent divisors are relatively prime.
has side lengths , , and . Select a point 
inside , and construct the incenters of 
, 
, and 
and denote them as 
, 
, 
. What is the maximum area of the triangle 
?
be a triangle with circumcentre 
. The points 
and 
are interior points of the sides 
and 
respectively. Let 
and 
be the midpoints of the segments 
and 
. respectively, and let 
be the circle passing through and . Suppose that the line is tangent to the circle . Prove that 
satisfying the following conditions:
for all 
Define the set ![\[S = \left\{ \left( \sum_{i=1}^{10} a_i x_i, \sum_{i=1}^{10} a_i y_i \right) : a_1, a_2, \cdots, a_{10} \in \{0, 1\} \right\},\]](http://latex.artofproblemsolving.com/d/e/b/deb50830b9fe10c86ea169c044c1b26f515b2514.png)
then 
，and any rectangular strip of width 1 covers at most two points of S.
be non-negative real numbers such that 
Prove that
consecutive positive integers were written on the board, where 
. Then some 
of the written numbers were erased, and it turned out that any two of the remaining numbers are coprime. Find the largest possible value of .
, diagonals 
and 
are equal, and they intersect at 
. Perpendicular bisectors of 
and 
intersect at point lying inside triangle 
, and perpendicular bisectors of 
and 
intersect at point 
lying inside triangle 
. Prove that 
.
be a positive integer. Egor has 
cards with the number “
” written on them, and cards with the number “
” written on them. Egor wants to paint each card red or blue so that no subset of cards of the same color has the sum of the numbers equal to . Find the greatest such that Egor will not be able to paint the cards in such a way.
. Prove that for every nonnegative integer , there is an odd number and a parabolic triangle with vertices at three distinct points with integer coordinates with area 
.
set the vertices at 
,
,
.
set the vertices at 
,
,
.
is odd for all integers 
,
,
.
,
,
and 
.
and 
.![$A = \frac{1}{2}(2^{2n+1})(2^{2n+1}+1)^{4} = [2^{n}(2^{2n+1}+1)^{2}]^{2}$](http://latex.artofproblemsolving.com/7/d/2/7d2fcd982bd4dd4ed41b4bd92c44bc7ca87d5f49.png)
.
,
,
.
.
instead of 
because it looks nicer 
which as area 1. For 
,
.
from now on.
By Shoelace, the area is 
When 
,
WLOG 
,
all perfect squares, so we need a Pythagorean triple. Amusingly, we can do the following:
with 
Then, assign 
as 
,
,
so the area would be 
so 
Therefore, we just need to show that for any positive integer 
of 
as one of the sides, and making the other sides 
which will both be odd for 
then
then 
then 
then 
?
,
is 
.
for each 
.
.
,
.
,
but multiply both 
.
,
,
,
and 
,
.
.
,
,
,
and 
we get 
.
in the first case, 
in the second case, and 
in the last case, we are done. (Whether we are given 
,
or 
.
,
,
with 
,
can be![\[(4^n, 4^{2n-1}+1, 4^{2n-1}-1).\]](http://latex.artofproblemsolving.com/2/e/e/2eeb37232b12e3b362324c6bbff9cf6a4e0f335d.png)
[for 
,
as the vertices]
![\[a(b^2-a^2)=4^n\cdot m^2\]](http://latex.artofproblemsolving.com/2/5/0/2507fc28e3413202d65a73cfbb1f67ed5cb82a07.png)
by Shoelace (actually this isn't necessary at all lol just use normal base x height). Let 
for “convenience”. So 
,
.
.
and 
,
,
.
holds, so we are done. 
has area 
.
and 
have areas 
and 
,
is a complete residue class modulo 3, hence, for given 
is a perfect square and 
,
.
Let the difference 
be equivalent to 
and let 
.
.
and 
for some odd integer 
.
.
,
.
a solution to 
It is well known that more solutions 
can be generated by raising 
to the power of a nonnegative integer, including the 
t
.
to the power of a power of 2. When we have a solution 
,
gives us the solution pair 
Note that this increases the 2-adic valuation of 
by one, and as 
,
.
,
,
,
.![$[XYZ]= \lvert 2a^3- 2ab^2\rvert$](http://latex.artofproblemsolving.com/d/c/4/dc411657bfd0dfcd22728b8615a4629afeb8342c.png)
“above” 
so that the area is ![$[XYZ]=2ab^2-2a^3$](http://latex.artofproblemsolving.com/d/6/b/d6b1daa3e571c7abea6b4af8177629f6c76b58d5.png)
Set 
and calculating, we get 
.
.
,
,
.
,
and the parabolic triangle with vertices 
has area
,
and the parabolic triangle with vertices 
has area
,
and the parabolic triangle with vertices 
has area
![\[Area = \dfrac{1}{2} \left|(ab^2 + bc^2 + ca^2) - (ba^2 + cb^2 + ac^2) \right|\]](http://latex.artofproblemsolving.com/0/5/c/05c774fe2270ec3e4ed8da0ad522f9193590213f.png)
which is equal to :- 
and 
= 
,
and 
= 
and so on is given by the determinant![\[\frac 12 \begin{vmatrix} x_1 & x_1^2 & 1 \\ x_2 & x_2^2 & 1 \\ x_3 & x_3^2 & 1 \end{vmatrix} = \frac 12(x_1-x_2)(x_2-x_3)(x_3-x_1).\]](http://latex.artofproblemsolving.com/d/e/f/def1dd5fd136c90809b6988b8f54b25a2e5be059.png)
Now take 
and 
,![\[\frac 12 (x_1-x_2)(x_2-x_3)(x_3-x_1) = m^2\left(\frac{m^2-2}2\right)^2 \left(\frac{m^2+2}2\right)^2.\]](http://latex.artofproblemsolving.com/a/1/9/a194504dcf92097bd934b761986d6893303b602e.png)
By varying 
,
.
![\[\left\{ \begin{array}{ll}
P=(2^{2n}(2^{2n+1}+1), 2^{4n}(2^{2n+1}+1)^2) \\
Q=(-2^{2n}(2^{2n+1}+1), 2^{4n}(2^{2n+1}+1)^2) \\
R=((2^{2n}+1)(2^{2n+1}+1), (2^{2n}+1)^2(2^{2n+1}+1)^2)
\end{array}\right. \]](http://latex.artofproblemsolving.com/1/6/7/16776fe8fee02bb87481fec488c2701d5ee6a57b.png)
to get ![$[PQR]=(2^n(2^{2n+1}+1))^2$](http://latex.artofproblemsolving.com/b/a/f/baf385b8652ca50fb4f8856209a16132dd0c982f.png)
,