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.
be an acute triangle with 
and let 
be its orthocenter. Let 
be a point on the perpendicular bisector of 
such that 
and and 
are on different sides of 
, 
a point on the perpendicular bisector of 
such that 
and 
a point on the perpendicular bisector of 
such that 
and 
lie on the opposite side of 
w.r.t 
. Prove that 
and are collinear.
square is filled with numbers 
.The numbers inside four 
squares is summed,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?
be a triangle with points 
lie on the perpendicular bisector of such that 
lie on a circle. Suppose 
are perpendicular to sides 
at points 
The tangent lines from points 
to the circumcircle of 
intersects at point 
Prove that: 
are parallel., 
, , 
be four points in the plane, with and on the same side of the line , such that 
and 
. Find the ratio![\[\frac{AB \cdot CD}{AC \cdot BD}, \]](http://latex.artofproblemsolving.com/f/3/c/f3cf9729b32b210faeea6e2d23da3582be2ae061.png)
and 
are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicuar. Thus, proving that the circumcircles of the triangles and are orthogonal is equivalent to proving that the tangents to the circumcircles of the triangles and at the point are perpendicular.)
such that for any positive real numbers 
and 
,
Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
of positive real numbers satisfies ![\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]](http://latex.artofproblemsolving.com/0/3/9/03909a5f60d944063182ec97cf73f283178e8d4d.png)
for every positive integer 
. Prove that 
for every 
.
, 
and 
be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of 
intersects 
a second time at point 
different from A. Define 
and 
analogously. Prove that the circumcenter of 
lies on the Euler line of ABC.
such that 
for all 
of the acute-angled triangle 
is tangent to its side 
at a point 
. Let 
be an altitude of triangle , and let 
be the midpoint of the segment . If 
is the common point of the circle and the line 
(distinct from ), then prove that the incircle and the circumcircle of triangle 
are tangent to each other at the point .
is called lucky if it has at least two distinct prime divisors and can be written in the form:![\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]](http://latex.artofproblemsolving.com/7/4/4/744a5ccaeb9476ebd7d999c395762cb6e99a7a71.png)
where 
are distinct prime numbers that divide . (Note: it is possible that has other prime divisors not among .) Prove that for every prime number 
, there exists a lucky number such that 
.
be a non-isosceles triangle with incenter 
Let 
be the smaller circle through 
tangent to 
and 
(the addition of indices being mod 3). Let 
be the second point of intersection of 
and 
Prove that the circumcentres of the triangles 
are collinear.
is not divisible by 
for any positive integer 
.
is even then 
but 
, impossible, so is odd, and clearly 
fails. Now we claim all 
fail. To see why, note that![\[-1 = \left( \frac{2^n+65}{5} \right) = \left( \frac{5}{2^n+65} \right) = \left( \frac{5^n}{2^n+65} \right),\]](http://latex.artofproblemsolving.com/5/f/8/5f870ab02b5740d8954f5634fb5eadf538ced09d.png)
this in turn equals![\[\left( \frac{3^n}{2^n+65} \right) = \left( \frac{3}{2^n+65} \right) = \left( \frac{2^n+65}{3} \right) = +1,\]](http://latex.artofproblemsolving.com/4/0/f/40fc6dae5bcae9e94e8065f24709306873e1fbe1.png)
. Since 
and 
, we get 
. Therefore, 
for some and , implying 
, so 
. Clearly, 
, so . However, we also have
contradiction. Therefore, 
for all .
. By taking modulo 
, is odd, and it is clear that fails, so assume 
. Therefore, if 
, 
must be a quadratic residue modulo . However, letting 
denote the Jacobi symbol, by quadratic reciprocity we have
contradiction. 
denotes the Jacobi symbol.
fails, assume that 
. Note that 
. even fails, as would divide 
but cannot divide 
. Now assume is odd.
is equal to 
(since 
), since is odd, this implies 
.![\[\left( \frac{5}{2^n + 65} \right) = \left( \frac{2^n + 65}{5} \right) = -1\]](http://latex.artofproblemsolving.com/e/f/8/ef8896e3aa4d52f445ba0f2436f104b7919b63da.png)
and ![\[\left( \frac{3}{2^n + 65} \right) = \left( \frac{2^n + 65}{3} \right) = \left( \frac{1}{3} \right) = 1,\]](http://latex.artofproblemsolving.com/f/d/2/fd2db7a2fddc11e59bb9ee0ff66b6d4f156b46c8.png)
absurd. is to let 
be a prime dividing . Then 
. Since this is true for all primes 
, we have the desired result.
. Assume for contradiction that there exists a solution for 
.
of the fractional expression, it is clear that must be odd (because otherwise divides the denominator but not the numerator). Let denote the Jacobi symbol henceforth.
, and as is odd, it is clear that 
must be a quadratic residue (mod ). But:
a contradiction, and we conclude. 
such that 
. If is even, then is a multiple of , impossible. So is odd. be a prime dividing . Obviously 
. Then![\[p\mid 2^n+65\mid 5^n-3^n\mid 15^n-9^n,\]](http://latex.artofproblemsolving.com/d/1/5/d15b48633e59d860c6db451aac45ba18cc2101f5.png)
so 
is a QR mod .![\[\left(\frac{p}{3}\right)\left(\frac{p}{5}\right)=\left(\frac{15}{p}\right)\left(\frac{p}{15}\right)=(-1)^{(p-1)/2}.\]](http://latex.artofproblemsolving.com/3/6/8/3684b0ac02ea9e40fa68976f9aab6c9ac7e2a05c.png)
Caseworking, we get that mod 
is one of 
. Under further inspection, we note that 
is closed under multiplication mod , so mod is also in . So 
mod is in![\[\{2,6,12,38,44,48,54,56\}.\]](http://latex.artofproblemsolving.com/6/9/7/6971591c7e67473776d844e2db9d6d79c0be1776.png)
Since it must be a multiple of 
and not a multiple of , it must be one of 
. But that means is or mod 
, impossible since is odd. 
and 
with arbitrary odd positive integers, say and , and see how well this problem generalizes. For even, you want 
, so 
; also you want 
. For , you just want 
. For odd with , applying Jacobi symbol, you want that
For a huge convenience, you also want 
, so
We can now pick 
and 
(as in the original problem). In summary..., , 
, , and , then 
does not divide 
for any 
., , , and , then does not divide for any .
such that 
is an integer. By taking mod , must be odd. Clearly, fails. Let 
denote the Jacobi symbol. We then have that 
and 
forces odd. We have for 
![\[\left(\frac{5\cdot 3^{-1}}{2^n+65}\right)=\left(\frac{5}{2^n+65}\right)\left(\frac{3}{2^n+65}\right)=\left(\frac{2^n+65}{5}\right)\left(\frac{2^n+65}{3}\right)=(-1)(1)=-1\]](http://latex.artofproblemsolving.com/8/6/b/86bcd52aecabb2ce6301131cdbe26a6c03e0360f.png)
a contradiction because we need 
.
is even, then 
but 
contradiction. Since 
we get 
odd number.
then we get 
mod 
Now let us introduce the jaccobi symbol. We get
mod 
we get 
mod 
we get 
mod 
and 
so 
a contradiction
is not divisible by for any positive integer .
is the Jacobi symbol such that
If 
Since 
is even:
Since is odd 
is a perfect square
does not divide for any positive integer has to be odd, because if it is even then 
which is a contradiction, 
works. does not work for size reasons. And also even does not work because we get 
, which I think is wrong as far as I know.![\[\left(\frac{5^n}{2^n+65}\right)=\left(\frac{3^n}{2^n+65}\right) \iff \left(\frac{5}{2^n+65}\right)=\left(\frac{3}{2^n+65}\right) \iff \left(\frac{2^n+65}5\right)=\left(\frac{2^n+65}3 \right) \iff \left(\frac{2^n}5 \right)=\left(\frac{2^n-1}3\right)\]](http://latex.artofproblemsolving.com/b/a/f/baf2a4bc9134af58cda52621f20ef9f93ca6550d.png)
which is obviously false for odd because it implies 
but 
, a contradiction.
is even we get a contradiction, so is odd and 
3., and.
is even then 
is divisible by which doesn't divide . is odd. This implies 
.
.
is odd, hence 
. As:![\[\left\{\begin{array}{cl}
2^n \equiv 1&n \mbox{ odd} \\
2^n \equiv 2&n \mbox{ even}
\end{array}\right.\]](http://latex.artofproblemsolving.com/d/c/d/dcdc566ae7b16289b31e9d45aeee702332eb4596.png)
and for 
, we must have odd.
, we have:![\[\left\{\begin{array}{cc}
2^n\equiv2&n\equiv1\pmod{4} \\
2^n\equiv4&n\equiv2\pmod{4} \\
2^n\equiv3&n\equiv3\pmod{4} \\
2^n\equiv1&n\equiv4\pmod{4} \\
\end{array}\right.\]](http://latex.artofproblemsolving.com/e/e/7/ee7b609b9b4db429b144556131c31331ca82a4e2.png)
and as is odd, 
, consequently 
, implying 
is 
.![\[\left( \frac{(\frac{5}{3})}{2^n +65}\right) \equiv \left( \frac{5}{3}\right)\left( \frac{5}{2^n+65}\right)\equiv \left( \frac{2^n+65}{5}\right)\equiv\left( \frac{\{2,3\}}{5}\right)\]](http://latex.artofproblemsolving.com/d/2/6/d26813ee5384d0bbbc1b3cfcd2a912b362dfd5fd.png)
but: ![\[\left( \frac{2}{5}\right), \left( \frac{3}{5}\right)\equiv -1\]](http://latex.artofproblemsolving.com/7/e/1/7e13d0bdd2db23ecc967fb180f329cb14896eae5.png)
therefore as: ![\[\left( \frac{(\frac{5}{3})}{2^n +65})\right)=\prod_{p\mid 2^n +65}\left(\frac{(\frac{5}{3})}{p}\right)\]](http://latex.artofproblemsolving.com/4/6/3/463736310d66caf77486cce83b8296e3640f19aa.png)
there exists 
prime such that 
isn't 
, contradicting the fact that is 
is even, then 
contradiction is odd. Specifically, we can get that 
This implies is a quadratic residue modulo 
However,
which is a contradiction.
works. is odd. is even. Then, 
, but . be a prime factor of . Then,![\[\left(\frac{5}{p}\right) = \left(\frac{5^n}{p}\right) = \left(\frac{3^n}{p}\right) = \left(\frac{3}{p}\right).\]](http://latex.artofproblemsolving.com/0/9/c/09c237c543ca946e56041906c1712aebd75bc730.png)
and 
be positive integers such that 
and every prime factor of is 
and every prime factor of is 
.
. Then, by quadratic reciprocity, 
.
. Then, by quadratic reciprocity, 
.
with 
prime.
, is even.![\[\left(\frac{b}{5}\right) = \prod_{i=1}^k \left(\frac{p_i}{5}\right) = \prod_{i=1}^k -\left(\frac{p_i}{3}\right) = \prod_{i=1}^k \left(\frac{p_i}{3}\right) = \left(\frac{b}{3}\right).\]](http://latex.artofproblemsolving.com/1/1/3/113553e1d14f526fd8aba1b6e163614db2fecd27.png)
By similar logic, 
. Multiplying, we get 
. is odd, the LHS is 
and the RHS is , a contradiction.
is even, and take both expressions modulo .![\[(1) \quad 2^n+65\equiv 1 +65\equiv 66\equiv 0 \pmod 3\]](http://latex.artofproblemsolving.com/4/4/f/44f160b0b70d3a66f88f280cabd0bc076a5f151c.png)
![\[(2) \quad 5^n-3^n\equiv 5^n\not\equiv 0\pmod 3\]](http://latex.artofproblemsolving.com/0/f/6/0f6c037779bacbe7449db71b2cba0b3465df2927.png)
Thus combining 
and 
yields a contradiction.
such that 
such that:![\[\left(\frac{5}{p}\right)\neq \left(\frac{3}{p}\right)\]](http://latex.artofproblemsolving.com/5/d/2/5d241b34ad30783c5de8fbcb6de45e40ff178b1a.png)
We get a contradiction as if and we have that:![\[5^n\equiv 3^n\pmod p\]](http://latex.artofproblemsolving.com/a/1/f/a1f8e5b7d99d04c32d412f29dea30d6f2b99bc63.png)
so we get:![\[\left(\frac{5^n}{p}\right)= \left(\frac{3^n}{p}\right)\]](http://latex.artofproblemsolving.com/0/f/6/0f6337e84dcb46a70644eabd54d30ef9f47b7536.png)
as is odd this means:![\[\left(\frac{5}{p}\right)= \left(\frac{3}{p}\right)\]](http://latex.artofproblemsolving.com/4/1/b/41bb06a290a68e7ce76b5183ef8dfa23e7988053.png)
Which is a contradiction. Now I will show such a exists. Using the jacobi symbol we get:![\[(1)\quad \left(\frac{5}{2^n+65}\right) = \left(\frac{2^n+65}{5}\right)=1\]](http://latex.artofproblemsolving.com/3/1/4/3141385a5fb64356364dd489947fdf3f77114d02.png)
![\[(2)\quad \left(\frac{3}{2^n+65}\right)=\left(\frac{2^n+65}{3}\right)=-1\]](http://latex.artofproblemsolving.com/c/4/9/c49353f476fb0d49d57ab606a13825a4fc97ce26.png)
These follow because is odd and 
. Now let 
.![\[\left(\frac{k}{2^n+65}\right)=\prod_{i=1}^{k}\left(\frac{k}{p_i}\right)^{\alpha_i}\]](http://latex.artofproblemsolving.com/8/3/c/83c6ac529022c2277ab1c07e22084fb7ba6fc01b.png)
If for all such 
we have that![\[\left(\frac{5}{p_i}\right)= \left(\frac{3}{p_i}\right)\]](http://latex.artofproblemsolving.com/f/1/5/f15a789243dd9c822887b108201c67308fb63628.png)
we get that:![\[\left(\frac{5}{2^n+65}\right)= \left(\frac{3}{2^n+65}\right)\]](http://latex.artofproblemsolving.com/2/1/1/2117e35d13e9430c29d5aed6f54cc6ab00206aca.png)
from the definition of the jacobi symbol. Thus this is a contradiction.
,
,
such that 
coprime, 
and 
.
while 
.
for an integer 
.
.![\[ 5^n-3^n\equiv 0\pmod{p}\implies \left(\frac{5}{3}\right)^{2m}\equiv \frac{3}{5}\pmod{p},\]](http://latex.artofproblemsolving.com/e/4/c/e4ccd7a33996cc999516fb455b5b27b5fadff042.png)
where fractions are modular inverses. Hence, 
is a quadratic residue modulo 
by a simple application of quadratic reciprocity. Thus there are 
such that![\[ 2^{2m+1} + 5\equiv 2^n + 65\equiv 7^a11^b17^c(-1)^d\pmod{60}.\]](http://latex.artofproblemsolving.com/2/8/b/28b4319321add23d54f10ab4186b1924c63bdefc.png)
Doing some bashing (it’s not that bad; note that the orders of 
are 
)![\[2^{2m+1} + 5\in\{1, 59, 43, 17, 49, 11, 7, 53\}\pmod{60}.\]](http://latex.artofproblemsolving.com/4/6/5/4650f1cfd1d69df9f5c6a57f072b26236cc20c87.png)
Hence, 
.
are 
,
,
.
is necessary. But it is clear that 
,
and obviously 
.
.
and 
are not quadratic residues modulo 
.
and 
,
,
for which 
.
,
.
,
,
,
be a prime divisor of 
because 
so 
implying that 
since 
.
as well, this implies that 
as well, where 
denotes Legendre symbol. Now, assume for the sake of contradiction that 
However, since ![\begin{align*}
1 &= \left(\frac{1}{p}\right) \\
&= \left(\frac{(5/3)^n}{p}\right) \\
&= \left[\left(\frac{5/3}{p}\right)\right]^n \\
&= (-1)^n \\
&= -1,
\end{align*}](http://latex.artofproblemsolving.com/9/0/d/90d83c9e2d864026414cae59f8dcae5983ed1d9c.png)
a contradiction. Thus, we must have that 
,
.
is a QR iff 
is (this follows since 
)
Thus, we have four cases. Note that 
if 
,
if 
(note that we have assumed that 
)
if 
,
(we have shown 
if 
and 
(we have shown 
.
,
from the above characterizations.
.
from the above characterizations.
.
,
types of odd primes 
such primes of the 
t
so we have that 
so it follows that 
is even. Since 
so we have that 
so 
.
so we have that 
so 
implying that 
is odd. However, since 
is even. Since 
is also even, it follows that 
is even as well, a contradiction.
,
,
.
.
.
,
clearly and 
by 
Now if 
then 
with 
Now we must have 
which is equivalent by QR to 
But 
for odd 
and we can check 
contradiction.
From here on, our solution will consider 
which implies 
or, using Jacobi's notation:
We can easily prove each of there: (1) is from 
= (-1), multiplicity (or rather, power-icity) and 
.
,
.
.![\[
\left(\frac{15}{2^n+65}\right)=\left(\frac{3}{2^n+65}\right) \left(\frac{5}{2^n+65}\right)=\left(\frac{2^n+65}{3}\right) \left(\frac{2^n+65}{5}\right)=\left(\frac{2^n+2}{3}\right) \left(\frac{2^n}{5}\right)=(-1)(1)=-1
\]](http://latex.artofproblemsolving.com/7/3/1/731cf74fac35f050e8a75de4b1ef6a0a0c6052cc.png)