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 and 
its orthocenter. Let 
be a point on the parallel through 
to 
such that 
. Define 
and 
as points on the parallels through 
to 
and through 
to 
similarly. If 
are positioned around the sides of as in the given configuration, prove that are collinear.
be an acute triangle with 
, its orthocenter and 
it's circumcenter. Let be a point on the parallel through to such that 
and and are on different sides of . Denote by 
the intersection of the circumcircle of and 
, where 
is the reflection of over , 
the midpoint of 
, the intersection of 
and the parallel through to 
, the intersection of 
and the perpendicular through to 
and 
a point on such that 
, where 
is the midpoint of 
. Prove that 
lie on a line.
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
, let 
be all positive integers smaller than that are coprime to . Find all 
such that 
for all 
is the largest positive integer that divides both 
and 
. Integers and are coprime if 
.
, which doesn't have right angle, let be its circumcenter. Circle 
intersects and at 
and 
for the second time, respectively. Similarly, circle 
intersects and 
at 
and 
, and circle 
intersects 
and 
at 
and 
for the second time, respectively. Let 
and 
be circumcenters of triangle 
and 
, respectively. Prove that 
are collinear.
of integers is rootiful if, for any positive integer 
and any 
, all integer roots of the polynomial 
are also in . Find all rootiful sets of integers that contain all numbers of the form 
for positive integers and .
be an isosceles triangle with 
. Let 
be a point on 
. Let 
be a point inside the triangle such that 
and![\[
CL \cdot BD = BL \cdot CD.
\]](http://latex.artofproblemsolving.com/1/6/7/16771838c95f86e92f79b8d049e46ab473e6287d.png)
Prove that the circumcenter of triangle 
lies on line 
.
satisfying 
and
for all 
.
be a cyclic quadrilateral with 
being the midpoints of segments 
respectively. Suppose 
is the intersection of diagonals 
of quadrilateral 
Define 
to be the isogonal conjugate point of point in 
Define 
similarly. Suppose 
intersects 
at a point 
Prove that: The Newton-Gauss line of quadrilateral bisects segment 
be an odd prime number. How many -element subsets 
of 
are there, the sum of whose elements is divisible by ?
. Then we get that: 
,so 
and hence 
, which is a contradiction by 
. Then we get that: 
With the same way as case 
we get that 
and hence 
, which is a contradiction by 
.
. Then we get that: 
, and hence 
, so equivalently we have that 
, which has only 
as a positive integer solution. So, in this case 
,which is the only solution of the given equation.
and 
.
.
.
but, 
which is contradiction.
. then, 
.
.
. obviously contradicition.
.
yields 
is even, so 
. Now if 
we have ![\[2x\geq \nu_{2}(2020^x+2^y)=\min\{2x, 2y_1\}=\nu_{2}(2024^z)=3z.\]](http://latex.artofproblemsolving.com/e/6/3/e631c0799f33d41fc92bc875e0e69e3f038c1ba8.png)
However, 
as 
. Therefore 
, and the equation becomes 
. Modulo 
implies 
is odd, at which point 
. Comparing the 
's of both sides gives 
, so 
, 
for some nonnegative integer 
. Clearly, 
leads to the solution 
. When 
we derive a contradiction from:![\[1>\frac{2020^{3t+1}}{2024^{2t+1}}=\frac{505}{506}\cdot \left(\frac{2020^3}{2024^2}\right)^t\geq \frac{505}{506}\cdot \left(\frac{2020^3}{2024^2}\right)>1.\]](http://latex.artofproblemsolving.com/c/b/2/cb24c7a9ef36eabb403e6fa4982babe2ac46b8b8.png)
, which obviously works. We now prove it is the only one.
.
to get 
. Clearly, we must have 
, so is even and 
is odd. Let 
, for a positive integer 
. Then, the given equation becomes 
. Note that 
is always odd. Since 
and 
are both even, it follows if 
, then either 
or 
, neither of which are odd. Thus, 
. Because 
, we must have 
, and our claim follows. is .
. Note that when , 
is divisible by all the prime factors of 
, which are 
, , and 
. If 
, by Zsigmondy, there exists some 
that divides 
, and hence it is impossible that 
. Thus, , and we are done. 
, which clearly works.
gives 
. Since 
, if divided either one of or , then we have that one of 
is 
, which is absurd. Hence 
.
are both not multiples of , so they cannot be equal to 
. If 
, then we would have 
, which is absurd, so 
, so 
. Now we have ![\[ 2020^x + 4^x = 2024^z \]](http://latex.artofproblemsolving.com/4/b/e/4be984ca8fd9e9a9078af51ca885348c8cc030f5.png)
If 
, then by Zsigmondy there exists a prime 
not dividing 
that divides 
, absurd. Hence 
must hold.
Working in modulo 
, we obtain:
is 
for all 
.
is 
is 
and 
.
, where 
have the same parity 
.
Working in modulo , we obtain:
is odd number 
is odd number.
.
by 
results:
.
is an even non-negative number and results: the last digit of 
is 
.
must be , hence 
and we obtain
.
in the last relation, results:
is solution of the equation 
.
. by 
results:
.
is odd number; 
and 
are even numbers, hence the last equation has no solutions.
..
, odd number (see 
).
, even number, hence the equation has no solutions.
, even number, hence the equation has no solutions.
.
, where 
(see ).
.
.
.
, contradiction since
(proved many times in the previous posts).
has the unique solution .
. Because of this, we know y must be even and z must be odd. and 
. Prime factorizing the original equation, 
is the smallest factor, then 
, impossible. Similarly, if 2x is the largest factor of 2, 
is impossible. This means 
and the factor of 2 on the LHS is 
since the v2 of 
is at 2. (since its 0 mod 2 and not 0 mod 4). So we have 
. since 
is an integer, set 
Plugging in and dividing by largest factor of 2 we get 
k=0 works, but any larger k dosent work since 
so the only solution is k=0, which represents 
even and 
odd
we get:
if even then: 
which is a contradiction so 
even we get that 
so our equation transforms into the following:
Claim: 
which is a contradiction because we found that 
Claim: 
on both sides and using Lifting the Exponent Lemma (LTE) we get:
Since 
we get that 
Let 
also 
Now by taking we get:
but this isn't true for 
so 
clearly 
so 
is the only solution 
.
is even
gives is odd
is same as 
nonsense
nonsense
again we get is odd
(for a non-negetive integer k)
for 
for 
, so is even and is odd. Now consider of both sides. The RHS is 
, so it's odd. The LHS is 
, and it's odd if and only if . Now write the LHS as 
, and taking mod 4 on the inside sum gives 
. Now we have the relationship 
. We can easily check that 
is a solution, and we see that for , the LHS is strictly greater than the RHS, so there are no solutions 
.
![\[
2020^x + 2^y = 2024^z
\]](http://latex.artofproblemsolving.com/1/e/5/1e5535fd71c16a98e46030aa8eae8404de4804b5.png)
![\[
v_2(2020^x + 2^y) = v_2(2024^z)
\]](http://latex.artofproblemsolving.com/4/0/3/403f86d0ea5bb9977b3bda19223adb3e9c62260f.png)
![\[
v_2(2020) = 2 \Rightarrow v_2(2020^x) = 2x, \quad v_2(2^y) = y, \quad v_2(2024) = 4 \Rightarrow v_2(2024^z) = 4z
\]](http://latex.artofproblemsolving.com/c/2/0/c20aac682f210f5a4fcc284430d5eee48e3bdf1f.png)
![\[
2020^x + 2^y = 2^y(a + 1), \quad \text{where } a \in \mathbb{N}, \text{ odd}
\]](http://latex.artofproblemsolving.com/0/f/2/0f2148ff5c9fa26d0a4fd71a250d11c8771c6e1d.png)
So,![\[
v_2(2020^x + 2^y) = y = 4z \Rightarrow 2x = 4z \Rightarrow x = 2z
\]](http://latex.artofproblemsolving.com/e/b/1/eb16f0372960c1aed9b43478173ad5131e365f96.png)
![\[
2^{2x} + 1 \equiv 0 \pmod{3} \Rightarrow 2^{2x} \equiv -1 \pmod{3}
\]](http://latex.artofproblemsolving.com/c/e/3/ce36c0bd7856b2d5c2c9807b7c408ccbdfc742ac.png)
![\[
2^{2x} \equiv 1 \pmod{3} \Rightarrow \text{Contradiction}
\]](http://latex.artofproblemsolving.com/d/1/7/d1704603ecc3666f2d0498c3f981936d10fd52c9.png)
![\[
2020^x + 2^y = 2^{2x}(a + 1), \quad \text{where } a \text{ is odd}
\Rightarrow v_2 = 2x
\Rightarrow 4z = 2x \Rightarrow z = \frac{x}{2}
\Rightarrow x \text{ must be even}
\]](http://latex.artofproblemsolving.com/c/e/f/cef5778173248f0b5ba255751ce2b51b2cf5bbe0.png)
![\[
2020^x + 2^{2x} = 2024^z
\Rightarrow 4^x(505^x + 1) = 2^{4z}(253^z)
\]](http://latex.artofproblemsolving.com/7/c/9/7c99831ed753bc3fee358d0b7b10c175d0fce540.png)
![\[
2^{2x}(505^x + 1) = 2^{4z}(253^z)
\Rightarrow 505^x + 1 = 2^{2z}(253^z)
\]](http://latex.artofproblemsolving.com/b/9/3/b93b102a69645ef2f1c7e9211a6bb9be85d1a752.png)
![\[
2020^1 + 2^2 = 2020 + 4 = 2024 = 2024^1 \Rightarrow \text{Valid solution}
\]](http://latex.artofproblemsolving.com/6/8/a/68a1da7eb19ad5b3baff0ef83acc0bb5907a636f.png)
![\[
x = 1,\quad y = 2x = 2,\quad 2x + 1 = 3z \Rightarrow z = \frac{2x + 1}{3} = \frac{3}{3} = 1
\]](http://latex.artofproblemsolving.com/7/6/9/769beb4f38b4409b15dd5e6f7a8373d3011f2812.png)
![\[
2020^2 + 2^4 = 4080400 + 16 = 4080416 \\
2024^2 = 4096576 \Rightarrow \text{LHS} < \text{RHS}
\]](http://latex.artofproblemsolving.com/8/a/a/8aa1b5c9ada5421a8ffcf3e48364666b6a29b81e.png)
![\[
2020^3 + 2^6 = 8204080000 + 64 = 8204080064 \\
2024^3 = 8283809024 \Rightarrow \text{LHS} < \text{RHS}
\]](http://latex.artofproblemsolving.com/f/3/6/f360a4f6f8552bf9d485a5f9c658131fa6ccb364.png)
![\[
\boxed{(x, y, z) = (1, 2, 1)}
\]](http://latex.artofproblemsolving.com/2/2/0/220c534c815104162b53ed786d25a370af5ccfdf.png)
Prove that
Prove that
Prove that
that satisfy the equation
we get:
![$2020^x=2024^z-8^z=2016[...]$](http://latex.artofproblemsolving.com/3/8/e/38eaa7c52b664c59165d40a2823dc5b2d0416c86.png)
contradiction since 
but not 
or 
we get that 
and then from inelyalites esily get 
,
and we claim there are no more solutions. Our strategy relies on that the powers of 
is the minimal among the three, then we have a contradiction via mod 
.
and 
.
.
,
.
(as 
for 
and 
for some positive integer 
.
,
,
.
.
is even, but not divisible by 
.
,
.
,
,
and we claim there are no more solutions. Have in mind that 
and 
.
:
and hence, by computations up to 
or similar, 
for some positive integer 
(we used 
by Fermat's Little Theorem) and hence, by computations up to 
or similar, 
:
,
.
.
now imply that 
and 
and later by 
)