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.
such that :
for all prime numbers 
.
for all positive integers 
such that 
, draw exterior equilateral triangles on sides 
and 
to obtain 
and 
, respectively. Let 
be the intersection of the altitude through 
and the median through 
. Let 
be the intersection of the altitude through 
and line 
. Let 
be the intersection of the median through and the line 
. Prove that , , and all lie on a line.
+1 w
be an even positive integer. A sequence of real numbers is called complete if for every integer 
with 
either the sum of the first terms of the sum or the sum of the last terms is integral. Determine numbers.
be a triangle inscribed in a circle, where the tangents to the circle at points and intersect at the point 
. Let 
be a point on the arc (not containing ) such that 
and 
. Let the lines 
and 
intersect at point 
. Let 
be the reflection of with respect to the line . The lines 
and 
intersect at point 
, and intersects the circumcircle of again at point 
. lies on the circumcircle of 
.
of positive integers with 
is called good if we can colour each integer red, blue, green or purple, in such a way that
of each 
consecutive integers at least one is coloured red;
of each 
consecutive integers at least one is coloured blue;
of each 
consecutive integers at least one is coloured green;
of each 
consecutive integers at least one is coloured purple.
are positive real numbers satisfied 
. Prove that:
is written on the blackboard. Every time, Carmela can erase the number on the black board and replace it with a new number , if and only if 
is a perfect square. Prove or disprove that all positive integers 
can be written exactly once on the blackboard.
be a square with sides length 
. Let 
be a path within which does not meet itself and which is composed of line segments 
with 
. Suppose that for every point on the boundary of there is a point of at a distance from no greater than 
. Prove that there are two points and of such that the distance between and is not greater than 
and the length of the part of which lies between and is not smaller than 
.
and the sets 
If 
has exactly three elements, prove that cannot be an integer.
be a positive integer not divisible by 3. Prove that there are infinitely many positive integers such that 
and 
are divisible by , where 
is the sum of digits of 
.
books are given as prizes to students (each students gets one book). In how many ways can the organisers give these prizes if they have copies of one book and 
other books each in one copy?
. Show that there exists 
such that![\[a_1x_1^2+a_2x_2^2+\ldots +a_nx_n^2\geqslant (a_1x_1+a_2x_2+\ldots +a_nx_n)^2.\]](http://latex.artofproblemsolving.com/4/d/2/4d269e5b682be6778e89237aa91da5e0399042a7.png)
so that the sum of its edges' lengths is minumum.
of positive integers, with prime, such that 
is a perfect square.
![\[p^a+a^4=k^2\iff p^a=(k-a^2)(k+a^2)\]](http://latex.artofproblemsolving.com/e/8/1/e81469624e6363ce2428ad6caee80ade5579af7f.png)
and 
![\[2a^2=p^{a-x}-p^x=p^x(p^{a-2x}-1)\]](http://latex.artofproblemsolving.com/7/a/b/7ab4e383cf69539a52ee61cd13c8f2697843f206.png)
Let 
where 
.![\[2b^2+1=p^{a-2x}=p^{p^{\frac{x}{2}}b-2x}\]](http://latex.artofproblemsolving.com/e/a/6/ea64238a167e99195ebeafe6abad07325d67a22d.png)
But![\[2b^2+1=p^{p^{\frac{x}{2}}b-2x}\geq p^{5^{\frac{x}{2}}b-2x}\geq p^b\geq 5^b>2b^2+1\]](http://latex.artofproblemsolving.com/f/8/1/f8171171c7276c18d5e4d7ae80c364757d28bc7a.png)
Since 
and 
which can be proved by induction.
![\[3^a=(k-a^2)(k+a^2)\]](http://latex.artofproblemsolving.com/7/b/4/7b49532179b6aa0478101edbc3450da95ec7e73f.png)
and 
![\[2a^2=3^x(3^{a-2x}-1)\]](http://latex.artofproblemsolving.com/3/3/3/3338a893ba2b33f60e8dc94f1c819a4da50a42e0.png)
Denote 
We have![\[2b^2+1=3^{3^{\frac{x}{2}}b-2x}\]](http://latex.artofproblemsolving.com/4/1/a/41aa77ca656cf35924fd602526fb4050cc4a9689.png)
If 
then 
which gives 
so we get 
. If 
then 
Thus we get 
.![\[3^b> b^2+1=3^{3^{\frac{x}{2}}b-2x}\implies 2x>b(3^{\frac{x}{2}}-1)\geq 3(3^{\frac{x}{2}}-1)\geq 6x\]](http://latex.artofproblemsolving.com/6/5/1/651dea063a9315e4f5bc216179407e791a4a6e4c.png)
Which is impossible.
Let 
![\[2a^2=2^x(2^{a-2x}-1\iff a^2=2^{x-1}(2^{a-2x}-1)\]](http://latex.artofproblemsolving.com/9/9/2/9922a000a982eb882fe5165f6ccddbf64a105e3d.png)
Let 
and 
![\[b^2+1=2^{2^mb-4m-2}\]](http://latex.artofproblemsolving.com/4/2/8/42801367d5fa53cd69af502b8fe5794455e3eed3.png)
For 
we have 
which gives no solution and for 
we have 
which gives no solution again. If 
then 
hence![\[2^{b+1}\geq b^2+1=2^{2^mb-4m-2}>2^{(4m+2)(b-1)}\geq 2^{12(b-1)}\implies b+1\geq 12(b-1)\implies b=1\]](http://latex.artofproblemsolving.com/0/4/0/040138cae30840152b87053b049e0cc1e7130421.png)
Which gives a contradiction.
then 
Then set 
where are nonnegative integers with 
and 
We also get 
so we are trying to solve 
We get ![\[2^{y-x}-1=\frac{(x+y)^2}{2^{x-1}}.\]](http://latex.artofproblemsolving.com/0/8/6/0865b71094f80a5d635139b7c2afd9caf28d916d.png)
Then 
gives 
is even, so 
is a square and an integer. Thus we have 
by Mihailescu, so plugging this in we have 
Now the right side is odd, so 
which doesn't work. Thus there are no solutions in this case.
we have 
By 
we have even, so 
and 
are both squares. However, 
Now it is well known that 
is a QR 
iff 
and 
is a QR iff 
Thus 
is a QR iff 
Thus 
we have ![\[8y^2=2(y+y)^2>2(x+y)^2=p^y-p^x\ge p^y-p^{y-1}\ge10p^{y-1}>8p^{y-1},\]](http://latex.artofproblemsolving.com/2/b/4/2b42de99359708532605727a8a7defe719f6407c.png)
so 
But for 
we have 
and 
grows faster than 
since 
but 
for 
Thus there are no solutions in this case.
We have ![\[8y^2>2(x+y)^2=3^y-3^x\ge 3^y-3^{y-1}=2\cdot 3^{y-1},\]](http://latex.artofproblemsolving.com/9/3/d/93d61c6d6e7f7ef7c14183f1e51add2a5ba6cffb.png)
so 
This is true for 
but false for 
and thus false for 
since 
for 
so the right side grows faster than the left. and 
Checking, we get only 
work. Using 
we get 
These work, so we are done.
: 
, 
, 
.
is equivalent to 
, hence (as is prime) 
and 
for some 
. Thus 
. In the right-hand side the first time decreases as increases, while the second term increases, hence![\[ 2a^2 \geq p^{a - \left \lfloor \frac{a-1}{2} \right\rfloor} - p^{\left \lfloor \frac{a-1}{2} \right\rfloor} = p^{\left \lfloor \frac{a-1}{2} \right\rfloor}(p^{a - 2\left \lfloor \frac{a-1}{2} \right\rfloor} - 1). \]](http://latex.artofproblemsolving.com/a/a/c/aacb0979922670f726eca81330861a4c2f4efd39.png)
, then in we can have only 
, so 
. If 
, then again and 
yield .
and 
. In the above inequality the right-hand side is at least 
, so necessarily 
. This is false for 
(direct check) and for 
by induction: the base case is checked directly, and if 
, then multiplying with 
(the right-hand side is indeed not more than 
) we obtain what we want. If 
in we have 
and 
, but none of and yields a solution (in the first case 
, in the second one divides the right-hand side, so 
which does not work). and . In if , then 
, impossible since the right-hand side is larger by induction. If 
, then 
divides 
and so is divisible by . For the main expression is 
, for 
is 
, for 
it is 
. For 
in the main inequality the right-hand side is at least 
, but actually 
by induction for 
(for 
we have 
and the step follows by 
). and . In the main inequality the right-hand side is at least 
, so 
, false for 
(by induction, since 
and 
). For 
the expression 
equals 
, 
, 
, 
, 
, 
, 
, 
, 
, 
, 
, 
, 
ends at , impossible for a square; 
and 
is not a square; 
ends at , impossible for a square; 
and is not a square; 
ends at , impossible for a square; 
, the second multiplier is divisible by 
, but not by 
; 
is divisible by , but not by 
; 
, where 
is not a square. Therefore there is no solution for .
by factoring out 
. The second bracket is strictly positive, so the first one must be as well. We now consider several cases for 
:
. Then 
. The right-hand side is odd, so 
. Now 
for 
(by induction, or taking the derivative of 
, etc.), and 
lead to the solutions 
.
and 
. Then 
and 
, so 
and 
. As 
, these imply 
and 
. Now if 
and 
, and 
, 
, then we can write the initial equation as: ![\[p^{a_1p^{A}} + a_1^4 p^{4A} = x_1^2 p^{2X}.\]](http://latex.artofproblemsolving.com/f/1/7/f17932be217bd7f6046c970ff2091d6aa0ca9f2c.png)
As 
is even, so must be 
. However, if we assume that we don't just have 
, then 
, 
, and 
, at which point 
, which is odd, contradiction. Therefore, 
and so![\[4A = \nu_p(p^a+a^4) = \nu_p(x^2) = 2X \Longrightarrow X = 2A.\]](http://latex.artofproblemsolving.com/a/b/0/ab0feecc73d90ef02d9bfee81317d705f1c3442f.png)
Dividing both sides by 
leaves 
. Now if 
, then 
as well, so 
and 
, contradiction with the assumption that and from before. Hence 
and so![\[p^{a_1p^{A} - 4A} = 2a_1^2+1.\]](http://latex.artofproblemsolving.com/b/4/6/b460f74d1070b4456c4bf399f36e91d12c0ec5d2.png)
If 
, then 
. Checking 
leads to the solutions 
and 
, respectively.
and 
. Here 
and 
for some positive integers 
, and so 
. Diving by until we can, we must get to 
for some 
. However, 
for 
, so we must have 
. This leads to 
, or 
. Plugging this into the original equation gives 
, so 
. At this point, is some power of two, say 
, but 
, which is impossible due to modulo issues again. Thus, there are no solutions in this case, and we're done.
.
be a pair of integers satisfying the equation 
for some integer 
. Then we have:![\[ p^a = (k - a^2)(k + a^2) = p^m \cdot p^{a-m} \]](http://latex.artofproblemsolving.com/0/c/7/0c7b1a7fd5c654b5931eebed262337ee804cca35.png)
where 
, 
, and 
is a non-negative integer.![\[ p^{a-m} - p^m = 2a^2 \]](http://latex.artofproblemsolving.com/e/6/0/e60b1ace58b5d24a9839ff0ad5010f7df4cbcce0.png)
![\[ p^m(p^{a-2m} - 1) = 2a^2 \]](http://latex.artofproblemsolving.com/8/6/6/866d7a08ac83e6bc66aa644702063c0f5915a731.png)
and 
, leading to 
, implying 
.
and 
for some positive integer 
and non-negative integer 
. Hence,![\[ \frac{p^{a-2m} - 1}{2} = c^2 = \frac{p^{c \cdot p^b - 4b} - 1}{2} \]](http://latex.artofproblemsolving.com/f/8/2/f82baaf31a633e204b579f3e9c8b0c7bf1576701.png)
, we have 
. Considering as an integer, the inequality 
is always true. Furthermore, since , we have 
, which results in:![\[ c \geqslant 1 \geqslant \frac{4b - 1}{p^b - 1} \implies cp^b - c \geqslant 4b - 1 \implies cp^b - 4b \geqslant c - 1 \]](http://latex.artofproblemsolving.com/7/1/9/719f3bef54f67065c3bef7a1dd10595c99c3b6ab.png)
, we obtain:![\[ \frac{1}{2}(p^{cp^b - 4b} - 1) \geqslant \frac{1}{2}(p^{c - 1} - 1) \implies c^2 \geqslant \frac{1}{2}(p^{c - 1} - 1) \geqslant \frac{1}{2}(3^{c - 1} - 1) \]](http://latex.artofproblemsolving.com/7/6/1/761688aa4b1e3637c1d0f853c0c62eb4119abb2b.png)
, the inequality 
holds true. are in the range 
. Let's consider each case:
If 
, then 
, which implies 
and 
. Thus, 
or 
. If , then 
, satisfying 
, a perfect square. If , then 
, satisfying 
, also a perfect square. If 
, then 
, implying and 
. Thus, or 
. If , then 
, satisfying 
, a perfect square. If , then 
, satisfying 
, also a perfect square. If 
, then 
leads to no integer solutions. If 
, then 
results in 
, but 
is not a power of any prime number, hence no solutions exist in this case.
that satisfy all conditions are 
. Each pair has been verified to meet all criteria.
.
so we have that![\[p^a = (x-a^2)(x+a^2) \implies x-a^2 = p^k, x+a^2 = p^{a-k} \implies a^2 = \frac{p^{a-k} - p^k}2.\]](http://latex.artofproblemsolving.com/e/9/0/e90d8850c1c3714b1126b05bfa9c426db8857894.png)
, we get the only solution is 
. We will proceed with casework.
.
where 
. If 
, we have 
unless or which we verify to not work. We have that 
where the last inequality holds unless 
and 
, but the remaining case can be checked to fail. So there are no solutions here..
unless 
and 
but this does not result in a solution. So we get that 
This reduces to a case check for 
which results in solutions 
..
, then 
which clearly has no solutions. So we can get that 
so 
is odd. Let 
and 
, we get that 
unless 
or 
. But these cases give no solution as the first implies 
and the second means 
and this does not give a solution.
. be a positive integer such that 
. We can set 
giving 
. We thus must have 
or 
. Thus we must have 
. Checking 
gives no solutions and one can finish with induction showing the LHS is larger.
and we must have 
for an non-negative 
. The case 
clearly gives no solutions. Then we must have 
Notice that for 
we have 
. Thus we must have 
. Thus we must have either 
or 
and . the equality simplifies to 
so we must have 
which has solutions at 
and 
giving as and 
. the equality simplifies to 
so we must have 
which has solutions at and 
giving as 
and 
.
. We have 
> If 
, then 
. We have that 
, for 
. Say 
for 
. Then, 
, and 
, so for 
, now we simply have to check 
, which yield 
as solutions.
. Then 
. Let 
and 
, where 
. Let:![\[
k - a^2 = p^{\gamma}, \quad k + a^2 = p^{a - \gamma}
\]](http://latex.artofproblemsolving.com/8/f/7/8f7ad5951a246940446f490855f2b44751d5cd8d.png)
First, note that 
. We have 
, and WLOG 
which gives 
. If 
, then 
since 
, a contradiction (as it's not equal to 1). Therefore 
, and 
.
. Note that 
, so 
. If 
, then 
as well which yields 
a contradiction therefore 
, resulting in 
, and 
, and since 
, we have 
. Since , 
, and therefore 
.
for odd primes and 
. If we prove 
, we would be done. Note that for 
, 
, so say we are done for 
. Then, 
, since 
, and 
. Therefore, we are simply reduced to a case check from 
.
and we are only considering odd at the moment, so for 
, not possible. For 
, which gives 
, which works. For 
, no odd primes divide it. For , 
and this fails. For 
and 
, so this works. For 
, and this fails. For 
, no odd primes divide it. For , we fail. So, we have the solutions 
as well.. We have 
. Since 
, we have 
, which means 
is a perfect square therefore 
is a perfect square as well. However, this is not possible if 
, as then 
, so 
, which gives 
, impossible by parity (clearly 
).
.
, or 
. Let 
and 
, since 
, 
.. Then, 
, 
is a perfect square, so 
, by taking modulo . Therefore, 
, which gives 
, or 
, which clearly has no solutions.. Then![\[p^\beta - p^\alpha = 2a^2 \ge (p - 1)p^{\frac{a - 1}{2}}.\]](http://latex.artofproblemsolving.com/3/1/d/31d178aaf01fab75d11d6ff6d47eed7886de5d64.png)
Now we proceed with a massive cases bash:: then 
. If , 
works. If 
, 
works. If , fails. If , 
fails. If , 
fails. If 
, works. If , 
fails. If , 
fails. If 
, 
works.: then 
. If , 
fails. If , 
fails. If , 
fails. If , 
fails.
: All fail.
, or 
. So both factors are powers of , i.e.![\[2a^2 = p^r - p^{a-r} \geq \begin{cases} p^{a/2 + 1} - p^{a/2 - 1} = p^{a/2 - 1} (p^2-1) & \text{$a$ even} \\ p^{(a+1)/2} - p^{(a-1)/2} = p^{(a-1)/2} (p-1) & \text{$a$ odd} \end{cases}\]](http://latex.artofproblemsolving.com/9/4/3/9435a04212294811d08ec6681386bc37c7cd0f92.png)
Okay, now time for the bash., the inequality holds only for 
, and no cases yield valid solutions. If , check 
, which yields and . If 
, no values of work., check 
; for , check 
, yielding and ; for , check ; for 
we have 
which clearly will not yield any solutions.
, 
, 
, 
in total.
we get that 
thus both 
and 
are powers of so:![\[n-a^2\mid n+a^2\]](http://latex.artofproblemsolving.com/9/3/8/9383df286eca61e06aa1986baabfcc82bb601d8d.png)
Thus:![\[n-a^2\mid n+a^2+n-a^2=2n\]](http://latex.artofproblemsolving.com/f/1/4/f1440a4cc182d818cd453ccb0752d91cee418b8a.png)
Thus if 
and 
we get 
. This means we must also have 
, thus 
, thus we can divide
to obtain 
we can repeat the above argument until we get values 
and 
where 
and where we have 
or . If we get 
, suppose 
and 
, thus we have that 
and we also have that 
for all so or , if we have 
which is clearly not possible so is the only possible case. If 
we have that 
and also that 
so 
thus when we get the solutions and and when we get the solutions and . If 
we get 
so 
this is clearly not possible because of catalans unless 
, if we get however this means 
which is clearly not possible.
Thus, 
and assume the factors are 
where we require 
Thus, we have 
Clearly, 
are the only possible values
We have Define the function 
where 
We have that 
Hence, 
is decreasing over the interval. Clearly note that 
and thus 
Thus, for integers 
we get 
for 
Observe that from taking modulo 
we get 
However,
which is a contradiction as the expression evaluates to for 
let 
satisfy 
Hence, we require 
However, by catalan, we require 
and thus 
However, we have that is a power of 2 while it is an odd number, and so 
This doesn't work by testing it.
We have 
(recall the lower bound we established earlier). However, 
for 
Simple testing tells us that 
work.
gives solution . Now we consider 
.
Note 
some power of since . Now, 
. Write 
where 
.
(since 
)
, contradiction.
but for 
and gives solution 
. If 
we get solution 
. If 
, we get solution . Now consider 
. If 
(note the relation cant be that of equality): trivially 
, a contradiction. So, let 
where 
. Trivially 
(trivially 
for 
) 
, a contradiction from 
.
. Let 
(since 
) 
, contradiction by zsigmondy
. Let 
where 
. So, 
. When 
this becomes subcase 1. Trivially, 
. If 
then trivially, except for 
, . Case working thru 
gives no solution.
