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 integer. We call a pair of integers 
good if ![\[0\leqslant a<k,\hspace{0.2cm} 0<b \hspace{1cm} \text{and} \hspace{1cm} (a+b)^2=ka+b\]](http://latex.artofproblemsolving.com/e/8/d/e8df562e024b8476d69fdaea15338c27bb6c4eb8.png)
Prove that the number of good pairs is a power of 
.
written on the first page and 
empty pages. Suppose in each of the next seconds, Alice can flip to the next page, and write down the sum or product of two numbers (possibly the same) which are already written in her grimoire.
be the largest possible number such that for any 
, Alice can write down the number 
on the last page of her grimoire. Prove that there exists a positive integer 
such that for all 
, we have that ![\[n^{0.99n}\leqslant F(n)\leqslant n^{1.01n}.\]](http://latex.artofproblemsolving.com/5/7/e/57e2b7e3d97564665d8de28a3afd74bdcf74ba5b.png)
be a triangle with incenter 
. Let segment 
intersect the incircle of triangle at point 
. Suppose that line 
is perpendicular to line 
. Let 
be a point such that 
. Point 
lies on segment such that the circumcircle of triangle 
is tangent to line 
. Point 
lies on line 
such that 
. Prove that 
.
, 
, 
, 
be distinct positive integers, 
. Prove that there exist distinct indices 
and 
such that 
does not divide any of the numbers 
, 
, , 
.
, let 
and 
be a pair of isogonal conjugate points. The line 
intersects 
at 
, and the line 
intersects at 
. Let the circumcircle of 
and the circumcircle of 
intersect again at 
(other than ). Prove that the line 
bisects 
.
is right angled (for clarity, 
). Define 
as a point on line such that 
is parallel to 
. Additionally, since 
, 
meaning 
, 
, and 
. Also, since 
, by SSS, we have 
meaning 
. Since 
, we have 
or 
and we are done.
from the reals to the reals such that![\[f\left(f(x)+y\right)=2x+f\left(f(y)-x\right)\]](http://latex.artofproblemsolving.com/e/b/4/eb4e0d6580d9d3d6a627e39837edb3d1943d4596.png)
.
is called 
if
is a perfect square, and
, 
is the smallest positive integer such that 
is a perfect square., there exists a positive integer such that 
for all integers 
. be a scalene triangle with incircle 
. Denote by the midpoint of arc in the circumcircle of , and by the point where the 
-excircle touches 
. Suppose the circumcircle of 
meets again at 
and intersects at two points , 
.
or 
is tangent to .
, with 
, for which the edges 
are
with 
, for which the edges 
are all coloured blue. be a positive integer. Alice and Bob play the following game. Alice considers a permutation 
of the set ![$[n]=\{1,2, \dots, n\}$](http://latex.artofproblemsolving.com/8/f/e/8fea25e7cf72aabc60f48c3b30f4d4818a084ae3.png)
and keeps it hidden from Bob. In a move, Bob tells Alice a permutation 
of ![$[n]$](http://latex.artofproblemsolving.com/0/d/e/0deff9086fd7840ef23860f5bc924f1d4d2d2310.png)
, and Alice tells Bob whether there exists an ![$i \in [n]$](http://latex.artofproblemsolving.com/e/6/7/e67206e47d5d5875105b339f177a3ab8f5151a5e.png)
such that 
. Bob wins if he ever tells Alice the permutation . Prove that Bob can win the game in at most 
moves.
denote the set of all real numbers. Find all functions 
such that
for all real numbers 
, 
.
for which there exist infinitely many positive integers 
such that ![\[ \frac{a^m+a-1}{a^n+a^2-1} \]](http://latex.artofproblemsolving.com/2/8/a/28a68e409451bbdcaee7c1286341f6aeea9ffdad.png)
is itself an integer.
such that 
divides 
.
is even, and 
has the same parity. are odd, we have that 
divides 
, but doesn't divide . are even, and let's say 
.
divides 
is odd, then 
divides but not , so is even.
.
divides 
(because both terms are even), and so 
.
we easily get 
, and for 
, we have that 
. Also, 
for all 
.
. We'll get 
, or 
as the only solution.
is even, and has the same parity. are odd, we have that divides , but doesn't divide . are even, and let's say 
.
divides 
is odd, then divides but not , so is even.
.
divides 
(because both terms are even), and so 
.
we easily get 
, and for 
, we have that 
. Also, 
for all 
.
. We'll get 
, or as the only solution.
is unnecessary, the solution in that case is a bit more messy, but similar. The motivation for this solution is using size issues in divisibility, similar to the above posts.
. Since 
have the same parity so this forces that either are both odd or are both even. We'll firstly work with the case when are both odd. Let 
. Notice that 
But 
a contradiction. Now let 
. Clearly we must have that 
. But since 
so we can estabilish that 
. This is equivalent to 
. But notice that from induction 
we have that 
and hence we have 
. But notice that 
hence we have to check for 
.
then we have that 
. But notice that 
for all 
hence we have to check till 
. But clearly none of the solution sets work. 
.
then we have 
. But again notice that we that 
for 
. Hence we check all the way uptil 
. But clearly none of the solution sets work again .
then notice that we have 
but notice that this is true only when 
. So we check till 
and we have a solution set that woks which is 
or that 
.
works, and we can easily verify that it does.
so 
implying 
This implies that 
so 
implying that 
and 
Say that 
and 
Then note 
so we must have 
implying that 
Let 
then we want to solve 
implying 
Note that for all 
we have 
and for all positive integers 
So the only possible solutions are natural numbers 
note that we must have![\[(7-9^y)(7+9^y)\mid 16(1+y^2),\]](http://latex.artofproblemsolving.com/5/c/3/5c31ce479f239b1a5e1f37f30262aeb5f50501bb.png)
or 
Note this fails for any 
and that 
yields equality, so 
is the only solution for 
note that we must have![\[(49-9^y)(49+9^y)\mid 16(16+y^2).\]](http://latex.artofproblemsolving.com/7/f/4/7f4a1b07e035fd6ce5b8008ac632b4d6d36d61dd.png)
We can easily verify that fails. For we want to ensure 
which fails for all 
note that we must have![\[(343-9^y)(343+9^y)\mid 16(81+y^2).\]](http://latex.artofproblemsolving.com/e/0/a/e0a935d3e8c90388e8577f44c1ebb16679c91f82.png)
We can easily verify that and 
fail by size reasons, noting 
and 
for 
For 
we want to ensure that 
which fails for all 
is 
corresponding to 
and 
don't have the same parity, then 
is even and 
is odd, absurd. Otherwise, 
so and are both even. Let 
. Then 
. Now, if 
and are the same parity, the LHS is divisible by 
and otherwise it is still divisible by , so either way 
and thus we can let 
so 
. Now, we use the crowbar known as "size": it is clear that the LHS's absolute value is at least 
, so at least one of 
and 
occurs. The former can never occur and the latter can only occur for 
. Now, at 
we get 
. But the value of that minimizes 
is 
, which is bad because 
. If we have 
, we get 
. Similarly to before, note 
is minimized at 
, so since 
, this is bad. So finally 
and 
. Equivalently, 
. Equality holds exactly at . Moreover, 
and 
, so by differentiation the result follows: the divisibility holds only when 
, which can easily be checked to be true.
. Hence and have the same parity. If and are both odd, then 
(mod 
) 
, which is contradictory to the congruence 
(mod ). Thus and are both even,let . The divisibility condition becomes 
. Since 
,we have 
. Then we can check case by case and get the unique solution .
). This solution is essentially the same as m.essein's, but posted for fun. reasons, that and are even. Taking also modulo , we get that and 
.
Thus, 
as 
. Note that 
for all 
and 
for all 
. Therefore we must only consider values 
. Note that if 
, then or . Thus, we only need to consider pairs 
. Only solution is 
..
must be even, as both terms are odd. This implies that is even, so and have the same parity. However, if and were both odd, then we would have 
as and are odd, and 
as an odd perfect square must be 
, which is a contradiction. Therefore and are both even.
and 
, then 
Then, since 
, we must also have 
in particular 
. We must have 
. The minimum of 
is 
at 
, and 
for 
, so 
. Similarly, the minimum of 
is 
at 
, and 
for 
, so 
. Checking the remaining cases, we get that 
is the only one that works, which corresponds to 
.
meaning that 
and therefore 
and consequently 
, writing 
, we have 
Notice that 
(by the Catalan Conjecture) therefore 
Notice that 
for all 
and 
for all 
.
and 
gives us the only pair 
.
, which works as 
divides 
. and have the same parity.
.
which means and are even.
, which further means 
.
and 
with even to obtain ![\[ \left( 7^x - 3^y \right)\left( 7^x + 3^y \right) \mid (2x)^4 + (2y)^2. \]](http://latex.artofproblemsolving.com/8/a/0/8a0f7a101edd9ab8316f904ccd6452ad524b717f.png)
To reduce to finite casework we contend:
once 
.
. 
. Next
for 
. 
as well.
and consequently if 
there is nothing to verify. These cases are marked as ``big'' in the table. ![\[ \begin{array}{rr|ccccc} && 7 & 49 & 343 & 2401 & 16807 \\ && x=1 & x=2 & x=3 & x=4 & x=5 \\ \hline 9 &y=2 & \text{OK} & 40 \cdot 58 \nmid 272 & \text{big} & \text{big} & \text{big} \\ 81 &y=4 & 74 \cdot 88 \nmid 48 & 32 \cdot 130 \nmid 320 & \text{big} & \text{big} & \text{big} \\ 729 &y=6 & \text{big} & \text{big} & \text{big} & \text{big} & \text{big} \\ 6561 &y=8 & \text{big} & \text{big} & \text{big} & \text{big} & \text{big} \\ \end{array} \]](http://latex.artofproblemsolving.com/7/2/b/72b473883e285501118d41310c72bf63a0865f01.png)
and have same parity. This implies 
, so and are even, which implies 
, so 
. and 
where is even.
.
, then 
. If 
, then 
. So if and , then 
, a contradiction.
.
, then 
. So 
. Now 
. Thus, 
. If or , then 
, which implies 
, so . Indeed, is valid.
, then 
. So 
. If 
, then LHS is 
and RHS is 
, contradiction. If 
, then LHS is 
and RHS is 
, contradiction. Oops I forgot is even. Much less casework. If , then we get 
, contradiction.
, then 
. So 
. If 
, then 
, contradiction. If , then 
, contradiction. If , then 
, contradiction.
, then 
. So . If , then 
, contradiction. If , then 
, contradiction. If , then 
, contradiction. and .
. So 
, a contradiction.
is even, so is even. Thus, 
have same parity which means 
have same parity.
odd then is divisible by but 
so 
are even. Let 
to get 
which implies 
so 
is even. Thus, let 
Note that the first factor is a difference of two odd but distinct integers, so it's at least two. Thus, 
and so 
implying that 
or 
Then, more bounding gives the solution 
Thus, is the only solution
and have the same parity. Then, suppose that 
. Then we have that 
while , a contradiction. Then we may set and to obtain that![\[7^{2x}-3^{2y}\mid 16x^4+4y^2\implies 7^x+3^y\mid 16x^4+4y^2\implies 7^x+3^y\le 16x^4+4y^2.\]](http://latex.artofproblemsolving.com/9/9/9/9995a74de969bcf781111fc2f0ada5a89801fcdf.png)
Then by bounding we obtain that and 
. Finally, plugging into our original equation gives that only works (we omit the calculations).
are the same parity. If they are both odd then 
and 
, a contradiction. Thus and then . Thus 
Next note that 
for all 
, so we have that 
. This inequality gives . If then we have that 
or 
, and we get the solution 
so . If , then 
, so 
upon checking gives no solution. If , then 
which gives 
which upon checking gives no solutions, so the sole solution is .
and 
.
is even, and must have the same parity. Thus 
and so 
. Then 
and so .
and 
. We can rewrite the original equation as 
or, 
Bound. This is left as an exercise to the reader. The answer is 
. 
is even. So, parity of must be same. Therefore, 
. Which gives, are even.
. So, 
is even and not 
.
. So, which gives . we get 
. And by checking all of the pairs we can see that only 
works. Which gives the only solution 
is even so we need the parities of to be the same. So we have that which implies that are even. We now let and so we have that 
this implies that else there is a contradiction so we have that 
and so the only pair is which means that QED
so 
if and have the same parity, so and and are all even
divides 
divides 
divides 
divides 
, 
because 
satisfies it and multiplying by 
is more than multiplying by 
so it will always satisfy
, 
because 
satisfies it and multiplying by 
is more than multiplying by 
so it will always satisfy
, 
, 
and 
and 
to find that only 
works and results in 
only, which clearly works., so the RHS must also be a multiple of , so and are even. However, this makes the LHS a multiple of , so is a multiple of . Let 
and 
. Then![\[(7^m-9^n)(7^m+9^n)\mid 16(m^4+n^2).\]](http://latex.artofproblemsolving.com/8/d/5/8d55114c28c42b17b7650b3166541f1f6337ae80.png)
Since the first factor is even, we have![\[7^m+9^n\mid 8(m^4+n^2)\rightarrow (7^m-8m^4)+(9^n-8n^2)\le 0.\]](http://latex.artofproblemsolving.com/e/0/b/e0b26975574f3adbd163c81e6d9587b9153803ca.png)
and 
. Then we can see that 
, 
, 
, 
, and 
for all positive integers 
.
, 
, 
, and 
for all positive integers 
.
.
, then 
which is not true.
, then 
which is not true.
, then 
which is not true.
, then 
which is not true.
, then . QED.
, which it's obvious LHS increases faster so by obvious bounding we can manually check solutions that's only 
.
is even, 
are of the same parity; on the other hand, this implies 
, which forces 
and 
for mod reasons.
. By comparing the and -terms, it suffices to check 
, of which only 
works.
both are even. Then mod yields 
Thus let 
Then this yields 
Note that the RHS increases faster than the LHS for 
If 
testing 
yields no solutions. Therefore 
Only works meaning that the only solution is 
is the only solution.. We get that clearly the LHS is even, so and have equal parity. However, if they are both odd, the LHS is divisible by but the RHS is not., . Then we get 
It must follow that since 
and 
is even, it follows that 
Now note that by this rule . gives the desired result.
have different parity. However, this implies that they also have to be even from considering modulo 4. By taking modulo 8, we see that 
Substitute 
Hence, 
Hence, 
Rearranging, we have that 
However, observe that 
when 
and 
for 
Hence, we only need to test 
However, only works. We are done.
if and both are odd; then ; which is a contradiction hence;
if 
is odd, then 
; but then 
is a contradiction. Hence let, 
; we get
.
is the only solution which corresponds to 
as the only possible solution.
then 
,
then 
,
then ![\[ (7^c-3^d)(7^c+3^d)|4(4c^4+d^2) \implies |7^c-3^d||7^c+3^d| \le 16c^4+4d^2 \]](http://latex.artofproblemsolving.com/7/4/0/740abbe7cb2cd41c1df108400b2bbd67f020e2d9.png)
clearly never occurs, so 
,
.
.![\[ 7^c-16c^4 \le 4d^2-3^d \]](http://latex.artofproblemsolving.com/b/4/c/b4c45c1f33c75e235ff1c97135b8a351ea93062d.png)
When 
this means 
.
or 
.
.
,
.
works.
,
.
,
.
.
imply 
?
so 
then 
.
so let 
.
so 
which implies that 
.
for all integers 
and 
for all integers 
.
for all integers 
and 
for all integers 
.
or 
.
so 
.
so 
so 
.
so 
.
.
.
& Since, 
.
& 
,
Note: 
for 
for 
.
.
.
,
Contradiction!
to get 
and 
for all 
and 
for all 
,
.
.
.
doesn't work,so the only possible solution is 
.
so 
and 
that 
which means that 
)
Intuitively, this only holds true for small 
for 
and 
for 
.
and 
however for odd 
,
and 
is a contradiction.
then 
and 
,
.
correspondingly 
.
& 
.
