AoPS Academy
be real numbers such that ![\[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]](http://latex.artofproblemsolving.com/3/d/0/3d015292c249a69a44d5f923092201d7a271b896.png)
Find the smallest possible value of 
.
. There are 
mutually non-attacking rooks placed on a grid 
. The grid is splitted into two connected parts, symmetric with respect to the center of the grid. What is the largest number of rooks that could lie in the same part?
such that the largest prime divisor of 
is equal to the largest prime divisor of 
.
denote the set of positive integers. Find all functions 
such that for positive integers 
and 
![\[f(a^2 + b^2) = f(a)f(b) \text{ and } f(a^2) = f(a)^2.\]](http://latex.artofproblemsolving.com/c/1/f/c1f8ffe04cfcc45b498f3931a4796d5c56dc04d0.png)
be the orthocenter of acute triangle 
, let 
be the foot of the altitude from 
to 
, and let 
be the reflection of across 
. Suppose that the circumcircle of triangle 
intersects line at two distinct points 
and 
. Prove that is the midpoint of 
.
be a quadrilateral inscribed in circle 
with 
. Let 
and be the reflections of 
over lines 
and , respectively, and let be the intersection of lines 
and 
. Suppose that the circumcircle of 
meets at and 
, and the circumcircle of 
meets at and 
. Show that 
.
such that among any positive real numbers 
with 
, there exist three that are the side lengths of an acute triangle.
be the set of integers, and let 
be a function. Prove that there are infinitely many integers 
such that the function 
defined by 
is not bijective. is bijective if for every integer 
, there exists exactly one integer such that 
.
. WLOG 
. Take a prime 
and allow 
and 
which exists by CRT. It follows that 
so 
. This means that 
but since 
and 
we have 
which means 
or 
as desired.
such that 
and 
for some ..
.
![\[ a^n + n \mid b^n - a^n. \]](http://latex.artofproblemsolving.com/0/7/c/07c04e62fa4c5d8cd29a10b1f7a4326b10ab0a78.png)
Taking this , we then get that ![\[ p \mid a^n + n \mid b^n - a^n \]](http://latex.artofproblemsolving.com/9/1/2/912d3895b9fdeef42533791a0a5d000a5ca601a8.png)
so 
, and since it follows that 
.
. For such a prime , we must have 
,as seen from say 
. such that 
, and 
, which exists by CRT. Choosing such an , we have 
. But since this is true for all , we have 
.
. Then, let 
. We see that 
. Therefore, 
. Since 
, 
. Therefore, 
. Therefore, .
we have ![\[\prod_{1\le i\le m}a_i\big|\operatorname{lcm}(a_1,\dots,a_m)\cdot\prod_{1\le i\le j\le m}\gcd(a_i,a_j).\]](http://latex.artofproblemsolving.com/1/a/4/1a42548650403bf6985cd99509e0788ca405f18d.png)
be a prime and WLOG 
. Then ![\[\nu_p(a_1)+\dots+\nu_p(a_{m-1})+\nu_p(a_m)=\nu_p(\gcd(a_1,a_2))+\dots+\nu_p(\gcd(a_{m-1},a_m))+\nu_p(\operatorname{lcm}(a_1,\dots,a_m))\]](http://latex.artofproblemsolving.com/7/8/d/78d855ca421c7a85d55aa110985b5f8449265e93.png)
which implies the claim.
. Let 
be odd. Fix prime and greater than and notice that for 
we have ![\[a^{dp}+dp\mid(b^{dp}+dp)-(a^{dp}+dp)=b^{dp}-a^{dp}\mid b^{kp}-a^{kp}.\]](http://latex.artofproblemsolving.com/0/5/6/0563b96602f452027dc3ecc6f79924f95a2706d4.png)
Additionally note that for distinct 
we have ![\[\gcd(a^{d_1p}+d_1p,a^{d_2p}+d_2p)\mid\gcd(a^{kp}+(d_1p)^{\frac k{d_1}},a^{kp}+(d_2p)^{\frac k{d_2}})\mid(d_1p)^{\frac k{d_1}}-(d_2p)^{\frac k{d_2}}\]](http://latex.artofproblemsolving.com/c/4/0/c40db3d6bafd939db2b56ead68866e9bf1722ab6.png)
and ![\[0<|(d_1p)^{\frac k{d_1}}-(d_2p)^{\frac k{d_2}}|<(kp)^k\]](http://latex.artofproblemsolving.com/2/8/1/281fec52d2198cba56849b32949bfee0db308a3a.png)
where the left inequality follows by 
.
so ![\[\frac{a^{\frac{\sigma(k)}k}}b<(kp)^{\frac1p\binom{\tau(k)}2}.\]](http://latex.artofproblemsolving.com/4/4/6/4461568e46b53738e5bcb847c450d6a91b0b0e00.png)
Now taking 
the RHS goes to 
so 
.
unbounded, for example by making the product of the smallest odd primes for arbitrarily large , since the sum of the reciprocals of primes diverges. This gives a contradiction, so we are done.
and 
where is prime. We can say 
Also 
There are infinite many primes we can make. CRT implies 
sufficiently large. Take 
, 
. We have:
This is a contradiction unless 
since is much larger than 
. We are done.
We know the RHS is non negative. Pick some 
Consider taking 
We have that the LHS has 
However, the RHS only simplifies to 
Thus, by taking an infinitely many large primes, we see that 
is a multiple of 
for all positive integers 
.
s.t. 
.
.
s.t. 
.
and so 
and so 
.
and so 
(we remember that 
but 
Could you explain it to me, please ?
that they are equal)
is any prime and we have any 
,
,
.
,
.
,
.
,
.
.
from chossing a suitable 
we would get a contradiction by choosing a big prime.
and
then 
so that for every natural, 
,
so that 
.
be relatively prime. Then, 
,
.
.
and 
.
be the order of 
.
,
.
.
.
,
.
,
.
for some integer 
,
.![\[ a^{x + m\varphi (p^{l + 1})} + x + m\varphi (p^{l + 1}) \equiv a^x + x + mp^l(p - 1)\equiv p^l(k + m(p - 1))\equiv 0\bmod p^{l + 1}
\]](http://latex.artofproblemsolving.com/3/4/9/349485f2435b423c83473a6d8d6cab394cab89b0.png)
which must exist by CRT. Then 
,![\[0 \equiv b^n + n \equiv b-a \pmod{p}.~\blacksquare\]](http://latex.artofproblemsolving.com/0/b/a/0baa92bc8f29139702d3854cb505fed491709054.png)
,
.
,
therefore for any prime 
for a large prime 
Then,![\[p \mid a^n+n \mid b^n - a^n \implies p \mid b - a \implies a = b\]](http://latex.artofproblemsolving.com/2/0/d/20d75b7e8ec732dfcfd1f2e74f2ab08d21022a83.png)
and 
This forces 
.
and 
.
.
is divisible by every prime 
.
by letting 
and 
but we know 
and 
so this means 
for all large 
.