AoPS Academy
of positive integers such that
, 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 
.
be an isosceles trapezoid such that 
. Suppose that there exists a point 
in such that 
and 
. Prove that 
and circumcircle of
are tangent to each other let that line L2 and in # we know that 
it means B and C are in 
or out of it
whit angel 
is down of L2
is among of 
and 
is among of 
be a circle tangent ton 
and 
at 
and 
;
be a circle tangent ton and at 
and 
;
be the midpoints of and respectively.
must be inside the circle .
must be inside . and must intersect each other at two distinct points 
. easy to see that 
passes throw . Thus if 
(see figure) then by AM-GM inquality:
![[asy]
import graph; size(12.42475239102794cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -5.724817209911015, xmax = 35.69993518111692, ymin = -13.45539462448323, ymax = 25.611128038591136; /* image dimensions */
pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666);
/* draw figures */
draw(circle((10.077897977003042,9.461099706682875), 4.944425615875747));
draw(circle((10.39364081793557,-0.6539477331910236), 8.529668362410135));
draw((9.64245549762135,23.41081056401747)--(5.401758015313216,11.067667392656867));
draw((5.401758015313216,11.067667392656867)--(2.3267941934540017,2.117555091954764));
draw((2.3267941934540017,2.117555091954764)--(18.271924907828524,2.6152849247334546));
draw((18.271924907828524,2.6152849247334546)--(14.644733170136668,11.3561883562411));
draw((14.644733170136668,11.3561883562411)--(9.64245549762135,23.41081056401747));
draw((5.401758015313216,11.067667392656867)--(14.644733170136668,11.3561883562411));
draw((16.458329038982598,6.985736640487278)--(14.318748831830447,6.918949298458), red);
draw((15.472614180225039,6.754642813958068)--(15.460120026196282,7.15490196266506), red);
draw((15.316957844616763,6.749783976280218)--(15.304463690588006,7.150043124987211), red);
draw((14.318748831830447,6.918949298458)--(6.003856311535769,6.659398584335082), blue);
draw((6.003856311535769,6.659398584335082)--(3.8642761043836087,6.592611242305816), red);
draw((5.018141452778205,6.428304757805879)--(5.005647298749449,6.828563906512872), red);
draw((4.8624851171699275,6.42344592012803)--(4.849990963141171,6.8237050688350225), red);
label("$x$",(15.054301336394083,8.34710663395235),SE*labelscalefactor);
label("$x$",(4.286535408758465,7.946652529205574),SE*labelscalefactor);
label("$y$",(9.892892875213374,6.522815712328148),SE*labelscalefactor);
/* dots and labels */
dot((5.401758015313216,11.067667392656867),linewidth(3.pt));
label("$A$", (4.242040508231045,11.105790466652362), NE * labelscalefactor);
dot((2.3267941934540017,2.117555091954764),linewidth(3.pt));
label("$D$", (0.860428068147132,1.9843358585313544), NE * labelscalefactor);
dot((18.271924907828524,2.6152849247334546),linewidth(3.pt));
label("$C$", (18.969852582807036,2.696254266970067), NE * labelscalefactor);
dot((14.644733170136668,11.3561883562411),linewidth(3.pt));
label("$B$", (15.143291137448923,11.773213974563655), NE * labelscalefactor);
dot((9.64245549762135,23.41081056401747),linewidth(3.pt) + uuuuuu);
dot((6.003856311535769,6.659398584335082),linewidth(3.pt) + uuuuuu);
label("$S$", (6.1553212309101015,7.3237239218217), NE * labelscalefactor,uuuuuu);
dot((14.318748831830447,6.918949298458),linewidth(3.pt) + uuuuuu);
label("$T$", (13.630464519516647,7.635188225513637), NE * labelscalefactor,uuuuuu);
dot((3.8642761043836087,6.592611242305816),linewidth(3.pt) + uuuuuu);
label("$N$", (2.729213890298768,6.700795314437826), NE * labelscalefactor,uuuuuu);
dot((16.458329038982598,6.985736640487278),linewidth(3.pt) + uuuuuu);
label("$M$", (16.92308715854572,7.234734120766861), NE * labelscalefactor,uuuuuu);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/1/8/9/1892cdefe4d4025ed88950cc04a6c45a8818c54f.png)
in such that: 
. and instead.
exists while this inequality is correct in the isosceles trapezoid. has to be inside both circles so we just have to prove that the inequality is correct while the two circles intersect at two points.
. The idea is to consider that the circle which includes points and is fixed and move the other circle to see what happens when the circles distance changes. (When the circle gets far from the other 
gets bigger and gets smaller while 
doesn't change)
, we can easily prove our guess
and 
