Download My solution for 5th mathlinks contest

My solution for 5th mathlinks contest
(Round 1)
Problem 1:find all pairs of positive integers x, y such that: x3  y3  2005( x 2  y 2 )
Solution:first we can easily know for all x  y is the solution of equation.
So we have to only find the pairs x  y ,assume x  y ,we
can
know: x 2  y 2  xy  2005( x  y) ,if x, y is a solution of the equation.the same
as: x 2  ( y  2005) x  y 2  2005 y  0 .
And
3( x  y)2
 2005( x  y)  ( x  y) 2 , 1503  x  y  2005 .with x  y ,we
4
know 0  y  1002 .then we consider   (3 y  2005)(2005  y) ,if x is an
integer,we can easy know   (3 y  2005)(2005  y) is also is integer.note
2005  y  a ,and we know
(8020  3a)a
is
a
square
number.assume
(8020  3a)a  b2 , (8020  3a, a)  (8020  3a,8020)  ( a,8020)  1, 2, 4,5,10, 20 (i
f a is not divide by 401).
(1)
(a,8020)
=1,we
get
8020-3a,a
are
square
numbers,so
8020  3a  1, 4(mod 5), a  1, 4(mod 5) ,contradiction.
(2) (a,8020) =2,assume a  2a1 ,we get 4010-3a1 ,a1
are square numbers,so
4010  3a1  1, 4(mod 5), a1  1, 4(mod 5) , contradiction.
(3) (a,8020) =4,we can also get the contradiction with the way above.
(4) (a,8020) =5,assume a  2a2 , we get 1604-3a 2 ,a 2
are square numbers,but
1604-3a 2  2(mod 3) , contradiction.
(5)
(a,8020) =10, assume
a  10a3 , we get 802-3a 3 ,a 3
are
square
numbers,so a3  1(mod 4),802  3a3  1(mod 4) , contradiction.
(6) (a,8020) =20, assume a  20a4 , we get 401-3a 4 ,a 4
are square numbers,so
401  3a4  2(mod 3) , contradiction.
If 401| a ,we can easily know a should be 1203,1604,each will have a
contradiction.
So we get the only solution for the equation is x  y  Z  .
(in the solution, a, a1 , a2 , a3 , a4 are all integers,and 401 is prime number.(a,b)
means gcd(a,b)).
Problem 2:find all positive integer n such that the residue of n when divided by
each prime number smaller than
n
.
2
Solution:first we consider n  10 .assume n stisfy the condition.then we can
easily know n must be a prime number or no power of p ( p >2)
So n must be a prime number or 2
k
 k  4 ,if
n  2 k ,we consider n-2,take the p
which is the prime factor of n-2.we can kow n  2  mod p  ,contradiction.
So n must be a prime number,of course n-2 is also a prime number.
So we know 3 | n  4 ,we can know the prime factor of n-4 is only 3.(if prime
number p>3 is the prime factor of n-4 we will easily have the contradiction.)
Assume n  4  3 .
k
Another we can 5 | n  6 or 5 | n  8 .
If 5 | n  6 ,we can know the prime factor of n-6 is only 5 in the same way.assume
n  6  5m .
m
We get 3  5  2 ,we can check 5  2  0(mod81) have no solution for m.so if
k
m
k  4 there is no solution.when k  3 ,we can directly the only solution is
k  3, m  2 , n  31 .
If 5 | n  8 , we can know the prime factor of n-8 is only 5 or 7 in the same
way.assume n  8  5 7 .
m
s
k
We get 3  5 7  4 ,with k  1 or m  1 (for n  10 ).we get 3  4(mod 5) or
k
3k  4(mod 7)
m
s
we
can
get
k
is
an
even.assume
k  2a
,then
(3a  2)(3a  2)  5m 7 s
,and
since
(3a  2,3a  2)  1
.we
a
m
a
m
3  2  5
3  2  5
or  a
get:  a
s
s
3  2  7
3  2  7
3a  2  5m
a
(1)  a
,if s  1 ,with 3  2  0(mod 5) we get a is odd.but with
s
3  2  7
3a  2  0(mod 7) we get a is even. so no solution.when s  0 we get a  1
and m  1 .we get the only solution n  13
a
m
3  2  5
a
m
(2)  a
with 3  2  5 we get
s
3  2  7
a  3, m  2 (the way is the same as the
first case)but it is not satisfy the second equation.so no solution.
With the statement above,we can know the only n satisfy the condition is 5,7,13,31.
(in the solution, k , m, a are all integers.and I don’t know whether we should
consider the number 1,2,3,4)_
Problem3.let ABC be a triangle and let A '  BC，B '  AC，C '  AB be three
collinear points.
a) prove that each pair of circles of diameters AA '，BB '，
CC ' has the same
radic al axis.
b) Prove that the circumcenter of the triangle formed by intersections of the line
and CC’ lies on the common radical axis found above.
AA’ ，BB’ ，
a) proof:note the circles of diameters
(1) I
AA '，BB '，CC ' be O1，O2，O3 .
proved that the intersection of O1，O2，is also in O3 ,note the intersection of
is P,so we get APA '  BPB ' =
O1，O2，

,and we draw the Perpendicular of
2
CP through P,and note the Perpendicular cut AB at C '' .we get
CPC '' 

2
,we
can
know
:
BA ' | sin BPA ' | | cos APB |


CA ' | sinCPA ' | | cos APC |
CB ' | cos BPC |
AC " | cos CPA |
,
,so we get


AB ' | cos BPA |
BC " | cos CPB |
,also
A ', B ', C " are three
collinear points.so C " and C ' are the same point.we have proved (1).
So we have proved a).(I think it is can be easily proved each pair of circles have
intersections.)
b)no proof.
Thank you for your problem.and so happy in take part in this contest.hope my solution
will be nice and right except 3(b).
And my usename in the forum is:zhaobin.
I was born in 1986.6. in china.but I am a unversity student now.only interest made me
to take part in it.thank you.
Best regards.