prove that if x,y,z are positive integers such that (x,y)=1 and x^2+y^2=z^4
then xy is divisible by 7
show that the condition (x,y)=1 is necessary/
(x,y) means their gcd.
Search found 41 matches
- Mon Mar 07, 2011 8:27 am
- Forum: Number Theory
- Topic: divisibility
- Replies: 1
- Views: 2251
- Sat Feb 12, 2011 12:06 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2011 National Olympiad: Problemsets
- Replies: 7
- Views: 20967
Re: BdMO 2011 National Olympiad: Problemsets
i have solved one :D its number 4. i have proved (1006)^2011 is greater.(just counting power) number 1 is rather easy.i did it by induct on n.but i am not sure wheather it is correct. however n=1 the statement can be varified as follow 1+4=2+3 suppose for some m we can fulfil the statement for 4m nu...
- Thu Feb 03, 2011 3:34 pm
- Forum: Higher Secondary Level
- Topic: find the ratio
- Replies: 0
- Views: 2014
find the ratio
triangle ABC has the property that all of its angles are equal to 60degre.
D is on side BC such that CD=2BD.
if CH is perpendicular to AD then whats the ratio of
<DBH/<DAB....?
D is on side BC such that CD=2BD.
if CH is perpendicular to AD then whats the ratio of
<DBH/<DAB....?
- Tue Feb 01, 2011 1:43 pm
- Forum: H. Secondary: Solved
- Topic: Dhaka Higher Secondary 2010/2 (Secondary 2010/9)
- Replies: 17
- Views: 22415
Re: Dhaka Higher Secondary 2010/2 (Secondary 2010/9)
yes moon perhaps stated the just statement
- Mon Jan 31, 2011 8:58 pm
- Forum: Secondary: Solved
- Topic: Dhaka Secondary 2011/9
- Replies: 9
- Views: 14931
Re: Dhaka Secondary 2011/9
my approach is this
suppose (x,3)=1 where (a,b) means their gcd
by euler x^2=1 mod 3
and again (2,3)=1
so 2^(2y)=1mod 3
let x=2y
so we need to sort out all those numbers coprime to 3 .but note problem occurs with 6.but in this case PIE (principle of inclusion exclusion) perhaps paves da way
suppose (x,3)=1 where (a,b) means their gcd
by euler x^2=1 mod 3
and again (2,3)=1
so 2^(2y)=1mod 3
let x=2y
so we need to sort out all those numbers coprime to 3 .but note problem occurs with 6.but in this case PIE (principle of inclusion exclusion) perhaps paves da way
- Mon Jan 31, 2011 8:41 pm
- Forum: H. Secondary: Solved
- Topic: Dhaka Higher Secondary 2011/9
- Replies: 11
- Views: 19594
Re: Dhaka Higher Secondary 2011/9
ups!!
sry.i perhaps misviewed the question.
after all the problem was quite captivating as well as challenging .
thank you..
sry.i perhaps misviewed the question.
after all the problem was quite captivating as well as challenging .
thank you..
- Mon Jan 31, 2011 8:35 pm
- Forum: H. Secondary: Solved
- Topic: Dhaka Higher Secondary 2011/2
- Replies: 10
- Views: 17795
Re: Dhaka Higher Secondary 2011/2
perhaps you tried its the credit.
- Mon Jan 31, 2011 3:32 pm
- Forum: H. Secondary: Solved
- Topic: Dhaka Higher Secondary 2011/9
- Replies: 11
- Views: 19594
Re: Dhaka Higher Secondary 2011/9
whats problem with this solution??
1971=S+32+16+2+1
1995=S+64+8+2+1
2011=S+64+16+8+2+1 where S=1024+512+256+128
as f(2^n)=f(2^[n+2])
set f(S)=H(S)
so f(1971)=H(S)+2(f(1)+f(2))
and f(1995)=H(S)+2(f(1)+f(2))
so f(1995)=f(1971)=-1.....
1971=S+32+16+2+1
1995=S+64+8+2+1
2011=S+64+16+8+2+1 where S=1024+512+256+128
as f(2^n)=f(2^[n+2])
set f(S)=H(S)
so f(1971)=H(S)+2(f(1)+f(2))
and f(1995)=H(S)+2(f(1)+f(2))
so f(1995)=f(1971)=-1.....
- Mon Jan 31, 2011 12:59 pm
- Forum: H. Secondary: Solved
- Topic: Dhaka Higher Secondary 2011/2
- Replies: 10
- Views: 17795
Re: Dhaka Higher Secondary 2011/2
whats problem with 111 and 109.see difference is 02...
- Sun Jan 23, 2011 7:12 am
- Forum: Higher Secondary Level
- Topic: diophantine equation
- Replies: 4
- Views: 3695
diophantine equation
determine all non negative integers (x,y) for which the relation hold
(xy-7)^2=(x^2+y^2)
indian math olympiad
hint:factorization you all need to hang about
(xy-7)^2=(x^2+y^2)
indian math olympiad
hint:factorization you all need to hang about