who can>hahahahaha

[phanbachloc]

solve this:
2x^2=y +1/y
and
2y^2=x+1/x

[phrodu]

x = 1
y = 1

[bugzpodder]

i solved your problem (see the post i made below)

[DiGiTaIErRoR]

There are lots of imaginary solutions.

[NotLKH]

Originally posted by bugzpodder
2x^2=y +1/y
and
2y^2=x+1/x

we establish and x>0 and y>0

2x^2=(y^2+1)/y^2
....

Are you sure?
I think you meant:

2x² = (y² + y)/(y²)

???
-Lou

[bugzpodder]

2x^2=y+1/y
2y^2=x+1/x

we establish that x>0,y>0

fine i'll subtract them:

2(x+y)(x-y)=y-x+1/y-1/x

2(x+y)(x-y)=-(x-y)+(x-y)/xy

(2x+2y+1-1/xy)(x-y)=0

so x=y or

2x+2y-1/xy+1=0

2xy+2y^2-1/x+y=0 (multiply by y)

replace 2y^2 with x+1/x

2xy+x+y=0

since x>0,y>0 therefore no solutions for this case.

we establish x=y

2x^2=x+1/x
2x^3=x^2+1

2x^3-x^2-1=0
since x=1 is a solution, then:

(x-1)(2x^2+x+1)=0

since no solutions for 2x^2+x+1,

so (x,y)=(1,1) is the only solution

[phanbachloc]

Originally posted by bugzpodder
i solved your problem (see the post i made below) but i bet you can't do this question, phan, hahahaha

given a function f(x) of degree n and some prime p.
f(0)=0
f(1)=1
f(x)=0 or 1 (mod p) for all integers x

prove that n >= p-1

note if a = b (mod c), then (a-b) is divisible by c

u want me to solve this ?
hehehe,if u bet i cant solve,why do u want to show me??KEEP WAITING!!!!!

[phanbachloc]

Originally posted by bugzpodder
2x^2=y+1/y
2y^2=x+1/x

we establish that x>0,y>0

fine i'll subtract them:

2(x+y)(x-y)=y-x+1/y-1/x

2(x+y)(x-y)=-(x-y)+(x-y)/xy

(2x+2y+1-1/xy)(x-y)=0

so x=y or

2x+2y-1/xy+1=0

2xy+2y^2-1/x+y=0 (multiply by y)

replace 2y^2 with x+1/x

2xy+x+y=0

since x>0,y>0 therefore no solutions for this case.

we establish x=y

2x^2=x+1/x
2x^3=x^2+1

2x^3-x^2-1=0
since x=1 is a solution, then:

(x-1)(2x^2+x+1)=0

since no solutions for 2x^2+x+1,

so (x,y)=(1,1) is the only solution

did u solve that by urself?

[bugzpodder]

yes of course i solved that by myself.