solve this:
2x^2=y +1/y
and
2y^2=x+1/x
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solve this:
2x^2=y +1/y
and
2y^2=x+1/x
x = 1
y = 1
i solved your problem (see the post i made below)
There are lots of imaginary solutions.
:p
Are you sure?Quote:
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
....
I think you meant:
2x2 = (y2 + y)/(y2)
???
-Lou
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
u want me to solve this ?Quote:
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
hehehe,if u bet i cant solve,why do u want to show me??KEEP WAITING!!!!!
Quote:
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?
yes of course i solved that by myself.