M M_B_S New member Joined Nov 12, 2013 Messages 16 Nov 15, 2013 #1 r²xy=xy^3+yx^3 , where r is the radius of the circle 8xy=(x+y)^4-(x-y)^4 circle with radius 1 <=> 1=x^2+y^2 q.e.d. M_B_S Last edited: Nov 15, 2013
r²xy=xy^3+yx^3 , where r is the radius of the circle 8xy=(x+y)^4-(x-y)^4 circle with radius 1 <=> 1=x^2+y^2 q.e.d. M_B_S
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Nov 15, 2013 #2 Hello, M_B_S! Do you have a question? Both statements are true (provided \(\displaystyle xy \ne 0\)).
Hello, M_B_S! Do you have a question? Both statements are true (provided \(\displaystyle xy \ne 0\)).
M M_B_S New member Joined Nov 12, 2013 Messages 16 Nov 15, 2013 #3 Does this proof the Beal Conjecture soroban said: Hello, M_B_S! Do you have a question? Click to expand... Yes! Does this proof the Beal conjecture ?
Does this proof the Beal Conjecture soroban said: Hello, M_B_S! Do you have a question? Click to expand... Yes! Does this proof the Beal conjecture ?
M M_B_S New member Joined Nov 12, 2013 Messages 16 Nov 16, 2013 #4 Denis said: Why are you asking that here? http://en.wikipedia.org/wiki/Beal's_conjecture Click to expand... Why not? 2³xy=(x+y)^4-(x-y)^4 Last edited: Nov 16, 2013
Denis said: Why are you asking that here? http://en.wikipedia.org/wiki/Beal's_conjecture Click to expand... Why not? 2³xy=(x+y)^4-(x-y)^4