Substituting for z,
z = 1 - (x+y)
z^5 = 1 - (x^5+y^5)
(1 - (x+y))^5 = 1 - (x^5+y^5)
(1 - (x+y))^5 =
-x^5-5x^4y+5x^4-10x^3y^2+20x^3y-10x^3-10x^2y^3+30x^2y^2-30x^2y+10x^2-5xy^4+20xy^3-30xy^2+20xy-5x-y^5+5y^4-10y^3+10y^2-5y+1
(I did not do this by hand, I used a polynomial calculator at symbolab.com)
Next, set all that equal to 1 - (x^5+y^5)
-5x^4y+5x^4-10x^3y^2+20x^3y-10x^3-10x^2y^3+30x^2y^2-30x^2y+10x^2-5xy^4+20xy^3-30xy^2+20xy-5x+5y^4-10y^3+10y^2-5y = 0
Other than factoring out a '5', symbolab was not able to factor this further.
Plugging this into a graphics calculator at desmos.com shows 3 lines:
x=1
y=1
and y = -x
So this exercise found no additional solutions beyond what we already had:
{a,b,c} = {0,1,-1}
or a = 1 AND b = -c
where {a,b,c} means "any permutation of {x,y,z}"
But I'm not sure that I have proved that no other solutions exist.
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Posted by Larry
on 2020-12-29 07:30:48 |
Second look
