Find all solutions to the system of equations $$a+b+c=1$$ $$a^2+b^2+c^2=2$$ $$a^4+b^4+c^4=3$$
By squaring the first equation and substituting value of $a+b+c$ we get $\sum ab=-\frac{1}{2}$
From first equation we get: $a+b=1-c$, on squaring both sides and substituting $a^2+b^2=1+c^2-2c-2ab$ in $2$nd equation we get $2c^2-2c-2ab-1=0$
How to proceed after this?
From squaring the first two equations we get $$a^2+b^2+c^2+2ab+2bc+2ac=1$$ $$a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4$$ We then can get that $$ab+bc+ac=-\frac{1}{2}$$ $$a^2b^2+b^2c^2+a^2c^2=\frac{1}{2}$$ Squaring the first of these 2 equations gives $$a^2b^2+b^2c^2+a^2c^2+2abc(a+b+c)=\frac{1}{4}$$ So, $$2abc(a+b+c)=-\frac{1}{4}$$ $$abc=-\frac{1}{8}$$ This means that $a,b,c$ are roots of the polynomial $$x^3-x^2-\frac{1}{2}x+\frac{1}{8}=0$$ $$8x^3-8x^2-4x+1=0$$ Substitute $y=2x$, $$y^3-2y^2-2y+1=0$$ $$(y+1)(y^2-y+1)-2y(y+1)=0$$ $$(y+1)(y^2-3y+1)=0$$ $$y=-1,\frac{3\pm\sqrt{5}}{2}$$ $$x=-\frac{1}{2},\frac{3\pm\sqrt{5}}{4}$$ So $a,b,c$ are some permutation of those $3$ roots.
