# To check the solution via radicals

Given equations are

$$x^5+y^5=a \tag 1$$

$$5xy(x^2+xy+y^2)=b \tag 2$$

Is it possible to find $x$ or $y$ via using radicals?

My attempt $$x^2+y^2=\frac{b}{5xy}-xy$$

$$x^4+y^4=\frac{b^2}{5^2x^2y^2}-x^2y^2-\frac{2b}{5}$$

$$(x^2+y^2)(x^4+y^4)=\left(\frac{b}{5xy}-xy\right)\left(\frac{b^2}{5^2x^2y^2}-x^2y^2-\frac{2b}{5}\right)$$

$$x^6+y^6+x^2y^2(x^2+y^2)=\left(\frac{b}{5xy}-xy\right)\left(\frac{b^2}{5^2x^2y^2}-x^2y^2-\frac{2b}{5}\right)$$

$$x^6+y^6+x^2y^2\left(\frac{b}{5xy}-xy\right)=\left(\frac{b}{5xy}-xy\right)\left(\frac{b^2}{5^2x^2y^2}-x^2y^2-\frac{2b}{5}\right)$$

$$x^6+y^6=\left(\frac{b}{5xy}-xy\right)\left(\frac{b^2}{5^2x^2y^2}-2x^2y^2-\frac{2b}{5}\right)$$

I haven't got $x^5+y^5$ in my way,it looks impossible to solve it via radicals but I need to proof.

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I'd see what $(x - y) (x^2 + x y + y^2) = x^3 - y^3$ yields... – vonbrand Feb 26 '13 at 18:03

No. First, notice that $$(x+y)^5 = (x^5+y^5) + 5 x y (x^2+xy+y^2) (x+y)$$ So your equations imply $$(x+y)^5 = a + b(x+y)$$ Putting $s = x+y$, if $x$ and $y$ could be expressed in terms of $a$ and $b$ using radicals, then so could the root of $s^5 -bs -a$. This quintic cannot be solved in radicals. (In fact, this quintic is in what is known as Bring-Jerrard normal form. Every quintic can, by a change of variables be put into this form so, if we could solve equations in this form by radicals, we could also solve general equations.)

So, how did I find this? I always try to use symmetries of the equations to reduce their degree before starting. In this case, the only symmetry I noticed was switching $x$ and $y$. So I set $s=x+y$ and $t=xy$, and used the fundamental theorem of symmetric functions to write: $$s^5 - 5 s^3 t + 5 s t^2 = a \quad 5 t (s^2-t) = b$$

The next thing I wanted to do was eliminate one of the variables. I wasn't smart enough to do it by hand, so I asked Mathematica:

(* Find an equation satisfied by t, containing no copies of s. *)
GroebnerBasis[{s^5 - 5 s^3 t + 5 s t^2 - a, 5 t (s^2 - t) - b}, {t}, {s}]


Mathematica output:

{b^5 - 25 b^4 t^2 + 125 b^3 t^4 - 3125 a^2 t^5 + 625 b^2 t^6 - 3125 b t^8 + 3125 t^10}


A degree $10$ equation. It probably isn't solvable by radicals, but it seems hard to prove. Let's try again:

(* This time, keep s and eliminate t *)
GroebnerBasis[{s^5 - 5 s^3 t + 5 s t^2 - a,  5 t (s^2 - t) - b}, {s}, {t}]


Output:

{-a - b s + s^5}


Ah. much simpler. And clearly not solvable by radicals, so neither is the original problem.

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PS I really hope this isn't homework. I decided it wasn't because (a) it seemed too hard for homework to me and (b) the tags didn't make sense; this would be assigned in a Galois theory class, not an abstract algebra class. If this was a homework problem, you could have gotten much better hints and guidance if you indicated the nature of the course and what you had learned. – David Speyer Feb 26 '13 at 20:53
Thanks a lot for your answer. This is not homework. I have not gotten a Galois theory lesson. I try to understand the relation why some polynomial equations (less than degree five) are can be solved, other cannot be solved by radicals. Could you please add more details in your answer why $s^5-bs-a=0$ cannot be solved via radicals? Thanks again for your help – Mathlover Feb 26 '13 at 22:39
It's a bit difficult to explain why $s^5-bs-a=0$ can't be solved in radicals, without giving a semester course in Galois Theory (though I'm sure that if anyone can do it, David Speyer can). – Gerry Myerson Feb 26 '13 at 23:30
Thanks for the vote of confidence! I certainly can't do it tonight. I give a talk to the Michigan math club which might turn into a blogpost at some point. In the meantime, I recommend Vladimir Arnold's book "Abel's theorem in problems and solutions". – David Speyer Feb 27 '13 at 1:17