# 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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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