# Identity in Ramanujan style

Is it possible to represent $$\sqrt {7\sqrt{20}-1} =\sqrt{A}+\sqrt{B}+\sqrt{C}$$ with rational $A,\,B,$ and $C?$

• Are you basing your question on $$\sqrt{7\sqrt{20} - 19} = \sqrt{\frac{5}{3}} - \sqrt{\frac{2}{3}}$$ and the similar identities in this post? – Tito Piezas III Apr 14 '15 at 3:07
• @Tito Piezas III: Thank you for the reference. I have not known those. I don't immediately see how to answer my question, using the methods from the references in this post. – user64494 Apr 14 '15 at 4:09
• After finding my answer, I think there is a connection between, $$\sqrt {-1+7\sqrt{20}} =\sqrt{\frac{16}{9}}+\sqrt{\frac{100}{9}}-\sqrt{\frac{5}{9}}\tag1$$ and $$\sqrt {-19+7\sqrt{20}} =\sqrt{\frac{4}{9}}+\sqrt{\frac{25}{9}}-\sqrt{\frac{80}{9}}\tag2$$ Where did you find the LHS of $(1)$? – Tito Piezas III Apr 14 '15 at 5:45
• @Tito Piezas III: Share you opinion. The question originates from math folklore. – user64494 Apr 14 '15 at 6:49

Yes.

$$\sqrt {-1+7\sqrt{20}} =\sqrt{\frac{16}{9}}+\sqrt{\frac{100}{9}}-\sqrt{\frac{5}{9}}\tag0$$

Solution: More generally, given the three roots $x_i$ of any cubic equation,

$$x^3+ax^2+bx+c=0\tag1$$

then sums involving the cube roots of the $x_i$ can be given in the simple form,

$$(u+x_1)^{1/3}+(u+x_2)^{1/3}+(u+x_3)^{1/3} = \big(w+3\,\sqrt[\color{blue}6]{d}\big)^{1/3}$$

where $u,w$ are the constants,

$$u = \frac{ab-9c+\sqrt{d}}{2(a^2-3b)}\tag2$$

$$w = -\frac{(2a^3-9ab+27c)+9\sqrt{d}}{2(a^2-3b)}\tag3$$

and $d$ is,

$$d = \tfrac{1}{27}\Bigl(4(a^2-3b)^3-(2a^3-9ab+27c)^2\Bigr)\tag4$$

Example: For your question, we have,

$$w=-1, \quad d =\frac{7^6\cdot20^2}{3^6}$$

and subbing these into $(3),(4)$, and using $Mathematica$ to simplify, we get $b,c$ for arbitrary $a$ as,

$$b= \tfrac{1}{9}(-343+3a^2)$$

$$c= \tfrac{1}{27}(-2058-343a+a^3)$$

Substituting $b,c,d$ into $(2)$ and $(1)$, we find that,

$$u=\tfrac{1}{9}(37+3a)$$

and $(1)$ factors as,

$$(7 + a + 3 x) (14 + a + 3 x) (-21 + a + 3 x) = 0\tag5$$

giving $x_1, x_2, x_3$. The expression $u+x_1$ simplifies as just $\frac{16}{9}$ and similarly for $x_2, x_3$, thus resolving into the numerical relation $(0)$ given above.

• thats a nice identity involving cube roots of roots of a cubic equation. +1 – Paramanand Singh Apr 14 '15 at 16:25
• @ParamanandSingh: Going to fifth powers, I have tried and have not found any example of $\sum_{n=1}^5 (u+x_n)^{1/5} = (w+\sqrt[k]{d})^{1/5}$ where the $x_i$ are five irrational roots of a quintic. – Tito Piezas III Apr 15 '15 at 2:33
• How do you get $a$? – Frank May 31 '16 at 15:30
• @Frank For OP's particular problem, then $a$ is arbitrary. – Tito Piezas III May 31 '16 at 15:56
• Sorry, but I am still not understanding what you mean by arbitrary $a$. Do you actually find the value of $a$? Or is it just a random number? (Both are definitions of arbitrary) – Frank May 31 '16 at 16:15