Simplifying the radicals

$s=\sqrt[5]{41+29*\sqrt{2}}-\sqrt[5]{29*\sqrt{2}-41}$

In the first term in R.H.S multiply and divide by $\sqrt[5]{41-29\sqrt{2}}$

$s=\frac{\sqrt[5]{41^{2}-2*29^{2}}}{\sqrt[5]{41-29\sqrt{2}}}+\sqrt[5]{41-29\sqrt{2}}$

$41^{2}-2*29^{2} = -1$ this can be easily calculated without a calculator.

$s= \frac{-1}{\sqrt[5]{41-29\sqrt{2}}}+ \sqrt[5]{41-29\sqrt{2}}$

let $a = \sqrt[5]{41-29\sqrt{2}}$

$s=\frac{-1}{a} +a$

$a^{2}-s*a-1=0$

so, $a=\frac{s\pm\sqrt{s^{2}+4}}{2}$

since, $a = \sqrt[5]{41-29\sqrt{2}} < 0$

so, $\sqrt[5]{41-29\sqrt{2}} =\frac{s-\sqrt{s^{2}+4}}{2}$

By taking a power of 5 to both sides

$32*(41-29\sqrt{2} )=(s-\sqrt{s^{2}+4} )^{5}$

We can expand the R.H.S using the binomial theorem

R.H.S $=s^{5} -5*s^{4} *\sqrt{s^{2}+4} +10*s^{3} *\sqrt{s^{2}+4} -10*s^{2} *(s^{2} +4)*\sqrt{s^{2}+4} +5*s*\sqrt{s^{2}+4} -(s^{2} +4)^{2} *\sqrt{s^{2}+4}$

By looking at the L.H.S we see it consists of two terms, one of them containing $\sqrt{2}$ so the term free from $\sqrt{2}$ should equal the summation of the terms free of $\sqrt{s^{2}+4}$ in the R.H.S

so by taking the terms free of $\sqrt{s^{2}+4}$ and simplifying then we get,

$16*s^{5} +80*s^{3} +80*s=32*41$

so,

$s*(s^{4} +4*s^{2} +4)=2*41$

2 and 41 are prime numbers.

so in the L.H.S one term should be 2 and the other should be 41.

obviously $s=2$ , and by substitution of $s=2$ in $(s^{4} +4*s^{2} +4)$ we get 41.

so

$s=2$

Wow..... some awesome solutions here.... My solution is a little primitive... We know 49*sqrt(2) ~ 41.006 (pen and paper) ...

So the First term is (81.006)^(1/5) and the second term is (0.006)^(1/5) Observing 2^5=32 and 3^5=243 , the first term must lie between (2,3).

For the second term , we observe that 0.5^5~0.031 (paper and pen)... Therefore, the second expression is a value between (0,0.5) .

Finally, the maximum and minimum bounds for the expression has to be 3 and 1.5 . So speculative range is (1.5,3) ... (open interval ---- as second term will yield a positive value).... Therefore, only integer value possible is 2.

First note that $S=\sqrt[5]{41+29\sqrt{2}}+\sqrt[5]{41-29\sqrt{2}}$ (after rewriting a little) is the sum of the real fifth roots of the solutions of some quadratic polynomial $P(z)$ . So, let $z=41+29\sqrt{2}$ . We will find the minimal polynomial, which also contains the conjugate $z=41-29\sqrt{2}$ :

$z=41+29\sqrt{2} \Longrightarrow (z-41)^2=1682 \Longrightarrow z^2-82z-1=0$

So, $P(z)=z^2-82z-1$ .

Now, let's save that result and try to find a general solution for the polynomial $P(x)=x^5+5ax^3+5a^2x+b=0$ . The identity $(u+v)^5-5uv(u+v)^3+5(uv)^2(u+v)-(u^5+v^5)=0$ will help us because we can compare coefficient by coefficient with the polynomial and obtain:

$x=u+v$

$-5uv=5a \Longrightarrow uv=-a \Longrightarrow u^5v^5=-a^5$

$-(u^5+v^5)=b \Longrightarrow u^5+v^5=-b$

Observe that by Vieta's formulas, $u^5$ and $v^5$ are the solutions for some quadratic polynomial $P(z)$ . But wait, we already know that polynomial!. By the first equation, clearly $S$ is a root of $P(x)$ . So let's compare again:

$u^5+v^5=-(-82) \Longrightarrow b=-82$

$u^5v^5=-1 \Longrightarrow a^5=1 \Longrightarrow a=1$

Substitute $a$ and $b$ in the polynomial $P(x)$ , which would be the minimal polynomial of $S$ :

$P(x)=x^5+5x^3+5x-82=0$

By the rational root test, it factors as:

$P(x)=(x-2)(x^4+2x^3+9x^2+18x+41)$

We don't need to check that the fourth degree polynomial doesn't have real roots, because the expressions inside the fifth roots is a real number, and there are other four complex combinations that actually are the remaining roots. (They are $\sqrt[5]{u}w^k+\sqrt[5]{v}w^{5-k}$ for $k=1,2,3,4$ , where $w$ is any primitive fifth root of unity).

Hence, $x=2$ and $\boxed{S=2}$ .

Because (( sqrt. 2)+ 1 )^5= (29* sqrt. 2)+ 41 and ((sqrt. 2)- 1)^5= (29* sqrt. 2)- 41 So, S= ((sqrt. 2)+ 1)- ((sqrt. 2)- 1) S= 1- (-1)= 2