Root sum of roots.

The minimal "integer-coefficiented" monic polynomial such that $P(\sqrt{a}+\sqrt{b}) = 0$ can be found by knowing that all of $\sqrt{a}+\sqrt{b}$ conjugates are also roots. It means

$P(x) = (x-\sqrt{a}-\sqrt{b})(x-\sqrt{a}+\sqrt{b})(x+\sqrt{a}-\sqrt{b})(x+\sqrt{a}+\sqrt{b})$ $\Rightarrow P(x) = x^4 -2(a+b)x^2 +(a-b)^2\$ .

From the problem statements, we have that the first expression $\alpha$ shows us $2(a+b) = 10(a-b)^2 \Rightarrow (a-b)^2 = \frac{a+b}{5}$ .

Also, we get the other expression, $\beta$ , shows $1 + -2(a+b) + (a-b)^2 = -2a-b \Rightarrow (a-b)^2 -b + 1 = 0$ .

Putting $\alpha$ inside $\beta$ , we get $\frac{a+b}{5} -b + 1 = 0 \Rightarrow \frac{a-4b}{5} = -1 \Rightarrow \boxed{(a-4b)^2 = 25.}$

Awesome exercise, Marco! Keep it up, man!

Let $c = \sqrt{a} + \sqrt{b}$ , so $c^2 = a+b + 2 \sqrt{ab}$ > $(c^2 -(a+b))^2 = 4ab$

$c^4 + (a+b)^2 -2(a+b)^2c^2 - 4ab = 0$

and so $\sqrt{a} + \sqrt{b}$ is root of the polynomial $x^4 -2(a+b)x^2 +(a-b)^2$

The coefficients are: $a_{4} = 1 ; a_{3} = 0 ; a_{2} = -2(a+b) ; a_{1}= 0 ; a_{0} = (a-b)^2$

So we have $-2(a+b) = -10(a-b)^2$

$\boxed {a+b = 5(a-b)^2}$

Now considering the sum: $-2a -b = \sum_{i=0}^4 a_{i} = 1-2(a+b) + (a-b)^2$

$1+ (a-b)^2 -2a -2b = -2a -b$

$(a-b)^2 -b +1 = 0$

$(a-b)^2 = b-1$

Sobstitute the last expression in the boxed expression up

$(a+b) = 5(b-1)$

$a-4b = -5$

And finally $(a-4b)^2 = (-5)^2 = \boxed{25}$

* NOTE: * $a ,b$ are in the form $a =4n +3 ; b = n+2$ and for every value of $n$ are coprime.