Let $a,c\in\Bbb R$ and $b,d \in \Bbb N$. Is there a way to determine if
$$a\sqrt b+c\sqrt d$$
can be written as
$$(e+f)^2,$$
with $e,f\in\Bbb R$? If so, is there a way to find what $e$ and $f$, given $a, b,c, $ and $d$?
For example, in my math textbook, it says that $$32-16\sqrt 3=(2\sqrt 6-2\sqrt 2)^2,$$ but they didn't explain how they got there. If someone can explain, that would be great. Thanks!
On
First, $x\sqrt y=\sqrt{x^2y}$ so we can simplify general case to $(\sqrt{a}+\sqrt{b})^2=c+2d$. Now we can obtain two equations to find $a$ and $b$: $a+b=c$ and $ab=d$. This system is easy to solve using substitution $a=c-b$ which leads to a quadratic equation $(c-b)b=d$. The system may or may not have a solution.
On
Let me show a method for finding some such roots, which should apply to your example. I'll instead apply the method to find $\sqrt{3+\sqrt{5}}$.
So, suppose $x = \sqrt{3 + \sqrt{5}}$. Then $x^2 = 3 + \sqrt{5}$; $x^2 - 3 = \sqrt{5}$; and so $(x^2 - 3)^2 = 5$. Expanding, we get: $$x^4 - 6x^2 + 4 = 0.$$
Now, let us observe that $(x^2 + 2)^2 = x^4 + 4x^2 + 4$ which is close to matching the left hand side. So, using the difference of squares formula, this is equivalent to $$(x^2 + 2)^2 - 10x^2 = (x^2 + 2)^2 - (x\sqrt{10})^2 = (x^2 + x\sqrt{10} + 2) (x^2 - x\sqrt{10} - 2) = 0.$$
However, by the quadratic formula, the roots of $x^2 + x \sqrt{10} + 2 = 0$ are $x = \frac{-\sqrt{10} \pm \sqrt{2}}{2}$, and the roots of $x^2 - x\sqrt{10} + 2 = 0$ are $x = \frac{\sqrt{10} \pm \sqrt{2}}{2}$. Thus, overall we have $x = \frac{\pm\sqrt{10} \pm\sqrt{2}}{2}$. It only remains to check which of these expressions actually is positive, and also gives $3 + \sqrt{5}$ when squared.
On
Well, in the reals there no point in talking about "perfect" squares as for any $x \ge 0$ then will be a $y$ so that $y^2 = x$ and $y = \sqrt x$.
So if you want $(e+f)^2 = a\sqrt{b}+ c\sqrt {d}>0$ then just let $y= \sqrt {a\sqrt b + c\sqrt d}$ and let $e= $ anything and $f= y-e$. That's it.
As to how the book got $32-16\sqrt 3=(2\sqrt 6-2\sqrt 2)^2$ that is very specific to those integers. (I'm more curious as to why rather than how) but:
Going from $2\sqrt 6 - 2\sqrt 2$ to noting that $(2\sqrt 6 - 2\sqrt 2)^2 = (2\sqrt 6)^2 - 2(2\sqrt 62\sqrt 2) + (2\sqrt 2)^2 = 4*6 - 8\sqrt {12} + 4*2=32 - 4\sqrt {4*3} = 32 - 16\sqrt 3$ is straightforward.
As for going from $32-16\sqrt 3$... well:
$\sqrt{32 - 16\sqrt 3}= 4(\sqrt {2 -\sqrt 3})$
.... so I guess the question is... how do you go from $\sqrt{a \pm \sqrt b}$ to $\sqrt x \pm \sqrt y$.
Okay.... if $\sqrt x \pm \sqrt y = \sqrt{2-\sqrt 3}$ then
$x + y \pm2\sqrt{xy}=x+y\pm \sqrt{4xy}= 2- \sqrt 3$.
So we need $x+y= 2$ and $4xy= 3$ or $xy = \frac 34$. And as we are subtracting we need $x > y$.
If we assume rational values so that $x =\frac ab$ and $y=\frac cd$ and so $\frac{ad+ cb}{bd} = 2$ and $\frac {ac}{bd}=\frac 34$ and as we need.
Trial and error but if $b=d =2$ then we need $ac = 3$ and $2a + 2c= 8$ or $a+c = 4$ and $a > c$. so $a =3$ and $c = 1$ and
$\sqrt{2 -\sqrt 3} = \sqrt {\frac 32} - \sqrt{\frac 12}$.
$\sqrt{32-16\sqrt 3} = 4(\sqrt {\frac 32} - \sqrt{\frac 12})=$
$2(\sqrt {\frac {4*3}2} - \sqrt{\frac {4}2})= 2(\sqrt 3- \sqrt {2})=$
$2\sqrt 6 - 2\sqrt 2$
The question is, what form are $e$ and $f$ to take? Otherwise, the problem is trivially solved since we can write any positive real number $r$ as $(\sqrt r)^2=(1+s)^2$ for some suitable $s,$ for example.
If we take a cue from the example given in your text, then you probably want $e,\,f$ to be the square roots of rationals, so that the question reduces to finding out when a binomial quadratic surd is the square root of another. That is, we want to find conditions on rational $p,\,q$ such that $$\sqrt{\sqrt p\pm\sqrt q}=\sqrt m\pm\sqrt n,$$ for some rational $m,\,n,$ which is standard fare. Square both sides to obtain $$\sqrt p\pm\sqrt q=m+n\pm2\sqrt{mn}.$$ Then we must have one of $\sqrt p$ or $\sqrt q$ to be rational, clearly, and this must be equal to $m+n.$ Secondly we must have $\pm2\sqrt{mn}$ equal to the other summand on LHS. You may examine these conditions further to deduce finer restraints.