Simplify a radical and integer sums

Question

I have recently encountered the problem of finding $\sqrt {3 + 2\sqrt 2}$. I have tried many things, such as squaring $3 + 2\sqrt 2$, which leads to $17 + 12\sqrt 2$, and multiplying it $\sqrt {3 - 2\sqrt 2}$, but none of these seem to get me close to the answer. How would one approach this type of question? I would not only like the solution for this problem, but for all problems similar to this.

Answer 1

The crux with these kinds of problems is to assume that a simplification is possible and that it takes the form of$$\sqrt{X\pm Y}=\sqrt{A}\pm\sqrt{B}$$ where $X,Y\in\mathbb{R}$. Square both sides to get$$X\pm Y=A+B\pm 2\sqrt{AB}$$

Now it's all a matter of solving the resulting quadratic, which arises when you compare the two parts separately. Doing all the work, which I leave up to you to figure out, we get that$$A=\frac {X+\sqrt{X^2-Y^2}}{2}$$$$B=\frac {X-\sqrt{X^2-Y^2}}2$$

Note that a denesting occurs when $X^2-Y^2$ is a perfect square.

Answer 2

We know that $$a^2+b^2+2.a.b=(a+b)^2.$$

$$3+2\sqrt {2}=1^2+(\sqrt {2})^2+2.1.\sqrt {2} $$ $$=(1+\sqrt {2})^2$$

thus

$$\sqrt {3+2\sqrt {2}}=\sqrt {(1+\sqrt {2})^2}=1+\sqrt {2} $$

Answer 3

$$\sqrt{3+2\sqrt{2}}$$

Note that we can write $\sqrt{3+2\sqrt{2}}$ as$$3+2\sqrt{2}+(\sqrt{2})^2-2$$ $$=(\sqrt{2})^2+2\sqrt{2}+1$$ $$=((\sqrt{2})^2+\sqrt{2})+(\sqrt{2}+1)$$ $$=\sqrt{2}(\sqrt{2}+1)+(\sqrt{2}+1)$$ $$=(\sqrt{2}+1)^2$$ $$\sqrt{3+2\sqrt{2}}=\sqrt{(\sqrt{2}+1)^2}=\sqrt{2}+1$$

Answer 4

If $3+2\sqrt{2}$ were a square of another number of the form $a+b\sqrt{2}$, then we would get the following: \begin{align*} 3+2\sqrt{2}&=(a+b\sqrt{2})^2 \\ 3 + 2\sqrt{2} &= a^2+2b^2 + 2ab\sqrt{2} \end{align*} So we need to solve the equations \begin{align*} a^2 + 2b^2 &= 3 \\ 2ab & =2 \end{align*} We get the obvious solutions $a=1, b=1$ and $a=-1, b=-1$ which tells us that \begin{align*} (1+\sqrt{2})^2 &= 3+2\sqrt{2} \\ (-1-\sqrt{2})^2 &= \left(-(1+\sqrt{2}) \right)^2 = 3+2\sqrt{2} \\ \end{align*}

Answer 5

Taking a lead from the title of the question, I want to focus here on numbers which may be formed from integers and quadratic radicals, specifically, of the form $a + b \sqrt n$, where $a, b \in \Bbb Z$ and $0 < n \in \Bbb Z$ is square-free, that is, $n$ may not be written in the form $n = m^2 k$, with $m \ne 1$. Furthermore, at the behest of our OP Zani Xu, as expressed in the comments to the answers given by Key Flex and Salahamam_Fatima, I would like to address, to at least some degree, the general techniques by which such problems may be approached, to cut down on the guesswork as much as possible.

In light of the above remarks, let us consider the following formulation of our problem: find the perfect squares in the ring

$\Bbb Z[\sqrt n] = \{ a + b\sqrt n \mid a, b \in \Bbb Z, 0 < n \in \Bbb Z, \; n \; \text{square-free} \}; \tag 1$

that is, for which $a + b\sqrt n \in \Bbb Z[\sqrt n]$ does there exist $c + d\sqrt n \in \Bbb Z[\sqrt n]$ with

$a + b\sqrt n = (c + d \sqrt n)^2. \tag 2$

It is clear that (2) entails

$a + b\sqrt n = c^2 + 2cd \sqrt n + nd^2 = (c^2 + nd^2) + 2cd \sqrt n, \tag 3$

whence

$a - (c^2 + nd^2) = (2cd - b)\sqrt n; \tag 4$

the left-hand side of for is an integer, but the right is irrational (since $n$ is square-free) unless they are both zero, so we have

$c^2 + nd^2 = a, \tag 5$

$2cd = b. \tag 6$

It is clear that (5), (6) strictly limit the set of pairs $(c, d) \in \Bbb Z \times \Bbb Z$ which may satisfy (2); first of all, (5) forces $a > 0$, so if perchance $a < 0$, we are done. We may also infer from (6) that $b$ must be even if a solution is to exist, and that furthermore,

$c, d \mid \dfrac{b}{2}. \tag 7$

We may utilize the above results to both eliminate manifestly impossible cases of $a + b\sqrt n$, since we must have $a > 0$, $b$ even if a solution is to exist, but also to strongly limit the region of $\Bbb Z \times \Bbb Z$ which must be searched if we are to try and find a solution via checking pairs $(c, d)$ manually, that is, by simply inserting $c$ and $d$ into (5) and (6) to see if a suitable pair $(c, d)$ exists.

Of course, if we don't insist on being integer-arithmetic purists and are willing to use the quadratic formula, we may home in on a solution, if there is one, immediately, and likewise rapidly determine whether one exists at all. For from (6) we obtain

$d = \dfrac{b}{2c}, \tag 8$

and if we substitute this into (6) we find

$c^2 + \dfrac{nb^2}{4c^2} = a, \tag 9$

or

$c^4 + \dfrac{nb^2}{4} = ac^2, \tag{10}$

or

$c^4 - ac^2 + \dfrac{nb^2}{4} = 0; \tag{11}$

we may apply the quadratic formula to (11):

$c^2 = \dfrac{1}{2}(a \pm \sqrt{a^2 - nb^2}), \tag{12}$

which shows that $a^2 - nb^2$ must be a perfect square if a solution is to be found. We further observe that

$a^2 - nb^2 = (a + b\sqrt n)(a - b\sqrt n), \tag{13}$

a quantity which is conventionally called the norm of the element $a + b\sqrt n \in \Bbb Z[\sqrt n]$, and denoted by

$N(a + b\sqrt n) = (a + b\sqrt n)(a - b\sqrt n) = a^2 - nb^2; \tag{14}$

we may thus write (12) in more conventional number-theoretic terms

$c^2 = \dfrac{1}{2}(a \pm \sqrt {N(a + b\sqrt n)}), \tag{15}$

which, though it may not significantly advance the present cause, makes clear its connection to the standard approach to similar issues. In any event, returning to (12), we see that not only must $N(a + b\sqrt n) = a^2 - b^2n$ be a perfect square, so also must the right -hand side of (15); fortunately for the present pursuit, we have

$\sqrt{N(a + b\sqrt n)} = \sqrt{a^2 - nb^2} \le a, \tag{16}$ so

$0 \le c^2 = \dfrac{1}{2}(a \pm \sqrt{N(a + b \sqrt n)}) \le a; \tag{17}$

we note that (17) holds whenever $N(a + b\sqrt n) \ge 0$, so our above-mentioned criterion that it be a perfect square covers this requirement. Now, assuming all the above criteria are met, we still need the right-hand side of (15) itself to be a perfect square if integral $c$ is to exist. Furthermore, we see from (7) that we must have $c \mid b/2$. We thus see that $c$ and $d$ must satisfy many criteria if $c + d \sqrt n$ is to be a solution to (2); having many tests at our disposal allows us to eliminate many cases relatively quickly in the event we must search for $c + d\sqrt n$ manually, that is, by trial and error.

Of course, we may use (5), (6) to derive an equation for $d$ instead of $c$ if we so desire; since

$c = \dfrac{b}{2d}, \tag{18}$

we have via (5)

$\dfrac{b^2}{4d^2} + nd^2 = a, \tag{19}$

or

$nd^4 + \dfrac{b^2}{4} = ad^2, \tag{20}$

or

$nd^4 - ad^2 + \dfrac{b^2}{4} = 0; \tag{21}$

applying the quadratic formula to (21):

$d^2 = \dfrac{1}{2n}(a \pm \sqrt{a^2 - nb^2}) = \dfrac{1}{2n}(a \pm \sqrt{N(a + b\sqrt n)}); \tag{22}$

if we compare (15) and (22), multiplying them out but choosing opposite signs for the $\sqrt{N(a + b\sqrt n})$ term, we see for example that

$c^2 d^2 = \left ( \dfrac{1}{2}(a \pm \sqrt{a^2 - nb^2}) \right ) \left ( \dfrac{1}{2n}(a \mp \sqrt{a^2 - nb^2}) \right )$ $= \dfrac{1}{4n}(a^2 - (a^2 - bn^2)) = \dfrac{nb^2}{4n} = \dfrac{b^2}{4}, \tag{23}$

from which (6) is recovered. We can in fact conduct an analysis of (22) similar to that we have done for (12), but as no surprises are expected along these lines, we leave such an undertaking to the sufficiently engaged reader.

Finally, we apply the above to the specific problem at hand, that is to finding $c + d \sqrt 2 \in \Bbb Z[\sqrt 2]$ with

$c + d \sqrt 2 = \sqrt{3 + 2 \sqrt 2}; \tag{24}$

here we have $a = 3$, $b = n = 2$; then from (12) we find

$c^2 = \dfrac{1}{2}(a \pm \sqrt{a^2 - nb^2}) = \dfrac{1}{2}(3 \pm \sqrt{3^2 - 2 \cdot 2^2})$ $= \dfrac{1}{2}(3 \pm \sqrt{9 - 8}) = \dfrac{1}{2}(3 \pm 1) = 1, 2; \tag{25}$

clearly there is no $c \in \Bbb Z$ with $c^2 = 2$, so we must have $c^2 = 1$ whence

$c = \pm 1; \tag{26}$

then from (8),

$d = \dfrac{b}{2c} = \dfrac{2}{2 \cdot \pm 1} = \pm 1; \tag{27}$

we see from (27) that $d$ is of the same sign as $c$; thus

$c + d\sqrt 2 = \pm (1 + \sqrt 2); \tag{27}$

we check:

$(\pm(1 + \sqrt 2))^2 = (1 + \sqrt 2)^2 = 1 + 2 + 2\sqrt 2 = 3 + 2\sqrt 2. \tag{28}$

We are done.