# Find the sum of radicals without squaring, Is that impossible?

Find the summation: $$\sqrt {3-\sqrt 5}+\sqrt {3+\sqrt 5}$$

My attempts: \begin{align*} &A = \sqrt{3-\sqrt{5}}+ \sqrt{3+\sqrt{5}}\\ \implies &A^2 = 3-\sqrt{5} + 3 + \sqrt{5} + 2\sqrt{9-5}\\ \implies &A^2 = 6+4 = 10\\ \implies &A = \sqrt{10} \end{align*} So I was wondering about a way to find this sum without squaring? It seems impossible, but I still want to ask.

• You could denest the radicals, but that would ultimately just amount to squaring the expression in its own right (or have about the same, or more, work despite being more general). To be honest, this expression is nice enough to get an answer in only a few lines; I don't see much hope for at least doing something "easier". Aug 18 at 20:18
• Answers without squaring for this very expression also here: math.stackexchange.com/questions/2656208/… Sep 14 at 0:46

Trick: $$\sqrt {3-\sqrt{5}}=\frac{\sqrt {6-2\sqrt{5}}}{\sqrt 2}=\frac{\sqrt {(\sqrt 5)^2-2\sqrt{5}+1}}{\sqrt 2}=\frac{\sqrt {(\sqrt 5-1)^2}}{\sqrt 2} =\frac{\sqrt 5-1}{\sqrt 2}.$$ Similarly we see $$\sqrt {3+\sqrt{5}} =\frac{\sqrt 5+1}{\sqrt 2}$$. Thus the sum is $$2\cdot \frac{\sqrt 5}{\sqrt 2}=\sqrt {10}$$.

A polynomial approach.

Note that $$a_1,a_2=\sqrt{3\pm\sqrt 5}$$ are two of the roots of $$x^4-6x^2+4=0.$$ The other roots are $$-a_1,-a_2.$$

So $$x^4-6x^2+4=(x-a_1)(x-a_2)(x+a_1)(x+a_2).$$

Now, $$a_1a_2=\sqrt{4}=2.$$ If $$S=a_1+a_2,$$ then this factorization becomes:

$$x^4-6x^2+4=(x^2-Sx+2)(x^2+Sx+2)=x^4+(4-S^2)x^2+4.$$

So $$4-S^2=-6,$$ or $$S^2=10.$$

I guess that's sort of squaring, but we never actually numerically square $$S.$$

More generally, given the roots of $$x^4-bx^2+c^2,$$ there are two roots $$a_1,a_2$$ with $$a_1a_2=c,$$ and then you get a similar result:

$$x^4-bx^2+c^2=(x^2-Sx+c)(x^2+Sx+c)=x^4+(2c-S^2)x^2+c^2,$$ so $$2c+b=S^2.$$

So this means, at least if $$c\geq 0,$$ that $$\sqrt{\frac{b+\sqrt{b^2-4c^2}}2}+\sqrt{\frac{b-\sqrt{b^2-4c^2}}2}=\pm\sqrt{2c+b}$$

Multiplying by $$\sqrt{2},$$ this gives:

$$\sqrt{b+\sqrt{b^2-4c^2}}+\sqrt{b-\sqrt{b^2-4c^2}}=\pm\sqrt{4c+2b}\tag1$$

You also get:

$$\sqrt{b+\sqrt{b^2-4c^2}}-\sqrt{b-\sqrt{b^2-4c^2}}=\pm\sqrt{2b-4c},\tag 2$$ since in this case $$a_1a_2=-c.$$

The case $$b=3,c=1$$ in (1) gives your result, since $$3\pm\sqrt 5$$ are both real and positive, so we know the result has to be positive.

In both $$(1)$$ and $$(2)$$ you get the positive sign if $$b,c$$ and all the square roots are real. If the square roots are complex, you are stuck figuring out the sign.

But if $$b,c$$ are real, and $$0\leq 2c\leq b,$$ we can solve $$(1)$$ and $$(2)$$ to get:

$$\sqrt{b+\sqrt{b^2-4c^2}}=\frac{\sqrt{2b+4c}+\sqrt{2b-4c}}{2}=\frac{\sqrt{b+2c}+\sqrt{b-2c}}{\sqrt 2}$$ and similarly: $$\sqrt{b-\sqrt{b^2-4c^2}}=\frac{\sqrt{b+2c}-\sqrt{b-2c}}{\sqrt 2}$$

If $$d=b^2-4c^2,$$ this gives us:

$$\sqrt{b+\sqrt{d}}=\frac{\sqrt{b+\sqrt{b^2-d}}+\sqrt{b-\sqrt{b^2-d}}}{\sqrt2}$$ when $$0\leq d\leq b^2.$$

Use Vieta's formulas:

Let, $$x_1=\sqrt {3-\sqrt 5},\,\,\,x_2=\sqrt {3+\sqrt 5}$$

$$x^2-px+2=0\\p=\frac{x^2+2}{x}$$

and we have,

$$p=\frac{5+\sqrt 5}{\sqrt {3+\sqrt 5}}=\frac{5-\sqrt 5}{\sqrt {3-\sqrt 5}}$$

Then, using the rule

$$p=\frac ab=\frac cd\implies p=\frac{a+c}{b+d}$$

We get

\begin{aligned}p&= \frac{5+\sqrt 5+5-\sqrt 5}{\sqrt {3+\sqrt 5}+\sqrt {3-\sqrt 5}}\\ &=\frac {10}{p}=p\\ &\implies p=\sqrt {10}. \end{aligned}

In this answer I tried to generalize the answer given above.

Generalization:

Let,

$$\sqrt {a\pm\sqrt b}=\sqrt m\pm\sqrt n,\,\,\,m\ge n$$

and

$$A=\sqrt {a+\sqrt b}+\sqrt {a-\sqrt b}=2\sqrt m$$

Then we have

$$\sqrt {a\pm\sqrt b}=\sqrt m\pm\sqrt n\\ \begin{cases} m+n=a\\mn =\frac {b}{4}\end{cases}\\ t^2-at+\frac b4=0\\ 4t^2-4at+b=0\\ m=\frac {a+\sqrt{a^2-b}}{2}\\ n=\frac {a-\sqrt{a^2-b}}{2}$$

This gives,

$$\sqrt {a\pm\sqrt b}=\sqrt{\frac {a+\sqrt{a^2-b}}{2}}\pm\sqrt{\frac {a-\sqrt{a^2-b}}{2}}$$

and

\begin{aligned}A&=2\sqrt m=\sqrt {4m}\\ &=\boxed{\sqrt {2\left(a+\sqrt {a^2-b}\right)}}.\end{aligned}

$$\text {QED}.$$

Write,

$$\sqrt {3\pm\sqrt 5}=\sqrt a\pm\sqrt b,\, a\ge b$$

and we obtain

$$A=\sqrt {3+\sqrt 5}+\sqrt {3-\sqrt 5}=2\sqrt a$$

Then, we want to find ratinal $$a,b$$ such that:

$$\sqrt {3\pm\sqrt 5}=\sqrt a\pm\sqrt b,\, a>b$$ holds.

We have

$$3\pm \sqrt 5=a+b+2\sqrt {ab}\\ \implies \begin{cases}a+b=3\\ ab=\frac 54\end{cases}$$

Using the Vieta's formulas, we have

$$t^2-3t+\frac 54=0\\ 4t^2-12t+5=0\\ t_{1,2}=\frac {6\pm4}{4}\\ \implies a=\frac 52\\ \implies b=\frac 12.$$

This gives,

$$\sqrt {3\pm\sqrt 5}=\sqrt \frac 52\pm \sqrt \frac 12$$

and

$$A=2\sqrt a=\sqrt {10}.$$

I want to solve the problem with a geometric approach. Consider a right triangle $$ABC$$ as below: If we suppose that $$BH = \sqrt{3 - \sqrt{5}}$$ and $$HC = \sqrt{3 + \sqrt{5}}$$, then we have $$AH^2 = BH.HC = \sqrt{9 - 5} = 2$$. So $$AH = \sqrt{2}$$. Now use Pythagorean theorem: \begin{align*} \overset{\triangle}{ABH}&: AB^2 = AH^2 + BH^2 = 2 + 3 - \sqrt{5} = 5 - \sqrt{5} \implies AB = \sqrt{5 - \sqrt{5}}\\ \overset{\triangle}{ACH}&: AC^2 = AH^2 + HC^2 = 2 + 3 + \sqrt{5} = 5 + \sqrt{5} \implies AC = \sqrt{5 + \sqrt{5}}\\ \overset{\triangle}{ABC}&: BC^2 = AB^2 + AC^2 = 5 - \sqrt{5} + 5 + \sqrt{5} = 10 \implies BC = \sqrt{10} \end{align*} Therefore: $$\bbox[5px, border: 2px solid magenta]{\sqrt{10} = BC = BH + HC = \sqrt{3 - \sqrt{5}} + \sqrt{3 + \sqrt{5}}}$$