I need to find an alternate form of $\sqrt{a} + \sqrt{b} + \sqrt{c} = d$ without square roots for a problem that I'm working on, but it's rather complicated to do. What we can do is

$\sqrt{a} + \sqrt{b} + \sqrt{c} = d \iff \sqrt{a} = d - \sqrt{b} - \sqrt{c} \iff a = (d - \sqrt{b} - \sqrt{c})^2$

and then calculate the right hand side, then iterate the process by putting a square root on one side, then proceed as above until all square roots are gone. However, this is a rather complicated process and I would like to do this with a computer but it exceeds WolframAlpha's server time. Can I get this done somewhere else? I would do this by hand myself but I will also need it for $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d} = e$ and perhaps even more complicated expressions, so if someone can show me an easy way to proceed with calculations - or point me to some computer program that can do this - I would appreciate it. Thanks in advance.

(note: not sure how to tag my question, feel free to change the tag if you can find something more appropriate).

Note: edited as per suggestions in the comments.

  • $\begingroup$ There are two problems: (1) the expressions are going to get large rapidly; a term with $n$ square roots will lead to a polynomial relation of degree $2^n$, and (2) your equivalences are running 'the wrong way' - it's not true that $a=b\Leftrightarrow a^2=\pm b^2$, but rather that $a^2=b^2\Leftrightarrow a=\pm b$, which is the opposite of what you want to make this work. In general, the polynomial relation will have $2^n$ distinct 'roots' for the one non-square-root variable of the original expression in terms of the other variables. $\endgroup$ – Steven Stadnicki Nov 14 '13 at 17:52
  • $\begingroup$ See the answers to this math StackExchange question: Rationalizing radicals $\endgroup$ – Dave L. Renfro Nov 14 '13 at 18:15

You are asking a special case of the following problem in abstract algebra: Suppose $x$ is a solution to $p(x) = 0$ and $y$ solves $q(y)=0$, for polynomials $p,q$ (with, say, integer coefficients); find a polynomial (with, again, integer or whatever coefficients) that has $x+y$ as a solution. In the "$\sqrt{a} + \sqrt{b} = c$" version of your question, you have $p(x) = x^2-a$ and $q(y) = y^2 - b$.

There is a general way to do this. Unfortunately in the version I described, the degrees of the polynomials get large. This is because of the following. Suppose that $p$ is of degree $\deg p$, and $q$ has degree $\deg q$. Then generically $p$ has $\deg p$ many complex solutions and $q$ has $\deg q$ many complex solutions. Thus (unless these solutions happen to satisfy some coincidences) there are $(\deg p)(\deg q)$ many possible values of $x+y$ if all you know is that $x$ solves $p(x)=0$ and $y$ solves $p(y)=0$. Whatever the polynomial is that $x+y$ solves, it must be solved by all of possible of these $(\deg p)(\deg q)$ numbers, since you have no way of telling it which solution you want. Therefore, this polynomial must have degree $(\deg p)(\deg q)$.

So in the example "$\sqrt a + \sqrt b = c$", we're looking for a polynomial solved by $c$ with "integer" (really, polynomial expressions in $a,b$) coefficients, and it necessarily will have degree $4$. Now adding on another square root means that in your case, your polynomial in $d$ will have degree $(4)(2) = 8$. In the next one, you get a polynomial of degree $16$. There's really nothing you can do about this.

Of course, there will be some patterns. Let me focus on your case of sums of square roots. The two solutions to $x^2-a = 0$ are, of course, $\pm \sqrt a$, and the two solutions to $y^2-b$ are $\pm \sqrt b$. Thus the four possible values of $x+y$ are $\pm \sqrt a \pm \sqrt b$. Therefore the polynomial we want is the degree-$4$ polynomial vanishing on these four points, namely:

$(z - \sqrt a - \sqrt b)(z - \sqrt a + \sqrt b)(z + \sqrt a - \sqrt b)(z + \sqrt a + \sqrt b) = ((z - \sqrt a)^2 - b)((z + \sqrt a)^2 - b) = (z^2 - a)^2 + b^2 - b((z - \sqrt a)^2 + (z + \sqrt a)^2) = z^4 - 2z^2a + a^2 + b^2 - 2z^2 b - 2ab = z^4 - 2(a+b)z^2 + (a-b)^2$

This illustrates, for example, that only even powers of the new variable (in your case, $d$) will appear — exactly because the set of solutions to this polynomial will necessarily be symmetric under $d \mapsto -d$. Such a polynomial is called "even".

In all cases, let me henceforth call the new variable $z$ — so you asked about $\sqrt a + \sqrt b + \sqrt c = z$ or $\sqrt a + \sqrt b + \sqrt c + \sqrt d = z$, and so far I've discussed $\sqrt a + \sqrt b = z$. Let's say there are $n$ terms on the left, so that we're looking for an even degree-$2^n$ polynomial in $z$; and I will set $a = a_1$, $b = a_2$, $c = a_3$, and on up to $a_n$. The above calculation also illustrates that the coefficient on $z^{2^n - 2k}$ will be a symmetric polynomial in the $a_i$, homogeneous of degree $k$. That it's symmetric is clear: the problem as posed is symmetric in the $a_i$. That it is homogeneous of degree $k$ follows from rescaling all $a_i$ to $\lambda a_i$; then the solutions $z$ uniformly rescale to $\sqrt \lambda z$.

So in the case that you originally asked about, with $a=3$, we're looking for:

$z^8 + p_1(a_1,a_2,a_3) z^6 + p_2(a_1,a_2,a_3) z^4 + p_3(a_1,a_2,a_3) z^2 + p_4(a_1,a_2,a_3)$

where each $p_j$ is a homogeneous symmetric polynomial of degree $j$. It is well-known, then, that $p_j$ is a polynomial in the polynomials $s_1 = a_1 + a_2 + a_3$, $s_2 = a_1^2 + a_2^2 + a_3^2$, $s_3 = a_1^3 + a_2^3 + a_3^3$, $\dots$, $s_j = a_1^j + a_2^j + a_3^j$. For example, $p_1 = \alpha(a_1 + a_2 + a_3)$ for some coefficient $\alpha$, and

$$ p_2 = \beta(a_1 + a_2 + a_3)^2 + \gamma(a_1^2 + a_2^2 + a_3^2) $$

There are three as-yet undetermined coefficients in $p_3 = \delta s_1^3 + \epsilon s_1 s_2 + \zeta s_3$, and five coefficients in $p_4 = \eta s_1^4 + \theta s_1^2 s_2 + \iota s_2^2 + \kappa s_1 s_3 + \lambda s_4$. All together, we have reduced the problem from computing some arbitrary degree-$8$ polynomial with polynomial coefficients to computing $11$ rational numbers.

Can we pair these down at all? When $a \neq 0$ but $b=c=0$, the two solutions need to be $z = \pm \sqrt a$. (Well, each of these is really for solutions, for "the two values of $\pm\sqrt{0}$".) So in this case our polynomial had better evaluate to $(z^2 - a)^4$. In particular, $\alpha = -4$, and $\beta + \gamma = 6$, $\delta + \epsilon + \zeta = -4$, and $\eta + \theta + \iota + \kappa + \lambda = 1$. This completely determines $\alpha$, and of the remaining $10$ unknowns, we have $3$ equations.

When $a = b$ and $c=0$, the solutions should be $\pm 2\sqrt a$ and $0$. (Actually, two each of the first two and four of the last one.) Thus the polynomial should be $(z^2 - 4a)^2z^4$. This will determine $\alpha$ again, and give you three more equations, one of which is enough to determine $\beta,\gamma$. So we've paired the space of unknowns down to being $4$-dimensional.

You can almost certainly continue in this way. For example, set $c=0$ and $b=4a$. Then $z = \pm3\sqrt a$ or $\pm \sqrt a$, each with multiplicity two, and so the polynomial in this case is $(z^2-a)^2(z^2-9a)^2$. I think this will give you two new equations when you combine it with what you already have, and I think it's enough to determine $p_3$.

One final thing to mention: the coefficient on $z^0$ is always the product (up to a sign, which is $+1$ since there are an even number of solutions) of all the solutions. Well, the product of the solutions is $(\sqrt a + \sqrt b + \sqrt c)^2(\sqrt a - \sqrt b + \sqrt c)^2(\sqrt a - \sqrt b - \sqrt c)^2(\sqrt a + \sqrt b - \sqrt c)^2$, which is a version of the $n=2$ case. Doing this gives a different way to get $p_4$.

Here's what I get, actually doing the arithmetic:

$$ z^8 - 4(a+b+c) z^6 + (2(a+b+c)^2 + 2(a^2+b^2+c^2))z^4 + (\frac43 (a+b+c)^3 - \frac83 (a^3+b^3+c^3))z^2 + (a^2 + b^2 + c^2 - 2(ab + bc + ac))^2 $$

Note that the fractions in the coefficient on $z^2$ are strong evidence that I made an arithmetic mistake there, but I think the coefficients on $z^6$, $z^4$, and $z^0$ are correct.

  • $\begingroup$ I tried inserting the values $a=4$, $b=4$, $c=9$ and $z=7$ (as $\sqrt{4} + \sqrt{4} + \sqrt{9} = 7$) into your last equation, it did not result in $0$ so I think you are right that there is some arithmetic mistake (your other equation seems to be right though). But no worries I'll repeat the calculations myself, now that I understand your method. Thanks a lot for giving me a detailed answer. As $a$, $b$ and $c$ are just placeholders for more complicated expressions, I have a lot more work to do on my problem. $\endgroup$ – Sid Nov 15 '13 at 4:22
  • $\begingroup$ Just realized that I have a question: I am more interested in the order of $a, b$ and $c$ than in the order of $z$. Can we know what order we will get for those if we have $n$ terms on the left? $\endgroup$ – Sid Nov 15 '13 at 4:28
  • $\begingroup$ Or actually it seems to me like you have answered that question, sorry. $\endgroup$ – Sid Nov 15 '13 at 5:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.