Some days ago i came across a question about writing $\sqrt {2001}$ as sum of two other square roots. I managed to prove that this is not possible unless one of them is zero and the other one is $2001$.

The proof was as following: $\sqrt{2001}=\sqrt a+\sqrt b$, $\sqrt{2001}-\sqrt a=\sqrt b$ so $2001+a-2\sqrt{2001a}=b$. This shows that $2\sqrt{2001a}$ is an integer so $2001*a$ is a perfect square.
We also know that $2001=3*23*29$ which is a square-free number. so $a$ must divide all of $3,23,29$ which means $a\geq2001$ so$\sqrt a\geq\sqrt{2001}$ and $\sqrt{b}\leq 0$ which means $b=0$.

With exact method we can prove that $\sqrt{s}=\sqrt a+\sqrt b$ does not have any natural solutions with $s$ being a square-free number. Then I tried to generalize the proof for $3$ or more square roots but i failed. The only thing I always get is $\sqrt {ab}+\sqrt {bc}+\sqrt {ac}$ is an integer which does not help at all.
For what numbers can we write the square root of a square-free number as sum of three or more non-zero square roots? I would appreciate any help.

  • $\begingroup$ I remember an excercise from a book that asked a proof of that if $p_n$ is the $n$th prime, then the field $\Bbb Q(\sqrt{p_1},\ldots,\sqrt{p_n})$ does not contain $\sqrt{p_{n+1}}$. I think that this statement must be related with your question. Sadly, I couldn't solve the exercise. $\endgroup$
    – ajotatxe
    Apr 3, 2014 at 8:40
  • $\begingroup$ @user2425 Thanks you, but i would prefer an elementary proof like the first case i proved :) $\endgroup$
    – CODE
    Apr 3, 2014 at 8:51
  • $\begingroup$ I don't think there will be an elementary solution. The general case has already been answered here: math.stackexchange.com/a/437374/43288 $\endgroup$ Apr 8, 2014 at 14:11
  • $\begingroup$ A related question: Integer solutions to $\sqrt{a} + \sqrt{b} = \sqrt{c}$. $\endgroup$ Sep 23, 2020 at 2:21

2 Answers 2


Lemma 1. If $m$ is a positive integer and $\sqrt m$ is rational, then $\sqrt m$ is an integer.

Proof. Easy.

Lemma 2. If $m,n$ are positive integers and $\sqrt m+\sqrt n$ is rational, then both $\sqrt m$ and $\sqrt n$ are integers.

Proof. Say $\sqrt m+\sqrt n=x\in\Bbb Q$. Then $$\sqrt m-\sqrt n=\frac{m-n}{x}$$ is rational and so is $$\sqrt m=\frac{(\sqrt m+\sqrt n)+(\sqrt m-\sqrt n)}{2}\ ,$$ and likewise $\sqrt n\,$. By lemma 1, $\sqrt m$ and $\sqrt n$ are integers.

Now suppose that $$\sqrt a+\sqrt b+\sqrt c=\sqrt s\ ,$$ where $a,b,c,s$ are positive integers and $s$ is squarefree. Squaring and rearranging, $$2\bigl(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\bigl)=s-a-b-c\ .$$ Now add to this equation the identity $2\sqrt a\sqrt a=2a$ and factorise to obtain $$2\sqrt{bc}+2\sqrt{as}=s+a-b-c\ .$$ By lemma 2, we see that $\sqrt{as}$ is an integer; since $s$ is squarefree, $a$ must be a square times $s$, say $a=p^2s$. Similarly $b=q^2s$ and $c=r^2s$, so $$p\sqrt s+q\sqrt s+r\sqrt s=\sqrt s\ ,$$ but as $p+q+r>1$, this is impossible.

  • $\begingroup$ Thank you! Very nice answer. But can we do the same for 4 or more square roots? $\endgroup$
    – CODE
    Apr 3, 2014 at 12:37
  • $\begingroup$ No real idea. I suspect it might be significantly more difficult. I guess the first thing would be to try to generalise lemma 2 to more than two square roots. $\endgroup$
    – David
    Apr 3, 2014 at 21:23

Note: Squaring only helps when you have 5/6 or fewer terms. Otherwise, you run the risk of introducing too many square roots. As such, you need something more powerful to deal with the general case.

Theorem. Let $SF$ be the set of positive integers that are not divisible by the square of any prime. $ SF = \{1, 2, 3, 5, 6, 7, 10, 11, 13, \ldots \} $. If $ \{a_i \} _{i=1}^n$ are distinct numbers from the set $SF$, and $ \{b_i\}_{i=1}^n$ are any integers, then $ S = \sum b_i \sqrt{a_i} = 0$ if and only if all $ b_i = 0 $.

Corollary No square free integer can be written as the sum of 3 or more non-zero square roots.

Proof. The simplest approach is to use Galois Theory, which might be beyond OP. I'd present an 'elementary' approach which I first saw from Feng Zuming.

Recall the idea of conjugates. Consider the linear expression $ L(x_1, x_2, \ldots, x_n) = a_1 x_1 + a_2 x_2 + \ldots + a_n x_n$. Consider the conjugate expressions, which have the form $ L' (x_1, x_2, \ldots, x_n) = a_1 x_1 \pm a_2 x_2 \pm \ldots \pm a_n x_n$. [There are $ 1 \times 2 \times \ldots \times 2 = 2^{n-1} $ such expressions.] Let $T$ be a variable and consider the polynomial

$$ F_{L(x_1, x_2, \ldots , x_n)} (T) = \prod_{L'} \big(T - L'(x_1, x_2, \ldots, x_N) \big). $$

Consider it as a polynomial in $x_i, i\neq 1$. Changing any of the signs of $x_i$ doesn't change $F$, since the set $ \{ L' \} $ stays the same. So,

$$ F_{L(x_1, x_2, \ldots , x_n)} (T) = F_{L(x_1, \pm x_2, \ldots ,\pm x_n)} (T) . $$

As such, the polynomial expansion only contains even powers of $x_i, i\neq 1$. It is clear that it can contain odd or even powers of $x_1$, hence we have

$$ F_{L(x_1, x_2, \ldots , x_n)} (T) = x_1 P ( x_1^2, x_2 ^2, \ldots, x_n ^2 , T) + Q( x_1^2, x_2 ^2, \ldots, x_n ^2 , T). $$

Since $F$ is a polynomial with integer coefficients, it follows that $P$ and $Q$ are also polynomials with integer coefficients.

We will show a slight variant of the original problem, namely that no non-zero integer $M$ can be represented as a nontrivial canonical integer sum of radicals. What this means, is that the square roots have all been simplified, so we're left with square free terms under the root sign. In this case, we are excluding the radical $ \sqrt{1} $, which is why we now include the non-zero integer $M$. Hence this problem is equivalent. We will prove this statement by induction.

Base case: Clearly, if $ b_1 \sqrt{a_1} = M$, then $ M^2 = b_1^2 a_1$, which means that $a_1$ must be a square, which is not possible.

Suppose that we have an expression of the form $ \sum b_i \sqrt{a_i} $ such that $ \sum b_i \sqrt{a_i} = M \neq 0 $. Then, the polynomial $ F_{L(\sqrt{a_1}, \sqrt{a_2}, \ldots , \sqrt{a_n})} (M) = 0 $. From the previous discussion, we have

$ 0 = \sqrt{a_1} P(a_1, a_2, \ldots, a_n, M) + Q(a_1, a_2, \ldots, a_n, M) $

Each of these polynomials has integer coefficients, and integer variables, hence when evaluated at an integer, is equal to an integer. By the base case, this shows that $ P(a_1, a_2, \ldots, a_n, M) = Q(a_1, a_2, \ldots, a_n, M) =0$. As such, this gives us $ 0 = - \sqrt{a_1} P(a_1, a_2, \ldots, a_n, M) + Q(a_1, a_2, \ldots, a_n, M) $

Now, consider the expression $ G_{L(x_1, x_2, \ldots, x_n)} (T) = \prod \big( T + L' (x_1, x_2, \ldots, x_n) \big)= - x_1 P ( x_1^2, x_2 ^2, \ldots, x_n ^2 , T) + Q( x_1^2, x_2 ^2, \ldots, x_n ^2 , T). $

We know that $ \prod(M + L') = 0$, and thus $ M = -b_1 \sqrt{a_1} \pm a_2 x_2 \pm \ldots \pm a_n x_n $, for some combination of signs. Adding this to $ M = b_1 \sqrt{a_1} + b_2 \sqrt{a_2} + \ldots + b_n \sqrt{a_n} $, we obtain that $2M = (b_2 \pm b_2) \sqrt{a_2} + \ldots + (b_n \pm b_n) \sqrt{a_n}$, which contradicts the induction hypothesis.


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