An example I can see using basic algebra is $\sqrt{2} + \sqrt{8} = \sqrt{18}$, but is there a general method to find integer solutions to the problem? Another question: say you are given the value of $c$; can you find the values of $a$ and $b$?

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    $\begingroup$ As a starting point, if $(\sqrt a+\sqrt b)^2\in\Bbb Z$, then $ab$ must be a square number, as $(\sqrt a+\sqrt b)^2=a+b+2\sqrt{ab}$. $\endgroup$ – Berci Dec 1 '12 at 14:20
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    $\begingroup$ Simplify by dividing out the least common divisor. Is it true that what remains will be integer square-roots only? In this example, divide out by 2, and get $\sqrt1+\sqrt4=\sqrt9$ $\endgroup$ – GEdgar Dec 1 '12 at 14:21
  • $\begingroup$ Other solutions are $a=0,\ b=c$ and $a=c,\ b=0$, and these are all of the solutions with one of the unknowns equal to zero. Also, they all have to be of the same sign if none of them is zero, and the negative solutions are in a bijective correspondence with the positive ones. So you can only consider $a,b,c>0$. $\endgroup$ – Bartek Dec 1 '12 at 14:23

Note that, since $a,b,c\in\mathbb N$, we have $$ \sqrt{a}+\sqrt{b}=\sqrt{c}\ \iff \ a+b+2\sqrt{ab}=c. $$ So a necessary and sufficient condition for $a,b,c$ to exist is that $2\sqrt{ab}\in\mathbb N$, i.e. $$ ab=\frac{n^2}4,\ \text{ for some }n\in\mathbb N; $$ as an integer is even exactly when its square is, this last condition occurs if and only if $$ ab=m^2\ \text{ for some }m\in\mathbb N. $$ Finding a solution given a $c$ is probably not easy; it can surely be done by exhaustion if $c$ is not too big.


If $a,b$ have a divisor $d$ in common, then $d$ is also a divisor of $c = a+b+2\sqrt{ab}=d\cdot\left(\frac ad+\frac bd+2\sqrt{\frac ad\frac bd}\right)$. Therefore, we can concentrate on the case that $\gcd(a,b)=1$.

Assume neither $a$ nor $b$ is a perfect square. Then $b$ also not a square in $\mathbb Q(\sqrt a)$ because $b=(u+v\sqrt a)^2$ would imply $2uv\sqrt a=b^2-u^2-av^2$, i.e. $\sqrt a$ rational (if $uv\ne0$) or $\gcd(a,b)>1$ (if $u=0$, $b^2=av^2$) or $\sqrt b=|u|\in\mathbb Q$ (if $v=0$). Therefore the field $\mathbb Q(\sqrt a, \sqrt b)$ allows four automorphisms, induced by sending $\sqrt a\to \pm\sqrt a$ and $\sqrt b\to \pm \sqrt b$ with independant choices of signs. The four different corresponding values of $\pm\sqrt a\pm\sqrt b$ must be roots of the polynomial $X^2-c$, but this polynomial has only two roots, contradiction.

Therefore at least one of $a,b$ is a perfect square. Wlog. $a=r^2$ with $r\in \mathbb N$. Then the equation becomes $r+\sqrt b=\sqrt c$, i.e. $r^2+2r\sqrt b+b=c$, whence $\sqrt b$ is rational and thus $b$ is also a perfect square.

Conclusion: Given $c=uv^2$ with $u$ squarefree, the solutions of $$\sqrt a+\sqrt b=\sqrt c$$ with $a,b\in\mathbb N$ are given by $a=ur^2$, $b=u(v-r)^2$ with $1\le r< v$. ($r\le \frac v2$ if we additionally want $a\le b$).

Example: If $c=18$, then $u=2$, $v=3$. The only possible choice with $1\le r<\frac v2$ is $r=1$ and yields $$\sqrt{2}+\sqrt 8=\sqrt{18}.$$

I knew I had answered a similar (but more general) question elsewhere before.

  • $\begingroup$ But the equation does not become $\sqrt b = c - r$, it becomes $\sqrt b = \sqrt c - r$. This looks to be an important step in your argument, too! $\endgroup$ – TonyK Dec 1 '12 at 16:50
  • $\begingroup$ @TonyK Thanks for the hint. Edited. $\endgroup$ – Hagen von Eitzen Dec 1 '12 at 22:38
  • $\begingroup$ @HagenvonEitzen, you assumed that $\gcd(a,b)=1$ but in your answer,$\gcd(a,b)=u$. Does that mean that $a,b,c$ must be have a common factor? $\endgroup$ – user97615 Oct 8 '14 at 21:00

Here's a somewhat more geometric solution. Excluding the trivial solution $(0,0,0)$, we can divide both sides by $\sqrt{c}$ to reduce the problem to the equivalent problem of finding all rational solutions to $$\sqrt{x}+\sqrt{y}=1.$$ The graph of this contains the point $(0,1)$. Given any other rational point $(a,b)$ on this graph, notice that the line connecting $(0,1)$ to $(a,b)$ will have rational slope $\le -1$. Conversely, I claim that if we draw a line through $(0,1)$ with rational slope $\le -1$, then it will hit this graph in another rational point.

To see this, let the line be $y=tx+1$ with $t$ rational. The intersection of the two graphs occurs at $$\sqrt{x}+\sqrt{tx+1}=1$$

Solving for $x$ gives $$x=\frac{4}{(t-1)^2}$$ and from $y=tx+1$ we get $$y=\frac{(t+1)^2}{(t-1)^2}.$$ These are all the rational points on this graph. To find all integers $(a,b,c)$ you can just plug in $t=\frac{r}{s}$ and clear denominators. You will get:

$$a=d(4s^2)$$ $$b=d(r+s)^2$$ $$c=d(r-s)^2$$

where $d\ge 1$.

The $d$ comes from the fact that if $\frac{x_1}{x_2}=\frac{y_1}{y_2}$ then $y_1=dx_1$ and $y_2=dx_2$ for some non-zero $d$.

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    $\begingroup$ Obviously this is overkill, I included it because the same method allows you to write down all pythagorean triples or all (a^2, b^2, c^2) in arithmetic progression. $\endgroup$ – Zach Dec 1 '12 at 23:15
  • $\begingroup$ This method is so elegant, for what values of d,r,s do I get 2,8,18? $\endgroup$ – hhsaffar Nov 14 '13 at 22:59
  • $\begingroup$ @Zach, does it mean that $a,b,c$ must have a common factor unless they are all perfect squares? $\endgroup$ – user97615 Oct 9 '14 at 1:08

Since the first part was answered, for the last part the answer is: Depends what you mean by it can be found.

First lets observe that the equation doesn't usually have unique solution. For example, if $c=n^2$, then $a=k^2, b=(n-k)^2$ is solution for all $0 \leq k \leq n$.

Similarly, for $c=n^2d$ then $a=k^2d, b=(n-k)^2d$ is solution for all $0 \leq k \leq n$. And in general, this equation will have some other solutions.

But, as pointed out, the equation has finitely many solutions (since $a,b \leq c$), thus all the possible solutions can be found.

  • $\begingroup$ The smallest possible value of $d$ gives all the possible solutions. $\endgroup$ – Henry Dec 1 '12 at 21:42

Let $c=d n^2$ with $n\ge 0$ and with $d \gt 0$ square-free, i.e. $d$ has no repeated prime factors. $d$ is the product of the individual prime factors of $c$ which occur an odd number of times.

Then $a=d k^2$ and $b=d (n-k)^2$ with $0 \le k \le n$ are the only solutions to $\sqrt a + \sqrt b = \sqrt c$, and there are $n+1$ such pairs.

This is similar to N.S.'s solution, except for the restriction on $d$.

$18 = 2^1 \times 3^2$ so $d=2$ and thus $n=3$. So there are $n+1=4$ solutions to $\sqrt a + \sqrt b = \sqrt{18}$ namely

  • $\sqrt{0^2 \times 2} + \sqrt{3^2 \times 2} = \sqrt{3^2 \times 2}$ i.e.$\sqrt{0} + \sqrt{18} = \sqrt{18}$
  • $\sqrt{1^2 \times 2} + \sqrt{2^2 \times 2} = \sqrt{3^2 \times 2}$ i.e.$\sqrt{2} + \sqrt{8} = \sqrt{18}$
  • $\sqrt{2^2 \times 2} + \sqrt{1^2 \times 2} = \sqrt{3^2 \times 2}$ i.e.$\sqrt{8} + \sqrt{2} = \sqrt{18}$
  • $\sqrt{3^2 \times 2} + \sqrt{0^2 \times 2} = \sqrt{3^2 \times 2}$ i.e.$\sqrt{18} + \sqrt{0} = \sqrt{18}$

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