Integer solutions to $\sqrt{x_1}+\cdots+\sqrt{x_k}=\sqrt{y}$

Question

I am trying to figure out integer solutions to the following equation

$$\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_k}=\sqrt{y},$$

where $x_1,...,x_k,y$ are variables and $k$ is a positive integer $\geq2$.

It is easy to see that if $a_1+a_2+\cdots+a_k=b$, then $(x_1,x_2,...,x_k,y)=(a_1^2,a_2^2,...,a_k^2,b^2)$ is a solution.

It is also easy to see that for all $a\in\mathbb{N}$, $(x_1,x_2,...,x_k,y)=(a,a,...,a,k^2a)$ is a solution.

Are these all the solutions we have?

A more generalized solution is: $(x_1,x_2,...,x_k,y)=(a_1^2c,a_2^2c,...,a_k^2c,b^2c)$ for some integer $c$. This covers both your stated solutions. — user3257842, Jun 05 '23 at 03:51
See How to prove the set ${\sqrt{n}:\textrm{$n$ is squarefree}}$ to be a linearly independent set? and links therein. — dxiv, Jun 05 '23 at 03:57
What kind of "variables" are your variables? Are we supposed to assume they are real numbers? Complex numbers? Integers? — Dan Asimov, Jun 05 '23 at 04:28
They are integers. So I am looking for integer solutions. I don't know if I understand this correctly, but by the link @dxiv suggested, the solutions from user3257842 are the only integer solutions? — Collier Gaiser, Jun 05 '23 at 04:30

score 2 · Accepted Answer · edited Jun 06 '23 at 20:08

As pointed out by @dxiv, the set $\{\sqrt{n}:n\text{ is squarefree}\}$ is $\mathbb{Q}$-linearly independent. This means that $\forall \lambda_1,...,\lambda_k\in\mathbb{Q},n_1,...,n_k\geq1$ squarefree $\lambda_1\sqrt{n_1}+...+\lambda_k\sqrt{n_k}=0$ necessarily implies that $\lambda_1=...=\lambda_k=0$. Let's now prove that the set of solutions of the diophantine equation $$\sqrt{x_1}+...+\sqrt{x_k}=\sqrt{y}$$ is $S=\{(ab_1^2,...,ab_k^2,a(b_1+...+b_k)^2)\in\mathbb{N}^{k+1}:a,b_1,...,b_k\in\mathbb{N}\}$ where I consider $0\in\mathbb{N}$. Note that, given a positive integer $x$ with prime decomposition $x=\prod_j p_j^{\nu_j}$, we can decompose it as $$x=s^2t\ \text{ with }\ s=\prod_{j}p_j^{\left\lfloor{\frac{\nu_j}{2}}\right\rfloor}\ \text{ and }\ t = \prod_{j}p_j^{\nu_j(\text{mod }2)}$$ where, clearly, $s^2$ is the greatest square divisor of $x$, e.g. $24 = 2^2\cdot 6$. This decomposition is unique. With this in mind, write $\forall1\leq j\leq k:x_j=s_j^2t_j$ with $t_j$ squarefree and $y=s^2t$ with $t$ squarefree as well. Then, the equation becomes $$s_1\sqrt{t_1}+...+s_k\sqrt{t_k}-s\sqrt{t}=0$$ So, since squarefree numbers are $\mathbb{Q}$-linearly independent, and $s$ and the $s_i$ are all positive, there are only two possibilities:

$\forall 1\leq j\leq k:t_j=t$ which means that $s=s_1+...+s_k$ and we can write the solution as $$(x_1,...,x_k,y)=(ts_1^2,...,ts_k^2,t(s_1+...+s_k)^2)\in S$$
$\exists 1\leq j_0\leq k:t_{j_0}\neq t$ which implies that $s_{j_0}=0$ and means that $(x_1,...,x_{j_0-1},x_{j_0+1},...,x_k,y)$ is a solution of $$\sqrt{x'_1}+...+\sqrt{x'_{k-1}}=\sqrt{y'}$$ Via induction, we see that $(x_1,..,x_k,y)\in S$ choosing $b_j=0$ for $x_j=ab_j^2$ if $x_j=0$. $\square$

Minor nitpick; it is not necessarily true that $t$ is the greatest squarefree divisor of $x$. For $x=12=2^2\cdot3$ you have $t=3$ but $6$ is a greater squarefree divisor. — Servaes, Jun 06 '23 at 20:02
Also you write "...$x_j=s_j^2t_j$ with $s_j$ squarefree...", but I think you mean "...with $t_j$ squarefree...". — Servaes, Jun 06 '23 at 20:05
I changed the variable $m$ to $s$. And I've added a small remark that was implicit and tripped me up. I also made explicit that $x$ is a positive integer I guess the argument goes through for negative integers as well. Feel free to revert if you disagree. — Servaes, Jun 06 '23 at 20:08

Integer solutions to $\sqrt{x_1}+\cdots+\sqrt{x_k}=\sqrt{y}$

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