For any $N \in \mathbb{N}$ there exist integers $x,y,z>N$ such that $\{\sqrt{x}\}+\{\sqrt{y}\}=1+\{\sqrt{z}\}$

Question

This problem is from the 2017 admission exam of the SNS (see also page 142 of this pdf for the original text in Italian)

The translation of the exercise is, more or less:

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1. Given $N$ positive integer show that there exists integers $x,y,z>N$ such that $$ \{\sqrt{x}\}+\{\sqrt{y}\}=1+\{\sqrt{z}\} $$ where $\{a\}=a-\lfloor a \rfloor$ is the fractional part of the real number $a$
2. Show that for any three numbers of this kind $z>4N$

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We can easily obtain that $$ \sqrt{x}+\sqrt{y}-\sqrt{z}=1+\lfloor \sqrt{x}\rfloor+\lfloor \sqrt{y}\rfloor-\lfloor \sqrt{z}\rfloor $$ Using the hint in the second part of the exercise i thought I can prove that, for any $N$ there is a squarefree integer $k>N$ such that $$ \sqrt{k} \ge \lfloor \sqrt{k} \rfloor+\frac{1}{2} $$ If this is true, by imposing $x=y=k$ and $z=4k$ we have a solution. I could not prove that that there are square free numbers that satisfies such relation and I cannot prove that the condition $z>4N$ is actually necessary.

By looking to the set of solution with Python (with $N<100$) i have seen that these are not the only solution. e.g $$(8, 8, 32) \quad (11, 99, 176) \quad (20, 245, 405)$$ are all solutions. Is my approach of focusing on squarefree numbers wrong? What would be a characterization of the triplets of solutions?

Answer 1

The equation is satisfied iff

1. $ \sqrt{ x} + \sqrt{y} - \sqrt{z} \in \mathbb{N}$
2. $\{\sqrt{x}\} + \{\sqrt{y} \} > 1 $

(Prove that this is the complete classification.)

OP's wishful thinking approach of $ x = y = k, z = 4k$ essentially gives us $ \sqrt{x} + \sqrt{y} - \sqrt{z} = 0 $, so condition 1 is satisfied.

To satisfy condition 2, we require $ \{ \sqrt{x} \} > 0.5$, which is true iff $ x \in [k^2 + k + 1 , k^2 + 2k ] $ for some positive integer $k$.
EG $ x = 3, 7, 8, \ldots$ are solutions, which could have been easily checked by OP.

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A classification of all such triplets will take a bit more work.

All solutions to $ \sqrt{x} + \sqrt{y} - \sqrt{z} = 0 $ are of the form $(x, y, z) = (a^2k,b^2k, c^2k)$ where $ a+b = c$. EG The solutions that OP listed out at the end.
It remains to check when $\{\sqrt{x}\} + \{\sqrt{y} \} > 1 $, which to me doesn't seem to have a simple classification. We could do things like separately require $ \{x \}, \{y \} > 0.5$, but it doesn't cover all cases.