$\{\sqrt{x}\}+\{\sqrt{y}\}=1+\{\sqrt{z}\}$, defined as $\lfloor a\rfloor+\{a\}=a $

Question

We define $\{a\}$ such that $\lfloor a\rfloor+\{a\}=a $, for $a\in R$. Show that, for every fixed $k\in N$, there exists a solution to the equation $\{\sqrt{x}\}+\{\sqrt{y}\}=1+\{\sqrt{z}\}$ with $x,y,z\in N$ and $x>k, y>k, z>k$.

My attempts:

I looked at some special cases trying to find a general relation/solution, and here is the closest I've got so far:

• Special case $\{\sqrt{z}\}=0$ implies that $\{\sqrt{x}\}+\{\sqrt{y}\}=1$ , so there exists some $m\in N$ such that $\sqrt{x}+\sqrt{y}=m$ while neither $x$ nor $y$ being a perfect square. Squaring both parts we have $x+y+2\sqrt{xy}=m^2$. Leaving just the root on the left side and quaring again, $4xy=m^4+x^2+y^2-2m^2x-2m^2y-2xy$, and thus $0=m^2(m^2-2x-2y)+(x-y)^2$. We now introduce $s=x+y$ and the equation transforms into $m^2(m^2-2s)+(s-2y)^2=0$. We now expand the square term and change the order to make it look like a quadratic equation in $s$: $s^2-2(m^2+y)s+y^2+m^4=0$. We solve for s and we get under the root $(m^2+y)^2-m^4-y^2=2m^2y$, which is a square if and only if $y=2u^2$ with $u$ an integer. Going back to $x+y+2\sqrt{xy}=m^2$, if $y=2u^2$, $x$ must be of the same form, $x=2w^2$. Plugging this in $\sqrt{x}+\sqrt{y}=m$ we get no integer solutions, and that's sad.

After, I tried some othe stuff like finding other solutions from one (supposedly) already given, multiplying $x,y$ and $z$ by 100 or something similar but found nothing particularly relevant or promising.

If anyone could give me hints/methods or solutions would be really appreciated. Cheers :)

Answer 1

There are infinitely many numbers whose square root has fractional part $>0.5.$

Just take any number that is $1$ less than a square number. For example:

$\{ \sqrt{15}\} = 0.872\ldots,\quad \sqrt{143}=0.958\ldots$

[We needn't take a number so close to a square number, but $1$ less does always work].

Now notice that if a number $x$ has fractional part $\{x\}>0.5,$ then $1+\{2x\} = \{x\} + \{x\}. $

So for example,

$\{ \sqrt{15}\} + \{ \sqrt{15}\} = 1 + \{ 2\sqrt{15}\} = 1 + \{ \sqrt{60}\}.$

This is the key to the problem.