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

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 :)

• Good question with a cool result that I never knew about. +1 Aug 1 at 12:19

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.