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 :)

  • 1
    $\begingroup$ Good question with a cool result that I never knew about. +1 $\endgroup$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.