Irrational number multiplied by its fractional part becomes rational (SOLVED)

Question

Here's a Korean middle school midterm problem I've been struggling for quite some time now.

"$X$ is an irrational number such that $X>0$, and $Y$ is fractional part of $X$. If $$X^2+Y^2=27$$, find $X$."

So if we say
$X=(Y+N)$, such that $N=|X|$, we get
$(Y+N)^2+Y^2=27$; which is
$(Y^2+2YN+N^2)+Y^2$
$=2(Y^2)+2NY+N^2=27$.
As both $N^2$ and $27$ are rational, we must prove that there exists $Y$ and $N$ such that
$2(Y^2)+2NY$ is rational, which also means
$Y^2+NY$
$=Y(N+Y)$
$=XY$ is rational.
I can't find such irrational number which becomes a rational number when multiplied with its fractional part.

Can anyone tell how to solve this problem, or at least prove whether this is solvable/unsolvable?

Answer 1

$X$ is between 5 and 6 because $Y^2$ is between 0 and 1. Write $X = 5 + Y$ and you get a quadratic in $Y$ which you should be able to solve.

Answer 2

For your internal question: let $A=\frac{1+\sqrt{3}}{2}$. Then $\{ A\} = \frac{-1+\sqrt{3}}{2}$ and $A\cdot\{A\}=\frac{1}{2}$.

In general, for any $n,a$ with $0<a<4n+4$, you have:

$$\left\{\frac{n+\sqrt{n^2+a}}{2}\right\} = \frac{-n+\sqrt{n^2+a}}{2}$$

and:

$$\frac{n+\sqrt{n^2+a}}{2}\cdot\left\{\frac{n+\sqrt{n^2+a}}{2}\right\}=\frac{a}{4}$$

[Here I'm using $\{x\}$ for the fractional part of $x$.]

Answer 3

2(Y^2)+2NY is an integer, which also means Y^2+NY is an integer

is not true.

Hint: from $2Y^2+2NY+N^2=27$, try different values for $0\le N<6$ and solve for $0 < Y < 1$.

Answer 4

$X=N+Y$ leads to $$\tag127=X^2+Y^2=N^2+2NY+2Y^2 $$ From $0<Y^2<1$ we conclude $26<X^2<27$, hence $\sqrt{26}<X<\sqrt{27}$ so that $N=\lfloor X\rfloor = 5$. Then $(1)$ becomes $$2Y^2+10Y+25=27$$ and can be solved for $Y$.

Answer 5

Since $0<Y<1$, the integer part of $X$ must be 5 (the largest integer whose square does not exceed 27. After some manipulations, one can get $XY=1$. From there one can solve a quadratic equation for $X$ and $Y$.