Just simple

2) Its equation of Hyperbola and locus tends infinity so infinite solutions.
I think there is a '+' sign instead of '-'.

Substitute x=2 in equation, you wont get LHS = RHS.

@bond001. For what you have posted for second question is wrong. Are all the points on hyperbola integers?

not all but they are many.... if it were an ellipse you have less number of solutions since its closed but since parabola like straight line ends up in infinity we can have lot of solutions...

so I'm not getting your point

\hspace{-16}(1)\; $Now Using Vivek Solution\\\\ $\left[x^2\right]=x+2\left\{x\right\}$\\\\ Now $x=[x]+\left\{x\right\}$\\\\ So $\left[x^2\right]=x+\left\{x\right\}+2.\left\{x\right\}$\\\\ $\underbrace{\left[x^2\right]-\left[x\right]}_{L.H.S}=\underbrace{3\left\{x\right\}}_{R.H.S}$\\\\ So Here $\mathbf{L.H.S}$ is an Integer So $\mathbf{R.H.S}$ is also an Integer\\\\ So $\left\{x\right\}=0\;,\frac{1}{3}\;,\frac{2}{3}$\\\\ $\bullet$ If $\left\{x\right\}=0\;,$ Then $x=\left[x\right]$\\\\ So $x^2-x=0\Leftrightarrow x(x-1)=0\Leftrightarrow \boxed{x=0\;,1}$\\\\ $\bullet$ If $\left\{x\right\}=\frac{1}{3}\;,$ Then $x=I+\frac{1}{3}$\\\\ Where $I\in\mathbb{Z}$\\\\ So $\left[\left(I+\frac{1}{3}\right)^2\right]-I=1$\\\\ $\left[\left(I+\frac{1}{3}\right)^2\right]=1+I$\\\\ Now Using $x-1<[x]\leq x$\\\\ So $\left(I+\frac{1}{3}\right)^2-1<\left[\left(I+\frac{1}{3}\right)^2\right]\leq \left(I+\frac{1}{3}\right)^2$\\\\ $\left(I+\frac{1}{3}\right)^2-1<I+1\leq \left(I+\frac{1}{3}\right)^2$

\hspace{-16}$Now Solve for $\left(I+\frac{1}{3}\right)^2-1<1+I$\\\\ $9I^2-3I-17<0$\\\\ $\frac{1}{6}.\left(1-\sqrt{69}\right)<I<\frac{1}{6}.\left(1+\sqrt{69}\right)$\\\\ So Here $I\in\mathbb{Z}$. So $\boxed{I=-1\;,0\;,+1}$\\\\ Similarly Solve for $\left(I+\frac{1}{3}\right)^2\geq 1+I$\\\\ So $9I^2-3I-*\geq 0$\\\\ So $I\geq \frac{1}{6}.\left(1+\sqrt{33}\right)$ and $I\leq \frac{1}{6}.\left(1-\sqrt{33}\right)$\\\\ So Again Here $I\in\mathbb{Z}$. So $\boxed{I\in\mathbb{Z}-\left\{0,1\right\}}$\\\\ So From These 2 Inequalities $I=-1$\\\\ So $\boxed{x=I+\frac{1}{3}=-1+\frac{1}{3}=-\frac{2}{3}}$\\\\ Similarly Calculate for\\\\ $\bullet$ If $\left\{x\right\}=\frac{2}{3}$\\\\ So $x=I+\frac{2}{3}$ and $I\in\mathbb{Z}$\\\\ In a Similar Manner You can Calculate It

can anyone have a better Solution................

2) take modulo 4

and see that both x,y has to be even

x= 2p, y= 2q

so p^2 - 2q^2 = 500

again take modulo 4

p= 2m, q = 2n

so m^2 - 2n^2 = 125

take modulo 16

RHS ≡ 13 (mod 16)

but LHS≡ 0,1,2,4,7,8,9,12,14,15, (mod 16)

so no sol.

once you have reduced it to

m^2-2n^2 = 125 you can analyse modulo 5 and see that m and n are both divisible by 5.

So it can be further reduced to p^2-2q^2 = 5

which again modulo 5 means p and q are divisible by 5 which means LHS is divisible by 25 and RHS is not

Yes thanks.

BTW the RMO problems were leaked like last year too...

As jagdish wrote {x}=0,1/3, 2/3

Now if {x} = 1/3, we can write the equation as [x^2] = [x]+1 = [x+1]

If x<-1, since [x²]≥0, the equation doesnt hold.

If -1<x<0, we see that x = -2/3 is a solution.

If 0<x<1, since x²<x, again the equation cannot hold

If 1<x<2, x=4/3 is not a solution.

If x>2, we have x^2 -x -2>0 \Rightarrow x^2 > (x+1)+1 \Rightarrow [x^2]>[x+1] and the equation doesnt hold

Similarly if {x}= 2/3 the equation is [x^2]=[x]+2 = [x+2]

Its easy to see that no solutions exist for x<1. x=5/3 is not a solution.

Further x²-x-3<0 means we need x<2.5. So no other solutions exist

So. I think the answer of the 1st q is -2/3, 0, 1..
and the 2nd q has no solution..

thanks to all... worthful post!

Bhatt Sir can u explain me how 2 calculate x=-2/3 if -1<x<0..