12) Its equation of Hyperbola and locus tends infinity so infinite solutions.
I think there is a '+' sign instead of '-'.
1) Solve for x:
[x2]= x +2{x} [ ] denotes the G.I.F.{ } is the fractional part.
2) Find the total number of integral solutions of x2-2y2 =2000.
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711. for the first one, write the right part as,
[x2] = [x] + 3{x}
[x2]-[x] = 3 {x}
Clearly the values that right side can take are 0 , 1 and 2. (Since the LHS is an integer)
For these 3 values solve to obtain the answer. Graphs might be helpful somewhere.
71@bond001. For what you have posted for second question is wrong. Are all the points on hyperbola integers?
1not 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
1708\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................
212) 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.
341once 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
341As 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
7So. I think the answer of the 1st q is -2/3, 0, 1..
and the 2nd q has no solution..