Finding the values of $x$ such that $[x^2]=[x]^2$ ($[cdot]$ denotes the floor function)

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I was trying to find the values of $x$ that makes the equation $[x^2]=[x]^2$ true where $[cdot]$ denotes the floor function. I tried doing the following:

Let $x=n+r$ where $n=[x]$ and $0leq r<1$. Then $x^2=n^2+2nr+r^2$ which implies that $[x^2]=[n^2+2nr+r^2]=n^2+[2nr+r^2]=[x]^2=n^2$. Subtracting both sides of the equation by $n^2$ I obtained $[2nr+r^2]=0$ if and only if $0leq 2nr+r^2<1$. I got stucked in here. Any suggestions or hints that can you give me?

elementary-number-theory floor-function

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edited Jul 27 at 21:59

asked Jul 27 at 4:43

user573497

2009

add a comment | 

up vote
6
down vote

favorite

1

I was trying to find the values of $x$ that makes the equation $[x^2]=[x]^2$ true where $[cdot]$ denotes the floor function. I tried doing the following:

Let $x=n+r$ where $n=[x]$ and $0leq r<1$. Then $x^2=n^2+2nr+r^2$ which implies that $[x^2]=[n^2+2nr+r^2]=n^2+[2nr+r^2]=[x]^2=n^2$. Subtracting both sides of the equation by $n^2$ I obtained $[2nr+r^2]=0$ if and only if $0leq 2nr+r^2<1$. I got stucked in here. Any suggestions or hints that can you give me?

elementary-number-theory floor-function

share|cite|improve this question

edited Jul 27 at 21:59

asked Jul 27 at 4:43

user573497

2009

add a comment | 

up vote
6
down vote

favorite

1

up vote
6
down vote

favorite

1

1

I was trying to find the values of $x$ that makes the equation $[x^2]=[x]^2$ true where $[cdot]$ denotes the floor function. I tried doing the following:

Let $x=n+r$ where $n=[x]$ and $0leq r<1$. Then $x^2=n^2+2nr+r^2$ which implies that $[x^2]=[n^2+2nr+r^2]=n^2+[2nr+r^2]=[x]^2=n^2$. Subtracting both sides of the equation by $n^2$ I obtained $[2nr+r^2]=0$ if and only if $0leq 2nr+r^2<1$. I got stucked in here. Any suggestions or hints that can you give me?

elementary-number-theory floor-function

share|cite|improve this question

edited Jul 27 at 21:59

asked Jul 27 at 4:43

user573497

2009

I was trying to find the values of $x$ that makes the equation $[x^2]=[x]^2$ true where $[cdot]$ denotes the floor function. I tried doing the following:

Let $x=n+r$ where $n=[x]$ and $0leq r<1$. Then $x^2=n^2+2nr+r^2$ which implies that $[x^2]=[n^2+2nr+r^2]=n^2+[2nr+r^2]=[x]^2=n^2$. Subtracting both sides of the equation by $n^2$ I obtained $[2nr+r^2]=0$ if and only if $0leq 2nr+r^2<1$. I got stucked in here. Any suggestions or hints that can you give me?

elementary-number-theory floor-function

share|cite|improve this question

edited Jul 27 at 21:59

asked Jul 27 at 4:43

user573497

2009

share|cite|improve this question

share|cite|improve this question

edited Jul 27 at 21:59

edited Jul 27 at 21:59

edited Jul 27 at 21:59

asked Jul 27 at 4:43

user573497

2009

asked Jul 27 at 4:43

user573497

2009

asked Jul 27 at 4:43

user573497

2009

2009

add a comment | 

add a comment | 

4 Answers
4

active

oldest

votes

up vote
5
down vote



accepted

$$0leq 2nr+r^2<1$$

Solve$$ 2nr+r^2-1=0$$ for $r$ in terms of $n$.

$$ r=-npm sqrt n^2+1 $$

For $$-n- sqrt n^2+1 <r< -n+sqrt n^2+1$$

We get $$0leq 2nr+r^2<1$$

Since $rge 0$, we just need $$ r< -n+sqrt n^2+1$$

For example if $n=10$, we have to have $r<sqrt 101-10 approx 0.049875$

share|cite|improve this answer

answered Jul 27 at 5:22

Mohammad Riazi-Kermani

27.3k41851

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The inequality $0 leqslant 2nr + r^2$ is not for free when $n < 0$, which makes the answer incomplete.
  – Adayah
  Jul 27 at 14:32
  

  
  
  
  

add a comment | 

up vote
8
down vote

Write $x=n+r$, where $ninmathbbZ$ and $0le rlt1$. Then we want
$$
leftlfloor n^2+2nr+r^2rightrfloor=leftlfloor(n+r)^2rightrfloor=lfloor n+rrfloor^2=n^2tag1
$$
If $r=0$, then $(1)$ is true. Thus, we have any $xinmathbbZ$ is a solution.

Now, assume $0lt rlt1$. $(1)$ requires
$$
0le2nr+r^2lt1tag2
$$
To satisfy the left-hand inequality of $(2)$, we need
$$
2n+rge0tag3
$$
So we need $nge0$. To satisfy the right-hand inequality of $(2)$, we need
$$
0lt rlt-n+sqrtn^2+1tag4
$$
which is equivalent to
$$
nlt xltsqrtn^2+1tag5
$$
Therefore, we get the complete solution set to be
$$
xinmathbbZcupbigcup_n=0^inftyleft(n,sqrtn^2+1right)tag6
$$

share|cite|improve this answer

answered Jul 27 at 5:53

robjohn♦

258k25297612

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I really liked your solution! Thank you very much!
  – user573497
  Jul 27 at 5:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can eq.6 be simplified to $$xinbigcup_n=0^inftybigg[n,sqrtn^2+1bigg)tag6$$?
  – AccidentalFourierTransform
  Jul 27 at 17:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AccidentalFourierTransform: that leaves out all the negative integers.
  – robjohn♦
  Jul 27 at 18:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @robjohn Ah, right. Silly me.
  – AccidentalFourierTransform
  Jul 27 at 18:28
  

  
  
  
  

add a comment | 

up vote
1
down vote

I'm not sure there is going to be a "clean" solution of this, you will have to consider various cases.

• If $r=0$ the inequalities are clearly true.


• If $r>0$ and $n<0$ then $2nle-2$ so $2n+r<0$ so $2nr+r^2<0$ and there is no solution.


• The remaining case is $r>0$ and $nge0$. Then $0le2nr+r^2$ is certainly true, so you have to solve $2nr+r^2<1$ to find $r$ in terms of $n$.

See if you can finish it from here.

share|cite|improve this answer

answered Jul 27 at 5:03

David

65.8k662124

add a comment | 

up vote
1
down vote

I think that you have done almost everything. In your notation you have the following inequalities: $r>=0$, $r<1$, $r^2+2nr-1<0$. The last one gives $r<sqrtn^2+1-n$. This gives possible values of $r$ for given $n>=0$. For negative $n$ your computation is not correct.

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answered Jul 27 at 5:08

V. Asnin

711

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote



accepted

$$0leq 2nr+r^2<1$$

Solve$$ 2nr+r^2-1=0$$ for $r$ in terms of $n$.

$$ r=-npm sqrt n^2+1 $$

For $$-n- sqrt n^2+1 <r< -n+sqrt n^2+1$$

We get $$0leq 2nr+r^2<1$$

Since $rge 0$, we just need $$ r< -n+sqrt n^2+1$$

For example if $n=10$, we have to have $r<sqrt 101-10 approx 0.049875$

share|cite|improve this answer

answered Jul 27 at 5:22

Mohammad Riazi-Kermani

27.3k41851

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The inequality $0 leqslant 2nr + r^2$ is not for free when $n < 0$, which makes the answer incomplete.
  – Adayah
  Jul 27 at 14:32
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

$$0leq 2nr+r^2<1$$

Solve$$ 2nr+r^2-1=0$$ for $r$ in terms of $n$.

$$ r=-npm sqrt n^2+1 $$

For $$-n- sqrt n^2+1 <r< -n+sqrt n^2+1$$

We get $$0leq 2nr+r^2<1$$

Since $rge 0$, we just need $$ r< -n+sqrt n^2+1$$

For example if $n=10$, we have to have $r<sqrt 101-10 approx 0.049875$

share|cite|improve this answer

answered Jul 27 at 5:22

Mohammad Riazi-Kermani

27.3k41851

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The inequality $0 leqslant 2nr + r^2$ is not for free when $n < 0$, which makes the answer incomplete.
  – Adayah
  Jul 27 at 14:32
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

up vote
5
down vote



accepted

$$0leq 2nr+r^2<1$$

Solve$$ 2nr+r^2-1=0$$ for $r$ in terms of $n$.

$$ r=-npm sqrt n^2+1 $$

For $$-n- sqrt n^2+1 <r< -n+sqrt n^2+1$$

We get $$0leq 2nr+r^2<1$$

Since $rge 0$, we just need $$ r< -n+sqrt n^2+1$$

For example if $n=10$, we have to have $r<sqrt 101-10 approx 0.049875$

share|cite|improve this answer

answered Jul 27 at 5:22

Mohammad Riazi-Kermani

27.3k41851

$$0leq 2nr+r^2<1$$

Solve$$ 2nr+r^2-1=0$$ for $r$ in terms of $n$.

$$ r=-npm sqrt n^2+1 $$

For $$-n- sqrt n^2+1 <r< -n+sqrt n^2+1$$

We get $$0leq 2nr+r^2<1$$

Since $rge 0$, we just need $$ r< -n+sqrt n^2+1$$

For example if $n=10$, we have to have $r<sqrt 101-10 approx 0.049875$

share|cite|improve this answer

answered Jul 27 at 5:22

Mohammad Riazi-Kermani

27.3k41851

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 5:22

Mohammad Riazi-Kermani

27.3k41851

answered Jul 27 at 5:22

Mohammad Riazi-Kermani

27.3k41851

answered Jul 27 at 5:22

Mohammad Riazi-Kermani

27.3k41851

27.3k41851

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The inequality $0 leqslant 2nr + r^2$ is not for free when $n < 0$, which makes the answer incomplete.
  – Adayah
  Jul 27 at 14:32
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The inequality $0 leqslant 2nr + r^2$ is not for free when $n < 0$, which makes the answer incomplete.
  – Adayah
  Jul 27 at 14:32
  

  
  
  
  

1

1

The inequality $0 leqslant 2nr + r^2$ is not for free when $n < 0$, which makes the answer incomplete.
– Adayah
Jul 27 at 14:32

The inequality $0 leqslant 2nr + r^2$ is not for free when $n < 0$, which makes the answer incomplete.
– Adayah
Jul 27 at 14:32

add a comment | 

up vote
8
down vote

Write $x=n+r$, where $ninmathbbZ$ and $0le rlt1$. Then we want
$$
leftlfloor n^2+2nr+r^2rightrfloor=leftlfloor(n+r)^2rightrfloor=lfloor n+rrfloor^2=n^2tag1
$$
If $r=0$, then $(1)$ is true. Thus, we have any $xinmathbbZ$ is a solution.

Now, assume $0lt rlt1$. $(1)$ requires
$$
0le2nr+r^2lt1tag2
$$
To satisfy the left-hand inequality of $(2)$, we need
$$
2n+rge0tag3
$$
So we need $nge0$. To satisfy the right-hand inequality of $(2)$, we need
$$
0lt rlt-n+sqrtn^2+1tag4
$$
which is equivalent to
$$
nlt xltsqrtn^2+1tag5
$$
Therefore, we get the complete solution set to be
$$
xinmathbbZcupbigcup_n=0^inftyleft(n,sqrtn^2+1right)tag6
$$

share|cite|improve this answer

answered Jul 27 at 5:53

robjohn♦

258k25297612

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I really liked your solution! Thank you very much!
  – user573497
  Jul 27 at 5:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can eq.6 be simplified to $$xinbigcup_n=0^inftybigg[n,sqrtn^2+1bigg)tag6$$?
  – AccidentalFourierTransform
  Jul 27 at 17:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AccidentalFourierTransform: that leaves out all the negative integers.
  – robjohn♦
  Jul 27 at 18:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @robjohn Ah, right. Silly me.
  – AccidentalFourierTransform
  Jul 27 at 18:28
  

  
  
  
  

add a comment | 

up vote
8
down vote

Write $x=n+r$, where $ninmathbbZ$ and $0le rlt1$. Then we want
$$
leftlfloor n^2+2nr+r^2rightrfloor=leftlfloor(n+r)^2rightrfloor=lfloor n+rrfloor^2=n^2tag1
$$
If $r=0$, then $(1)$ is true. Thus, we have any $xinmathbbZ$ is a solution.

Now, assume $0lt rlt1$. $(1)$ requires
$$
0le2nr+r^2lt1tag2
$$
To satisfy the left-hand inequality of $(2)$, we need
$$
2n+rge0tag3
$$
So we need $nge0$. To satisfy the right-hand inequality of $(2)$, we need
$$
0lt rlt-n+sqrtn^2+1tag4
$$
which is equivalent to
$$
nlt xltsqrtn^2+1tag5
$$
Therefore, we get the complete solution set to be
$$
xinmathbbZcupbigcup_n=0^inftyleft(n,sqrtn^2+1right)tag6
$$

share|cite|improve this answer

answered Jul 27 at 5:53

robjohn♦

258k25297612

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I really liked your solution! Thank you very much!
  – user573497
  Jul 27 at 5:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can eq.6 be simplified to $$xinbigcup_n=0^inftybigg[n,sqrtn^2+1bigg)tag6$$?
  – AccidentalFourierTransform
  Jul 27 at 17:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AccidentalFourierTransform: that leaves out all the negative integers.
  – robjohn♦
  Jul 27 at 18:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @robjohn Ah, right. Silly me.
  – AccidentalFourierTransform
  Jul 27 at 18:28
  

  
  
  
  

add a comment | 

up vote
8
down vote

up vote
8
down vote

Write $x=n+r$, where $ninmathbbZ$ and $0le rlt1$. Then we want
$$
leftlfloor n^2+2nr+r^2rightrfloor=leftlfloor(n+r)^2rightrfloor=lfloor n+rrfloor^2=n^2tag1
$$
If $r=0$, then $(1)$ is true. Thus, we have any $xinmathbbZ$ is a solution.

Now, assume $0lt rlt1$. $(1)$ requires
$$
0le2nr+r^2lt1tag2
$$
To satisfy the left-hand inequality of $(2)$, we need
$$
2n+rge0tag3
$$
So we need $nge0$. To satisfy the right-hand inequality of $(2)$, we need
$$
0lt rlt-n+sqrtn^2+1tag4
$$
which is equivalent to
$$
nlt xltsqrtn^2+1tag5
$$
Therefore, we get the complete solution set to be
$$
xinmathbbZcupbigcup_n=0^inftyleft(n,sqrtn^2+1right)tag6
$$

share|cite|improve this answer

answered Jul 27 at 5:53

robjohn♦

258k25297612

Write $x=n+r$, where $ninmathbbZ$ and $0le rlt1$. Then we want
$$
leftlfloor n^2+2nr+r^2rightrfloor=leftlfloor(n+r)^2rightrfloor=lfloor n+rrfloor^2=n^2tag1
$$
If $r=0$, then $(1)$ is true. Thus, we have any $xinmathbbZ$ is a solution.

Now, assume $0lt rlt1$. $(1)$ requires
$$
0le2nr+r^2lt1tag2
$$
To satisfy the left-hand inequality of $(2)$, we need
$$
2n+rge0tag3
$$
So we need $nge0$. To satisfy the right-hand inequality of $(2)$, we need
$$
0lt rlt-n+sqrtn^2+1tag4
$$
which is equivalent to
$$
nlt xltsqrtn^2+1tag5
$$
Therefore, we get the complete solution set to be
$$
xinmathbbZcupbigcup_n=0^inftyleft(n,sqrtn^2+1right)tag6
$$

share|cite|improve this answer

answered Jul 27 at 5:53

robjohn♦

258k25297612

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 5:53

robjohn♦

258k25297612

answered Jul 27 at 5:53

robjohn♦

258k25297612

answered Jul 27 at 5:53

robjohn♦

258k25297612

258k25297612

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I really liked your solution! Thank you very much!
  – user573497
  Jul 27 at 5:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can eq.6 be simplified to $$xinbigcup_n=0^inftybigg[n,sqrtn^2+1bigg)tag6$$?
  – AccidentalFourierTransform
  Jul 27 at 17:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AccidentalFourierTransform: that leaves out all the negative integers.
  – robjohn♦
  Jul 27 at 18:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @robjohn Ah, right. Silly me.
  – AccidentalFourierTransform
  Jul 27 at 18:28
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I really liked your solution! Thank you very much!
  – user573497
  Jul 27 at 5:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can eq.6 be simplified to $$xinbigcup_n=0^inftybigg[n,sqrtn^2+1bigg)tag6$$?
  – AccidentalFourierTransform
  Jul 27 at 17:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AccidentalFourierTransform: that leaves out all the negative integers.
  – robjohn♦
  Jul 27 at 18:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @robjohn Ah, right. Silly me.
  – AccidentalFourierTransform
  Jul 27 at 18:28
  

  
  
  
  

I really liked your solution! Thank you very much!
– user573497
Jul 27 at 5:54

I really liked your solution! Thank you very much!
– user573497
Jul 27 at 5:54

Can eq.6 be simplified to $$xinbigcup_n=0^inftybigg[n,sqrtn^2+1bigg)tag6$$?
– AccidentalFourierTransform
Jul 27 at 17:09

Can eq.6 be simplified to $$xinbigcup_n=0^inftybigg[n,sqrtn^2+1bigg)tag6$$?
– AccidentalFourierTransform
Jul 27 at 17:09

@AccidentalFourierTransform: that leaves out all the negative integers.
– robjohn♦
Jul 27 at 18:11

@AccidentalFourierTransform: that leaves out all the negative integers.
– robjohn♦
Jul 27 at 18:11

@robjohn Ah, right. Silly me.
– AccidentalFourierTransform
Jul 27 at 18:28

@robjohn Ah, right. Silly me.
– AccidentalFourierTransform
Jul 27 at 18:28

add a comment | 

up vote
1
down vote

I'm not sure there is going to be a "clean" solution of this, you will have to consider various cases.

• If $r=0$ the inequalities are clearly true.


• If $r>0$ and $n<0$ then $2nle-2$ so $2n+r<0$ so $2nr+r^2<0$ and there is no solution.


• The remaining case is $r>0$ and $nge0$. Then $0le2nr+r^2$ is certainly true, so you have to solve $2nr+r^2<1$ to find $r$ in terms of $n$.

See if you can finish it from here.

share|cite|improve this answer

answered Jul 27 at 5:03

David

65.8k662124

add a comment | 

up vote
1
down vote

I'm not sure there is going to be a "clean" solution of this, you will have to consider various cases.

• If $r=0$ the inequalities are clearly true.


• If $r>0$ and $n<0$ then $2nle-2$ so $2n+r<0$ so $2nr+r^2<0$ and there is no solution.


• The remaining case is $r>0$ and $nge0$. Then $0le2nr+r^2$ is certainly true, so you have to solve $2nr+r^2<1$ to find $r$ in terms of $n$.

See if you can finish it from here.

share|cite|improve this answer

answered Jul 27 at 5:03

David

65.8k662124

add a comment | 

up vote
1
down vote

up vote
1
down vote

I'm not sure there is going to be a "clean" solution of this, you will have to consider various cases.

• If $r=0$ the inequalities are clearly true.


• If $r>0$ and $n<0$ then $2nle-2$ so $2n+r<0$ so $2nr+r^2<0$ and there is no solution.


• The remaining case is $r>0$ and $nge0$. Then $0le2nr+r^2$ is certainly true, so you have to solve $2nr+r^2<1$ to find $r$ in terms of $n$.

See if you can finish it from here.

share|cite|improve this answer

answered Jul 27 at 5:03

David

65.8k662124

I'm not sure there is going to be a "clean" solution of this, you will have to consider various cases.

• If $r=0$ the inequalities are clearly true.


• If $r>0$ and $n<0$ then $2nle-2$ so $2n+r<0$ so $2nr+r^2<0$ and there is no solution.


• The remaining case is $r>0$ and $nge0$. Then $0le2nr+r^2$ is certainly true, so you have to solve $2nr+r^2<1$ to find $r$ in terms of $n$.

See if you can finish it from here.

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answered Jul 27 at 5:03

David

65.8k662124

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share|cite|improve this answer

answered Jul 27 at 5:03

David

65.8k662124

answered Jul 27 at 5:03

David

65.8k662124

answered Jul 27 at 5:03

David

65.8k662124

65.8k662124

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up vote
1
down vote

I think that you have done almost everything. In your notation you have the following inequalities: $r>=0$, $r<1$, $r^2+2nr-1<0$. The last one gives $r<sqrtn^2+1-n$. This gives possible values of $r$ for given $n>=0$. For negative $n$ your computation is not correct.

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answered Jul 27 at 5:08

V. Asnin

711

add a comment | 

up vote
1
down vote

I think that you have done almost everything. In your notation you have the following inequalities: $r>=0$, $r<1$, $r^2+2nr-1<0$. The last one gives $r<sqrtn^2+1-n$. This gives possible values of $r$ for given $n>=0$. For negative $n$ your computation is not correct.

share|cite|improve this answer

answered Jul 27 at 5:08

V. Asnin

711

add a comment | 

up vote
1
down vote

up vote
1
down vote

I think that you have done almost everything. In your notation you have the following inequalities: $r>=0$, $r<1$, $r^2+2nr-1<0$. The last one gives $r<sqrtn^2+1-n$. This gives possible values of $r$ for given $n>=0$. For negative $n$ your computation is not correct.

share|cite|improve this answer

answered Jul 27 at 5:08

V. Asnin

711

I think that you have done almost everything. In your notation you have the following inequalities: $r>=0$, $r<1$, $r^2+2nr-1<0$. The last one gives $r<sqrtn^2+1-n$. This gives possible values of $r$ for given $n>=0$. For negative $n$ your computation is not correct.

share|cite|improve this answer

answered Jul 27 at 5:08

V. Asnin

711

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 5:08

V. Asnin

711

answered Jul 27 at 5:08

V. Asnin

711

answered Jul 27 at 5:08

V. Asnin

711

711

add a comment | 

add a comment | 

 

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