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Prove there is no $x, y in mathbb Z^+ text satisfying fracxy +fracy+1x=4$
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Prove that there is no $x, y in mathbb Z^+$ satisfying
$$fracxy +fracy+1x=4$$
I solved it as follows but I seek better or quicker way:
$text Assume x, y in mathbb Z^+\ 1+fracy+1y+fracxy +fracy+1x=1+fracy+1y+4 \
Rightarrow left(1+fracxyright)left(1+fracy+1xright)=6+frac1y\
Rightarrow (x+y)(x+y+1)=x(6y+1)\
Rightarrow xmid (x+y) ;text or ; xmid (x+y+1)\
Rightarrow xmid y ;text or ; xmid (y+1)\
textPut; y=nx ,n in mathbb Z^+;Rightarrow; fracxnx +fracnx+1x=4 ;Rightarrow; frac1n +frac1x=4-n ;rightarrow;(1)\
textBut; frac1n +frac1x gt 0 ;Rightarrow; 4-n gt 0 ;Rightarrow; n lt 4\
textAlso; frac1n,frac1xle 1 ;Rightarrow; frac1n +frac1x le 2 ;Rightarrow; 4-n le 2 ;Rightarrow; n ge 2\
;Rightarrow; n=2 ;text or ; 3, ;textsubstituting in eq. (1), we find no integral values for x.\ textThe same for the other case.\
$
So is there any other better or intelligent way to get this result?
number-theory vieta-jumping
edited Jul 16 at 19:53
Will Jagy
97.2k594196
asked Jul 16 at 17:54
Wolfdale
24919
add a comment |Â
up vote
4
down vote
favorite
Prove that there is no $x, y in mathbb Z^+$ satisfying
$$fracxy +fracy+1x=4$$
I solved it as follows but I seek better or quicker way:
$text Assume x, y in mathbb Z^+\ 1+fracy+1y+fracxy +fracy+1x=1+fracy+1y+4 \
Rightarrow left(1+fracxyright)left(1+fracy+1xright)=6+frac1y\
Rightarrow (x+y)(x+y+1)=x(6y+1)\
Rightarrow xmid (x+y) ;text or ; xmid (x+y+1)\
Rightarrow xmid y ;text or ; xmid (y+1)\
textPut; y=nx ,n in mathbb Z^+;Rightarrow; fracxnx +fracnx+1x=4 ;Rightarrow; frac1n +frac1x=4-n ;rightarrow;(1)\
textBut; frac1n +frac1x gt 0 ;Rightarrow; 4-n gt 0 ;Rightarrow; n lt 4\
textAlso; frac1n,frac1xle 1 ;Rightarrow; frac1n +frac1x le 2 ;Rightarrow; 4-n le 2 ;Rightarrow; n ge 2\
;Rightarrow; n=2 ;text or ; 3, ;textsubstituting in eq. (1), we find no integral values for x.\ textThe same for the other case.\
$
So is there any other better or intelligent way to get this result?
number-theory vieta-jumping
edited Jul 16 at 19:53
Will Jagy
97.2k594196
asked Jul 16 at 17:54
Wolfdale
24919
I think there is a solution. Look at the formula. artofproblemsolving.com/community/c3046h1046841___
– individ
Jul 16 at 18:04
infinitely many solutions with $x,y < 0$
– Will Jagy
Jul 16 at 19:29
infinitely many rational solutions with $x,y > 0,$ we can take denominator as $2$
– Will Jagy
Jul 16 at 20:39
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Prove that there is no $x, y in mathbb Z^+$ satisfying
$$fracxy +fracy+1x=4$$
I solved it as follows but I seek better or quicker way:
$text Assume x, y in mathbb Z^+\ 1+fracy+1y+fracxy +fracy+1x=1+fracy+1y+4 \
Rightarrow left(1+fracxyright)left(1+fracy+1xright)=6+frac1y\
Rightarrow (x+y)(x+y+1)=x(6y+1)\
Rightarrow xmid (x+y) ;text or ; xmid (x+y+1)\
Rightarrow xmid y ;text or ; xmid (y+1)\
textPut; y=nx ,n in mathbb Z^+;Rightarrow; fracxnx +fracnx+1x=4 ;Rightarrow; frac1n +frac1x=4-n ;rightarrow;(1)\
textBut; frac1n +frac1x gt 0 ;Rightarrow; 4-n gt 0 ;Rightarrow; n lt 4\
textAlso; frac1n,frac1xle 1 ;Rightarrow; frac1n +frac1x le 2 ;Rightarrow; 4-n le 2 ;Rightarrow; n ge 2\
;Rightarrow; n=2 ;text or ; 3, ;textsubstituting in eq. (1), we find no integral values for x.\ textThe same for the other case.\
$
So is there any other better or intelligent way to get this result?
number-theory vieta-jumping
edited Jul 16 at 19:53
Will Jagy
97.2k594196
asked Jul 16 at 17:54
Wolfdale
24919
Prove that there is no $x, y in mathbb Z^+$ satisfying
$$fracxy +fracy+1x=4$$
I solved it as follows but I seek better or quicker way:
$text Assume x, y in mathbb Z^+\ 1+fracy+1y+fracxy +fracy+1x=1+fracy+1y+4 \
Rightarrow left(1+fracxyright)left(1+fracy+1xright)=6+frac1y\
Rightarrow (x+y)(x+y+1)=x(6y+1)\
Rightarrow xmid (x+y) ;text or ; xmid (x+y+1)\
Rightarrow xmid y ;text or ; xmid (y+1)\
textPut; y=nx ,n in mathbb Z^+;Rightarrow; fracxnx +fracnx+1x=4 ;Rightarrow; frac1n +frac1x=4-n ;rightarrow;(1)\
textBut; frac1n +frac1x gt 0 ;Rightarrow; 4-n gt 0 ;Rightarrow; n lt 4\
textAlso; frac1n,frac1xle 1 ;Rightarrow; frac1n +frac1x le 2 ;Rightarrow; 4-n le 2 ;Rightarrow; n ge 2\
;Rightarrow; n=2 ;text or ; 3, ;textsubstituting in eq. (1), we find no integral values for x.\ textThe same for the other case.\
$
So is there any other better or intelligent way to get this result?
number-theory vieta-jumping
edited Jul 16 at 19:53
Will Jagy
97.2k594196
asked Jul 16 at 17:54
Wolfdale
24919
edited Jul 16 at 19:53
Will Jagy
97.2k594196
edited Jul 16 at 19:53
Will Jagy
97.2k594196
edited Jul 16 at 19:53
Will Jagy
97.2k594196
97.2k594196
asked Jul 16 at 17:54
Wolfdale
24919
asked Jul 16 at 17:54
Wolfdale
24919
asked Jul 16 at 17:54
Wolfdale
24919
24919
I think there is a solution. Look at the formula. artofproblemsolving.com/community/c3046h1046841___
– individ
Jul 16 at 18:04
infinitely many solutions with $x,y < 0$
– Will Jagy
Jul 16 at 19:29
infinitely many rational solutions with $x,y > 0,$ we can take denominator as $2$
– Will Jagy
Jul 16 at 20:39
add a comment |Â
I think there is a solution. Look at the formula. artofproblemsolving.com/community/c3046h1046841___
– individ
Jul 16 at 18:04
infinitely many solutions with $x,y < 0$
– Will Jagy
Jul 16 at 19:29
infinitely many rational solutions with $x,y > 0,$ we can take denominator as $2$
– Will Jagy
Jul 16 at 20:39
I think there is a solution. Look at the formula. artofproblemsolving.com/community/c3046h1046841___
– individ
Jul 16 at 18:04
I think there is a solution. Look at the formula. artofproblemsolving.com/community/c3046h1046841___
– individ
Jul 16 at 18:04
infinitely many solutions with $x,y < 0$
– Will Jagy
Jul 16 at 19:29
infinitely many solutions with $x,y < 0$
– Will Jagy
Jul 16 at 19:29
infinitely many rational solutions with $x,y > 0,$ we can take denominator as $2$
– Will Jagy
Jul 16 at 20:39
infinitely many rational solutions with $x,y > 0,$ we can take denominator as $2$
– Will Jagy
Jul 16 at 20:39
add a comment |Â
4 Answers
4
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oldest
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accepted
For $x,yin Bbb Z^+$ we have $$frac xy+frac y+1x=4implies x^2-4xy+y^2+y=0implies x=2ypm sqrt 3y^2-yimplies$$ $$implies exists zin Bbb Z^+;( z^2=3y^2-y=y(3y-1))implies$$ $$ impliesexists a,b in Bbb Z^+;( y=a^2 land 3y-1=b^2)implies$$ $$impliesexists a,bin Bbb Z^+;(3a^2-1=b^2)implies$$ $$ implies exists bin Bbb Z^+;(b^2equiv -1 mod 3).$$ The 1st (displayed) line uses the Quadratic Formula.
In the 2nd line, $zne 0$ because $yin Bbb Z^+implies y(3y-1)>0.$
In the 3rd line, $a$ and $b$ exist because $y$ and $3y-1$ are co-prime members of $Bbb Z^+$ and their product is the square of the positive integer $z.$
answered Jul 21 at 2:51
DanielWainfleet
31.7k31644
Many thx for your answer, this is a nice complete solution.
– Wolfdale
Jul 21 at 3:12
My first impulse is to get rid of denominators. And whenever I see a quadratic form I apply the Quadratic Formula to see what can be found with it.
– DanielWainfleet
Jul 21 at 3:20
Actually, this is a normal action with quadratic eq. but I stopped at the last step like another solution here. But it was that simple memorizing that the square of any number is either divisible by 3 or has a remainder of 1 :D
– Wolfdale
Jul 21 at 3:23
add a comment |Â
up vote
6
down vote
Hint
$$fracxy +fracy+1x=4to fracxy+fracyx+frac1x=4$$
Call $x/y=tin Bbb Q$.
$$t+frac1t+frac1x=4to xt^2+t(1-4x)+x=0$$
By Rational Roots Theorem the candidates to be a rational root, $t$, are $pm1,pm x,pm 1/x$.
Now, test every root and check the value of $x$ you get.
Can you finish?
answered Jul 16 at 18:03
Arnaldo
18k42146
thx for the hint, but I have no idea of Rational Roots Theorem
– Wolfdale
Jul 16 at 18:06
@Wolfdale:en.wikipedia.org/wiki/Rational_root_theorem
– Arnaldo
Jul 16 at 18:07
@Wolfdale: It is simple and very useful.
– Arnaldo
Jul 16 at 18:17
1
The rational root theorem doesn't help as much as you suggest unless $x$ is prime; for example, so far as it's concerned, $t$ could be any factor of $x$. It's probably better to note that your quadratic has discriminant $1-4x$. So if $x$ is a positive integer the quadratic cannot have rational solutions.
– Micah
Jul 16 at 18:23
1
@Micah, I can't get this discriminant value, it should be $1-8x+12x^2=(2x-1)(6x-1)$
– Wolfdale
Jul 16 at 18:40
 |Â
show 6 more comments
up vote
2
down vote
Rewrite the equation: $x^2 - 4yx + y^2+y = 0impliestriangle'=(-2y)^2-1(y^2+y)= 4y^2-y^2-y = 3y^2-y=k^2 implies y(3y-1)=k^2$ . Observe that $textgcd(y,3y-1) = 1$ since if $d = textgcd implies d mid y, d mid 3y-1 implies y = md, 3y-1 = ndimplies 1 = 3y -nd= 3md - nd = d(3m-n)implies d = 1implies y = u^2, 3y-1 = v^2, uv = kimplies 3u^2-v^2=1$ . This is a Pell equation and it either has infinitely many solutions or no solutions at all. Please check its status via google.
answered Jul 16 at 18:10
DeepSea
69k54284
1
infinitely many solutions with $x,y < 0$
– Will Jagy
Jul 16 at 19:17
add a comment |Â
up vote
1
down vote
I include a proof for the original question. It is between the first two pictures.
There are infinitely many solutions with both $x,y leq 0.$ This comes under the heading of Vieta Jumping. Given a solution to $x^2 - 4xy + y^2 + y = 0,$ we can create new solutions by alternately taking mappings on the hyperbola,
$$ (x,y) mapsto (x, 4x-y - 1) ; ; ; , $$
$$ (x,y) mapsto (4x-y,y) ; ; ; . $$
The solutions on the hyperbola branch that is (mostly) in the third quadrant begin
$$ (0,0); ; (0,-1); ; (-4,-1); ; (-4,-16); ; (-60,-16); ; (-60,-225); ; ldots $$
There are infinitely many rational solutions with both $x,y > 0.$
$$ left(frac12,frac12right); ; ; left(frac32,frac12right); ; ; left(frac32,frac92right); ; ; left(frac332,frac92right); ; ; left(frac332,frac1212right); ; ; left(frac4512,frac1212right); ; ; left(frac4512,frac16812right); ; ; ldots $$
The $y$ values above are of the form $fracb_n^22,$ where $b_n+2 = 4 b_n+1 - b_n ; , ;$ giving $1,3,11,41,153,571,...$
If there were any integer solutions in the first quadrant, these same mappings would take us to an integer solution with both $x,y$ fairly small obeying certain explicit inequalities.
The following diagram goes with the inequality part:
If there were an integer solution with $x,y > 0$ and $x+y geq 10,$ we are going to show that one of the mappings given reduces $x+y$ by at least two; therefore a finite number of steps would take us to an integer solution with $x,y > 0$ and $x+y < 10.$ One can quickly inspect and find that there are no integer solutions with such small numbers. So that is it.
If $x+y geq 10$ and both positive, one case, closer to the positive $x$ axis, has $x > y$ and $y > 2.$ From the quadratic formula and the inequalities we find $x = 2y + sqrt3y^2 - y.$ Note that $3y^2 - y > 1.$ At this point, we have $x+y = 3y + sqrt3y^2 - y.$ Now, after we apply the mapping $ colormagenta (x,y) mapsto (4x-y,y) ; ; ; , $ the replacement value for $x+y$ is $3y - sqrt3y^2 - y,$ so it has shrunk by at least $2.$
The second case is closer to the positive $y$ axis, $y > x$ and $x > 2.$ We find $y = frac4x-1 + sqrt12x^2 - 8x+12.$ After applying the other mapping $ colormagenta (x,y) mapsto (x,4x-1-y) ; ; ; , $ The new $y$ value is $y = frac4x-1 - sqrt12x^2 - 8x+12.$ Therefore the sum we keep calling $x+y$ has decreased by $sqrt12x^2 - 8x+1$ which is larger than $2$ when $x > 2.$ Again, we have shrunken $x+y$ by at least $2.$
That's it. There cannot be any integer solutions with $x,y > 0$ because there are none with $x+y leq 10.$
$$ bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc $$
I found out how to draw line segments in desmos. One can put in a short table of points and have it connect them in order. Here is the hyperbola in the first quadrant, showing the first four of the half integer points and how a stairway connects them, beginning an infinite process. The next point on this staircase, which goes up and to the right, would be $left(frac332,frac1212right);$
Alright, here is the hyperbola with the first few points in the third quadrant that have integer coordinates, joined up with green line segments, beginning with the origin and $(0,-1).$ The next point in this downward stairway would be $(-60,-16)$
answered Jul 16 at 19:09
Will Jagy
97.2k594196
thx for this great analysis.
– Wolfdale
Jul 17 at 1:09
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
For $x,yin Bbb Z^+$ we have $$frac xy+frac y+1x=4implies x^2-4xy+y^2+y=0implies x=2ypm sqrt 3y^2-yimplies$$ $$implies exists zin Bbb Z^+;( z^2=3y^2-y=y(3y-1))implies$$ $$ impliesexists a,b in Bbb Z^+;( y=a^2 land 3y-1=b^2)implies$$ $$impliesexists a,bin Bbb Z^+;(3a^2-1=b^2)implies$$ $$ implies exists bin Bbb Z^+;(b^2equiv -1 mod 3).$$ The 1st (displayed) line uses the Quadratic Formula.
In the 2nd line, $zne 0$ because $yin Bbb Z^+implies y(3y-1)>0.$
In the 3rd line, $a$ and $b$ exist because $y$ and $3y-1$ are co-prime members of $Bbb Z^+$ and their product is the square of the positive integer $z.$
answered Jul 21 at 2:51
DanielWainfleet
31.7k31644
Many thx for your answer, this is a nice complete solution.
– Wolfdale
Jul 21 at 3:12
My first impulse is to get rid of denominators. And whenever I see a quadratic form I apply the Quadratic Formula to see what can be found with it.
– DanielWainfleet
Jul 21 at 3:20
Actually, this is a normal action with quadratic eq. but I stopped at the last step like another solution here. But it was that simple memorizing that the square of any number is either divisible by 3 or has a remainder of 1 :D
– Wolfdale
Jul 21 at 3:23
add a comment |Â
up vote
1
down vote
accepted
For $x,yin Bbb Z^+$ we have $$frac xy+frac y+1x=4implies x^2-4xy+y^2+y=0implies x=2ypm sqrt 3y^2-yimplies$$ $$implies exists zin Bbb Z^+;( z^2=3y^2-y=y(3y-1))implies$$ $$ impliesexists a,b in Bbb Z^+;( y=a^2 land 3y-1=b^2)implies$$ $$impliesexists a,bin Bbb Z^+;(3a^2-1=b^2)implies$$ $$ implies exists bin Bbb Z^+;(b^2equiv -1 mod 3).$$ The 1st (displayed) line uses the Quadratic Formula.
In the 2nd line, $zne 0$ because $yin Bbb Z^+implies y(3y-1)>0.$
In the 3rd line, $a$ and $b$ exist because $y$ and $3y-1$ are co-prime members of $Bbb Z^+$ and their product is the square of the positive integer $z.$
answered Jul 21 at 2:51
DanielWainfleet
31.7k31644
Many thx for your answer, this is a nice complete solution.
– Wolfdale
Jul 21 at 3:12
My first impulse is to get rid of denominators. And whenever I see a quadratic form I apply the Quadratic Formula to see what can be found with it.
– DanielWainfleet
Jul 21 at 3:20
Actually, this is a normal action with quadratic eq. but I stopped at the last step like another solution here. But it was that simple memorizing that the square of any number is either divisible by 3 or has a remainder of 1 :D
– Wolfdale
Jul 21 at 3:23
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
For $x,yin Bbb Z^+$ we have $$frac xy+frac y+1x=4implies x^2-4xy+y^2+y=0implies x=2ypm sqrt 3y^2-yimplies$$ $$implies exists zin Bbb Z^+;( z^2=3y^2-y=y(3y-1))implies$$ $$ impliesexists a,b in Bbb Z^+;( y=a^2 land 3y-1=b^2)implies$$ $$impliesexists a,bin Bbb Z^+;(3a^2-1=b^2)implies$$ $$ implies exists bin Bbb Z^+;(b^2equiv -1 mod 3).$$ The 1st (displayed) line uses the Quadratic Formula.
In the 2nd line, $zne 0$ because $yin Bbb Z^+implies y(3y-1)>0.$
In the 3rd line, $a$ and $b$ exist because $y$ and $3y-1$ are co-prime members of $Bbb Z^+$ and their product is the square of the positive integer $z.$
answered Jul 21 at 2:51
DanielWainfleet
31.7k31644
For $x,yin Bbb Z^+$ we have $$frac xy+frac y+1x=4implies x^2-4xy+y^2+y=0implies x=2ypm sqrt 3y^2-yimplies$$ $$implies exists zin Bbb Z^+;( z^2=3y^2-y=y(3y-1))implies$$ $$ impliesexists a,b in Bbb Z^+;( y=a^2 land 3y-1=b^2)implies$$ $$impliesexists a,bin Bbb Z^+;(3a^2-1=b^2)implies$$ $$ implies exists bin Bbb Z^+;(b^2equiv -1 mod 3).$$ The 1st (displayed) line uses the Quadratic Formula.
In the 2nd line, $zne 0$ because $yin Bbb Z^+implies y(3y-1)>0.$
In the 3rd line, $a$ and $b$ exist because $y$ and $3y-1$ are co-prime members of $Bbb Z^+$ and their product is the square of the positive integer $z.$
answered Jul 21 at 2:51
DanielWainfleet
31.7k31644
answered Jul 21 at 2:51
DanielWainfleet
31.7k31644
answered Jul 21 at 2:51
DanielWainfleet
31.7k31644
answered Jul 21 at 2:51
DanielWainfleet
31.7k31644
31.7k31644
Many thx for your answer, this is a nice complete solution.
– Wolfdale
Jul 21 at 3:12
My first impulse is to get rid of denominators. And whenever I see a quadratic form I apply the Quadratic Formula to see what can be found with it.
– DanielWainfleet
Jul 21 at 3:20
Actually, this is a normal action with quadratic eq. but I stopped at the last step like another solution here. But it was that simple memorizing that the square of any number is either divisible by 3 or has a remainder of 1 :D
– Wolfdale
Jul 21 at 3:23
add a comment |Â
Many thx for your answer, this is a nice complete solution.
– Wolfdale
Jul 21 at 3:12
My first impulse is to get rid of denominators. And whenever I see a quadratic form I apply the Quadratic Formula to see what can be found with it.
– DanielWainfleet
Jul 21 at 3:20
Actually, this is a normal action with quadratic eq. but I stopped at the last step like another solution here. But it was that simple memorizing that the square of any number is either divisible by 3 or has a remainder of 1 :D
– Wolfdale
Jul 21 at 3:23
Many thx for your answer, this is a nice complete solution.
– Wolfdale
Jul 21 at 3:12
Many thx for your answer, this is a nice complete solution.
– Wolfdale
Jul 21 at 3:12
My first impulse is to get rid of denominators. And whenever I see a quadratic form I apply the Quadratic Formula to see what can be found with it.
– DanielWainfleet
Jul 21 at 3:20
My first impulse is to get rid of denominators. And whenever I see a quadratic form I apply the Quadratic Formula to see what can be found with it.
– DanielWainfleet
Jul 21 at 3:20
Actually, this is a normal action with quadratic eq. but I stopped at the last step like another solution here. But it was that simple memorizing that the square of any number is either divisible by 3 or has a remainder of 1 :D
– Wolfdale
Jul 21 at 3:23
Actually, this is a normal action with quadratic eq. but I stopped at the last step like another solution here. But it was that simple memorizing that the square of any number is either divisible by 3 or has a remainder of 1 :D
– Wolfdale
Jul 21 at 3:23
add a comment |Â
up vote
6
down vote
Hint
$$fracxy +fracy+1x=4to fracxy+fracyx+frac1x=4$$
Call $x/y=tin Bbb Q$.
$$t+frac1t+frac1x=4to xt^2+t(1-4x)+x=0$$
By Rational Roots Theorem the candidates to be a rational root, $t$, are $pm1,pm x,pm 1/x$.
Now, test every root and check the value of $x$ you get.
Can you finish?
answered Jul 16 at 18:03
Arnaldo
18k42146
thx for the hint, but I have no idea of Rational Roots Theorem
– Wolfdale
Jul 16 at 18:06
@Wolfdale:en.wikipedia.org/wiki/Rational_root_theorem
– Arnaldo
Jul 16 at 18:07
@Wolfdale: It is simple and very useful.
– Arnaldo
Jul 16 at 18:17
1
The rational root theorem doesn't help as much as you suggest unless $x$ is prime; for example, so far as it's concerned, $t$ could be any factor of $x$. It's probably better to note that your quadratic has discriminant $1-4x$. So if $x$ is a positive integer the quadratic cannot have rational solutions.
– Micah
Jul 16 at 18:23
1
@Micah, I can't get this discriminant value, it should be $1-8x+12x^2=(2x-1)(6x-1)$
– Wolfdale
Jul 16 at 18:40
 |Â
show 6 more comments
up vote
6
down vote
Hint
$$fracxy +fracy+1x=4to fracxy+fracyx+frac1x=4$$
Call $x/y=tin Bbb Q$.
$$t+frac1t+frac1x=4to xt^2+t(1-4x)+x=0$$
By Rational Roots Theorem the candidates to be a rational root, $t$, are $pm1,pm x,pm 1/x$.
Now, test every root and check the value of $x$ you get.
Can you finish?
answered Jul 16 at 18:03
Arnaldo
18k42146
thx for the hint, but I have no idea of Rational Roots Theorem
– Wolfdale
Jul 16 at 18:06
@Wolfdale:en.wikipedia.org/wiki/Rational_root_theorem
– Arnaldo
Jul 16 at 18:07
@Wolfdale: It is simple and very useful.
– Arnaldo
Jul 16 at 18:17
1
The rational root theorem doesn't help as much as you suggest unless $x$ is prime; for example, so far as it's concerned, $t$ could be any factor of $x$. It's probably better to note that your quadratic has discriminant $1-4x$. So if $x$ is a positive integer the quadratic cannot have rational solutions.
– Micah
Jul 16 at 18:23
1
@Micah, I can't get this discriminant value, it should be $1-8x+12x^2=(2x-1)(6x-1)$
– Wolfdale
Jul 16 at 18:40
 |Â
show 6 more comments
up vote
6
down vote
up vote
6
down vote
Hint
$$fracxy +fracy+1x=4to fracxy+fracyx+frac1x=4$$
Call $x/y=tin Bbb Q$.
$$t+frac1t+frac1x=4to xt^2+t(1-4x)+x=0$$
By Rational Roots Theorem the candidates to be a rational root, $t$, are $pm1,pm x,pm 1/x$.
Now, test every root and check the value of $x$ you get.
Can you finish?
answered Jul 16 at 18:03
Arnaldo
18k42146
Hint
$$fracxy +fracy+1x=4to fracxy+fracyx+frac1x=4$$
Call $x/y=tin Bbb Q$.
$$t+frac1t+frac1x=4to xt^2+t(1-4x)+x=0$$
By Rational Roots Theorem the candidates to be a rational root, $t$, are $pm1,pm x,pm 1/x$.
Now, test every root and check the value of $x$ you get.
Can you finish?
answered Jul 16 at 18:03
Arnaldo
18k42146
edited Jul 16 at 18:06
answered Jul 16 at 18:03
Arnaldo
18k42146
answered Jul 16 at 18:03
Arnaldo
18k42146
answered Jul 16 at 18:03
Arnaldo
18k42146
18k42146
thx for the hint, but I have no idea of Rational Roots Theorem
– Wolfdale
Jul 16 at 18:06
@Wolfdale:en.wikipedia.org/wiki/Rational_root_theorem
– Arnaldo
Jul 16 at 18:07
@Wolfdale: It is simple and very useful.
– Arnaldo
Jul 16 at 18:17
1
The rational root theorem doesn't help as much as you suggest unless $x$ is prime; for example, so far as it's concerned, $t$ could be any factor of $x$. It's probably better to note that your quadratic has discriminant $1-4x$. So if $x$ is a positive integer the quadratic cannot have rational solutions.
– Micah
Jul 16 at 18:23
1
@Micah, I can't get this discriminant value, it should be $1-8x+12x^2=(2x-1)(6x-1)$
– Wolfdale
Jul 16 at 18:40
 |Â
show 6 more comments
thx for the hint, but I have no idea of Rational Roots Theorem
– Wolfdale
Jul 16 at 18:06
@Wolfdale:en.wikipedia.org/wiki/Rational_root_theorem
– Arnaldo
Jul 16 at 18:07
@Wolfdale: It is simple and very useful.
– Arnaldo
Jul 16 at 18:17
1
The rational root theorem doesn't help as much as you suggest unless $x$ is prime; for example, so far as it's concerned, $t$ could be any factor of $x$. It's probably better to note that your quadratic has discriminant $1-4x$. So if $x$ is a positive integer the quadratic cannot have rational solutions.
– Micah
Jul 16 at 18:23
1
@Micah, I can't get this discriminant value, it should be $1-8x+12x^2=(2x-1)(6x-1)$
– Wolfdale
Jul 16 at 18:40
thx for the hint, but I have no idea of Rational Roots Theorem
– Wolfdale
Jul 16 at 18:06
thx for the hint, but I have no idea of Rational Roots Theorem
– Wolfdale
Jul 16 at 18:06
@Wolfdale:en.wikipedia.org/wiki/Rational_root_theorem
– Arnaldo
Jul 16 at 18:07
@Wolfdale:en.wikipedia.org/wiki/Rational_root_theorem
– Arnaldo
Jul 16 at 18:07
@Wolfdale: It is simple and very useful.
– Arnaldo
Jul 16 at 18:17
@Wolfdale: It is simple and very useful.
– Arnaldo
Jul 16 at 18:17
1
1
The rational root theorem doesn't help as much as you suggest unless $x$ is prime; for example, so far as it's concerned, $t$ could be any factor of $x$. It's probably better to note that your quadratic has discriminant $1-4x$. So if $x$ is a positive integer the quadratic cannot have rational solutions.
– Micah
Jul 16 at 18:23
The rational root theorem doesn't help as much as you suggest unless $x$ is prime; for example, so far as it's concerned, $t$ could be any factor of $x$. It's probably better to note that your quadratic has discriminant $1-4x$. So if $x$ is a positive integer the quadratic cannot have rational solutions.
– Micah
Jul 16 at 18:23
1
1
@Micah, I can't get this discriminant value, it should be $1-8x+12x^2=(2x-1)(6x-1)$
– Wolfdale
Jul 16 at 18:40
@Micah, I can't get this discriminant value, it should be $1-8x+12x^2=(2x-1)(6x-1)$
– Wolfdale
Jul 16 at 18:40
 |Â
show 6 more comments
up vote
2
down vote
Rewrite the equation: $x^2 - 4yx + y^2+y = 0impliestriangle'=(-2y)^2-1(y^2+y)= 4y^2-y^2-y = 3y^2-y=k^2 implies y(3y-1)=k^2$ . Observe that $textgcd(y,3y-1) = 1$ since if $d = textgcd implies d mid y, d mid 3y-1 implies y = md, 3y-1 = ndimplies 1 = 3y -nd= 3md - nd = d(3m-n)implies d = 1implies y = u^2, 3y-1 = v^2, uv = kimplies 3u^2-v^2=1$ . This is a Pell equation and it either has infinitely many solutions or no solutions at all. Please check its status via google.
answered Jul 16 at 18:10
DeepSea
69k54284
1
infinitely many solutions with $x,y < 0$
– Will Jagy
Jul 16 at 19:17
add a comment |Â
up vote
2
down vote
Rewrite the equation: $x^2 - 4yx + y^2+y = 0impliestriangle'=(-2y)^2-1(y^2+y)= 4y^2-y^2-y = 3y^2-y=k^2 implies y(3y-1)=k^2$ . Observe that $textgcd(y,3y-1) = 1$ since if $d = textgcd implies d mid y, d mid 3y-1 implies y = md, 3y-1 = ndimplies 1 = 3y -nd= 3md - nd = d(3m-n)implies d = 1implies y = u^2, 3y-1 = v^2, uv = kimplies 3u^2-v^2=1$ . This is a Pell equation and it either has infinitely many solutions or no solutions at all. Please check its status via google.
answered Jul 16 at 18:10
DeepSea
69k54284
1
infinitely many solutions with $x,y < 0$
– Will Jagy
Jul 16 at 19:17
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Rewrite the equation: $x^2 - 4yx + y^2+y = 0impliestriangle'=(-2y)^2-1(y^2+y)= 4y^2-y^2-y = 3y^2-y=k^2 implies y(3y-1)=k^2$ . Observe that $textgcd(y,3y-1) = 1$ since if $d = textgcd implies d mid y, d mid 3y-1 implies y = md, 3y-1 = ndimplies 1 = 3y -nd= 3md - nd = d(3m-n)implies d = 1implies y = u^2, 3y-1 = v^2, uv = kimplies 3u^2-v^2=1$ . This is a Pell equation and it either has infinitely many solutions or no solutions at all. Please check its status via google.
answered Jul 16 at 18:10
DeepSea
69k54284
Rewrite the equation: $x^2 - 4yx + y^2+y = 0impliestriangle'=(-2y)^2-1(y^2+y)= 4y^2-y^2-y = 3y^2-y=k^2 implies y(3y-1)=k^2$ . Observe that $textgcd(y,3y-1) = 1$ since if $d = textgcd implies d mid y, d mid 3y-1 implies y = md, 3y-1 = ndimplies 1 = 3y -nd= 3md - nd = d(3m-n)implies d = 1implies y = u^2, 3y-1 = v^2, uv = kimplies 3u^2-v^2=1$ . This is a Pell equation and it either has infinitely many solutions or no solutions at all. Please check its status via google.
answered Jul 16 at 18:10
DeepSea
69k54284
edited Jul 16 at 18:49
answered Jul 16 at 18:10
DeepSea
69k54284
answered Jul 16 at 18:10
DeepSea
69k54284
answered Jul 16 at 18:10
DeepSea
69k54284
69k54284
1
infinitely many solutions with $x,y < 0$
– Will Jagy
Jul 16 at 19:17
add a comment |Â
1
infinitely many solutions with $x,y < 0$
– Will Jagy
Jul 16 at 19:17
1
1
infinitely many solutions with $x,y < 0$
– Will Jagy
Jul 16 at 19:17
infinitely many solutions with $x,y < 0$
– Will Jagy
Jul 16 at 19:17
add a comment |Â
up vote
1
down vote
I include a proof for the original question. It is between the first two pictures.
There are infinitely many solutions with both $x,y leq 0.$ This comes under the heading of Vieta Jumping. Given a solution to $x^2 - 4xy + y^2 + y = 0,$ we can create new solutions by alternately taking mappings on the hyperbola,
$$ (x,y) mapsto (x, 4x-y - 1) ; ; ; , $$
$$ (x,y) mapsto (4x-y,y) ; ; ; . $$
The solutions on the hyperbola branch that is (mostly) in the third quadrant begin
$$ (0,0); ; (0,-1); ; (-4,-1); ; (-4,-16); ; (-60,-16); ; (-60,-225); ; ldots $$
There are infinitely many rational solutions with both $x,y > 0.$
$$ left(frac12,frac12right); ; ; left(frac32,frac12right); ; ; left(frac32,frac92right); ; ; left(frac332,frac92right); ; ; left(frac332,frac1212right); ; ; left(frac4512,frac1212right); ; ; left(frac4512,frac16812right); ; ; ldots $$
The $y$ values above are of the form $fracb_n^22,$ where $b_n+2 = 4 b_n+1 - b_n ; , ;$ giving $1,3,11,41,153,571,...$
If there were any integer solutions in the first quadrant, these same mappings would take us to an integer solution with both $x,y$ fairly small obeying certain explicit inequalities.
The following diagram goes with the inequality part:
If there were an integer solution with $x,y > 0$ and $x+y geq 10,$ we are going to show that one of the mappings given reduces $x+y$ by at least two; therefore a finite number of steps would take us to an integer solution with $x,y > 0$ and $x+y < 10.$ One can quickly inspect and find that there are no integer solutions with such small numbers. So that is it.
If $x+y geq 10$ and both positive, one case, closer to the positive $x$ axis, has $x > y$ and $y > 2.$ From the quadratic formula and the inequalities we find $x = 2y + sqrt3y^2 - y.$ Note that $3y^2 - y > 1.$ At this point, we have $x+y = 3y + sqrt3y^2 - y.$ Now, after we apply the mapping $ colormagenta (x,y) mapsto (4x-y,y) ; ; ; , $ the replacement value for $x+y$ is $3y - sqrt3y^2 - y,$ so it has shrunk by at least $2.$
The second case is closer to the positive $y$ axis, $y > x$ and $x > 2.$ We find $y = frac4x-1 + sqrt12x^2 - 8x+12.$ After applying the other mapping $ colormagenta (x,y) mapsto (x,4x-1-y) ; ; ; , $ The new $y$ value is $y = frac4x-1 - sqrt12x^2 - 8x+12.$ Therefore the sum we keep calling $x+y$ has decreased by $sqrt12x^2 - 8x+1$ which is larger than $2$ when $x > 2.$ Again, we have shrunken $x+y$ by at least $2.$
That's it. There cannot be any integer solutions with $x,y > 0$ because there are none with $x+y leq 10.$
$$ bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc $$
I found out how to draw line segments in desmos. One can put in a short table of points and have it connect them in order. Here is the hyperbola in the first quadrant, showing the first four of the half integer points and how a stairway connects them, beginning an infinite process. The next point on this staircase, which goes up and to the right, would be $left(frac332,frac1212right);$
Alright, here is the hyperbola with the first few points in the third quadrant that have integer coordinates, joined up with green line segments, beginning with the origin and $(0,-1).$ The next point in this downward stairway would be $(-60,-16)$
answered Jul 16 at 19:09
Will Jagy
97.2k594196
thx for this great analysis.
– Wolfdale
Jul 17 at 1:09
add a comment |Â
up vote
1
down vote
I include a proof for the original question. It is between the first two pictures.
There are infinitely many solutions with both $x,y leq 0.$ This comes under the heading of Vieta Jumping. Given a solution to $x^2 - 4xy + y^2 + y = 0,$ we can create new solutions by alternately taking mappings on the hyperbola,
$$ (x,y) mapsto (x, 4x-y - 1) ; ; ; , $$
$$ (x,y) mapsto (4x-y,y) ; ; ; . $$
The solutions on the hyperbola branch that is (mostly) in the third quadrant begin
$$ (0,0); ; (0,-1); ; (-4,-1); ; (-4,-16); ; (-60,-16); ; (-60,-225); ; ldots $$
There are infinitely many rational solutions with both $x,y > 0.$
$$ left(frac12,frac12right); ; ; left(frac32,frac12right); ; ; left(frac32,frac92right); ; ; left(frac332,frac92right); ; ; left(frac332,frac1212right); ; ; left(frac4512,frac1212right); ; ; left(frac4512,frac16812right); ; ; ldots $$
The $y$ values above are of the form $fracb_n^22,$ where $b_n+2 = 4 b_n+1 - b_n ; , ;$ giving $1,3,11,41,153,571,...$
If there were any integer solutions in the first quadrant, these same mappings would take us to an integer solution with both $x,y$ fairly small obeying certain explicit inequalities.
The following diagram goes with the inequality part:
If there were an integer solution with $x,y > 0$ and $x+y geq 10,$ we are going to show that one of the mappings given reduces $x+y$ by at least two; therefore a finite number of steps would take us to an integer solution with $x,y > 0$ and $x+y < 10.$ One can quickly inspect and find that there are no integer solutions with such small numbers. So that is it.
If $x+y geq 10$ and both positive, one case, closer to the positive $x$ axis, has $x > y$ and $y > 2.$ From the quadratic formula and the inequalities we find $x = 2y + sqrt3y^2 - y.$ Note that $3y^2 - y > 1.$ At this point, we have $x+y = 3y + sqrt3y^2 - y.$ Now, after we apply the mapping $ colormagenta (x,y) mapsto (4x-y,y) ; ; ; , $ the replacement value for $x+y$ is $3y - sqrt3y^2 - y,$ so it has shrunk by at least $2.$
The second case is closer to the positive $y$ axis, $y > x$ and $x > 2.$ We find $y = frac4x-1 + sqrt12x^2 - 8x+12.$ After applying the other mapping $ colormagenta (x,y) mapsto (x,4x-1-y) ; ; ; , $ The new $y$ value is $y = frac4x-1 - sqrt12x^2 - 8x+12.$ Therefore the sum we keep calling $x+y$ has decreased by $sqrt12x^2 - 8x+1$ which is larger than $2$ when $x > 2.$ Again, we have shrunken $x+y$ by at least $2.$
That's it. There cannot be any integer solutions with $x,y > 0$ because there are none with $x+y leq 10.$
$$ bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc $$
I found out how to draw line segments in desmos. One can put in a short table of points and have it connect them in order. Here is the hyperbola in the first quadrant, showing the first four of the half integer points and how a stairway connects them, beginning an infinite process. The next point on this staircase, which goes up and to the right, would be $left(frac332,frac1212right);$
Alright, here is the hyperbola with the first few points in the third quadrant that have integer coordinates, joined up with green line segments, beginning with the origin and $(0,-1).$ The next point in this downward stairway would be $(-60,-16)$
answered Jul 16 at 19:09
Will Jagy
97.2k594196
thx for this great analysis.
– Wolfdale
Jul 17 at 1:09
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I include a proof for the original question. It is between the first two pictures.
There are infinitely many solutions with both $x,y leq 0.$ This comes under the heading of Vieta Jumping. Given a solution to $x^2 - 4xy + y^2 + y = 0,$ we can create new solutions by alternately taking mappings on the hyperbola,
$$ (x,y) mapsto (x, 4x-y - 1) ; ; ; , $$
$$ (x,y) mapsto (4x-y,y) ; ; ; . $$
The solutions on the hyperbola branch that is (mostly) in the third quadrant begin
$$ (0,0); ; (0,-1); ; (-4,-1); ; (-4,-16); ; (-60,-16); ; (-60,-225); ; ldots $$
There are infinitely many rational solutions with both $x,y > 0.$
$$ left(frac12,frac12right); ; ; left(frac32,frac12right); ; ; left(frac32,frac92right); ; ; left(frac332,frac92right); ; ; left(frac332,frac1212right); ; ; left(frac4512,frac1212right); ; ; left(frac4512,frac16812right); ; ; ldots $$
The $y$ values above are of the form $fracb_n^22,$ where $b_n+2 = 4 b_n+1 - b_n ; , ;$ giving $1,3,11,41,153,571,...$
If there were any integer solutions in the first quadrant, these same mappings would take us to an integer solution with both $x,y$ fairly small obeying certain explicit inequalities.
The following diagram goes with the inequality part:
If there were an integer solution with $x,y > 0$ and $x+y geq 10,$ we are going to show that one of the mappings given reduces $x+y$ by at least two; therefore a finite number of steps would take us to an integer solution with $x,y > 0$ and $x+y < 10.$ One can quickly inspect and find that there are no integer solutions with such small numbers. So that is it.
If $x+y geq 10$ and both positive, one case, closer to the positive $x$ axis, has $x > y$ and $y > 2.$ From the quadratic formula and the inequalities we find $x = 2y + sqrt3y^2 - y.$ Note that $3y^2 - y > 1.$ At this point, we have $x+y = 3y + sqrt3y^2 - y.$ Now, after we apply the mapping $ colormagenta (x,y) mapsto (4x-y,y) ; ; ; , $ the replacement value for $x+y$ is $3y - sqrt3y^2 - y,$ so it has shrunk by at least $2.$
The second case is closer to the positive $y$ axis, $y > x$ and $x > 2.$ We find $y = frac4x-1 + sqrt12x^2 - 8x+12.$ After applying the other mapping $ colormagenta (x,y) mapsto (x,4x-1-y) ; ; ; , $ The new $y$ value is $y = frac4x-1 - sqrt12x^2 - 8x+12.$ Therefore the sum we keep calling $x+y$ has decreased by $sqrt12x^2 - 8x+1$ which is larger than $2$ when $x > 2.$ Again, we have shrunken $x+y$ by at least $2.$
That's it. There cannot be any integer solutions with $x,y > 0$ because there are none with $x+y leq 10.$
$$ bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc $$
I found out how to draw line segments in desmos. One can put in a short table of points and have it connect them in order. Here is the hyperbola in the first quadrant, showing the first four of the half integer points and how a stairway connects them, beginning an infinite process. The next point on this staircase, which goes up and to the right, would be $left(frac332,frac1212right);$
Alright, here is the hyperbola with the first few points in the third quadrant that have integer coordinates, joined up with green line segments, beginning with the origin and $(0,-1).$ The next point in this downward stairway would be $(-60,-16)$
answered Jul 16 at 19:09
Will Jagy
97.2k594196
I include a proof for the original question. It is between the first two pictures.
There are infinitely many solutions with both $x,y leq 0.$ This comes under the heading of Vieta Jumping. Given a solution to $x^2 - 4xy + y^2 + y = 0,$ we can create new solutions by alternately taking mappings on the hyperbola,
$$ (x,y) mapsto (x, 4x-y - 1) ; ; ; , $$
$$ (x,y) mapsto (4x-y,y) ; ; ; . $$
The solutions on the hyperbola branch that is (mostly) in the third quadrant begin
$$ (0,0); ; (0,-1); ; (-4,-1); ; (-4,-16); ; (-60,-16); ; (-60,-225); ; ldots $$
There are infinitely many rational solutions with both $x,y > 0.$
$$ left(frac12,frac12right); ; ; left(frac32,frac12right); ; ; left(frac32,frac92right); ; ; left(frac332,frac92right); ; ; left(frac332,frac1212right); ; ; left(frac4512,frac1212right); ; ; left(frac4512,frac16812right); ; ; ldots $$
The $y$ values above are of the form $fracb_n^22,$ where $b_n+2 = 4 b_n+1 - b_n ; , ;$ giving $1,3,11,41,153,571,...$
If there were any integer solutions in the first quadrant, these same mappings would take us to an integer solution with both $x,y$ fairly small obeying certain explicit inequalities.
The following diagram goes with the inequality part:
If there were an integer solution with $x,y > 0$ and $x+y geq 10,$ we are going to show that one of the mappings given reduces $x+y$ by at least two; therefore a finite number of steps would take us to an integer solution with $x,y > 0$ and $x+y < 10.$ One can quickly inspect and find that there are no integer solutions with such small numbers. So that is it.
If $x+y geq 10$ and both positive, one case, closer to the positive $x$ axis, has $x > y$ and $y > 2.$ From the quadratic formula and the inequalities we find $x = 2y + sqrt3y^2 - y.$ Note that $3y^2 - y > 1.$ At this point, we have $x+y = 3y + sqrt3y^2 - y.$ Now, after we apply the mapping $ colormagenta (x,y) mapsto (4x-y,y) ; ; ; , $ the replacement value for $x+y$ is $3y - sqrt3y^2 - y,$ so it has shrunk by at least $2.$
The second case is closer to the positive $y$ axis, $y > x$ and $x > 2.$ We find $y = frac4x-1 + sqrt12x^2 - 8x+12.$ After applying the other mapping $ colormagenta (x,y) mapsto (x,4x-1-y) ; ; ; , $ The new $y$ value is $y = frac4x-1 - sqrt12x^2 - 8x+12.$ Therefore the sum we keep calling $x+y$ has decreased by $sqrt12x^2 - 8x+1$ which is larger than $2$ when $x > 2.$ Again, we have shrunken $x+y$ by at least $2.$
That's it. There cannot be any integer solutions with $x,y > 0$ because there are none with $x+y leq 10.$
$$ bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc bigcirc $$
I found out how to draw line segments in desmos. One can put in a short table of points and have it connect them in order. Here is the hyperbola in the first quadrant, showing the first four of the half integer points and how a stairway connects them, beginning an infinite process. The next point on this staircase, which goes up and to the right, would be $left(frac332,frac1212right);$
Alright, here is the hyperbola with the first few points in the third quadrant that have integer coordinates, joined up with green line segments, beginning with the origin and $(0,-1).$ The next point in this downward stairway would be $(-60,-16)$
answered Jul 16 at 19:09
Will Jagy
97.2k594196
edited Jul 21 at 1:38
answered Jul 16 at 19:09
Will Jagy
97.2k594196
answered Jul 16 at 19:09
Will Jagy
97.2k594196
answered Jul 16 at 19:09
Will Jagy
97.2k594196
97.2k594196
thx for this great analysis.
– Wolfdale
Jul 17 at 1:09
add a comment |Â
thx for this great analysis.
– Wolfdale
Jul 17 at 1:09
thx for this great analysis.
– Wolfdale
Jul 17 at 1:09
thx for this great analysis.
– Wolfdale
Jul 17 at 1:09
add a comment |Â
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I think there is a solution. Look at the formula. artofproblemsolving.com/community/c3046h1046841___
– individ
Jul 16 at 18:04
infinitely many solutions with $x,y < 0$
– Will Jagy
Jul 16 at 19:29
infinitely many rational solutions with $x,y > 0,$ we can take denominator as $2$
– Will Jagy
Jul 16 at 20:39