Integer solutions to $x^2(y-1)+y^2(x-1)=1$ [duplicate]
This question already has an answer here:
Find all integer roots of: $x^2(y-1)+y^2(x-1)=1$
5 answers
Find all values of $(x,y)$ where $x,y$ belongs to integers if:
$$x^2(y-1)+y^2(x-1)=1$$
I m a beginner so I do need some help.
number-theory elementary-number-theory divisibility diophantine-equations
edited Dec 27 '18 at 18:58
greedoid
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asked Dec 27 '18 at 18:06
user160040
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Dec 27 '18 at 19:11
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Find all integer roots of: $x^2(y-1)+y^2(x-1)=1$
5 answers
Find all values of $(x,y)$ where $x,y$ belongs to integers if:
$$x^2(y-1)+y^2(x-1)=1$$
I m a beginner so I do need some help.
number-theory elementary-number-theory divisibility diophantine-equations
edited Dec 27 '18 at 18:58
greedoid
38.2k114795
asked Dec 27 '18 at 18:06
user160040
21
marked as duplicate by Dietrich Burde number-theory
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Dec 27 '18 at 19:11
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Find all integer roots of: $x^2(y-1)+y^2(x-1)=1$
5 answers
Find all values of $(x,y)$ where $x,y$ belongs to integers if:
$$x^2(y-1)+y^2(x-1)=1$$
I m a beginner so I do need some help.
number-theory elementary-number-theory divisibility diophantine-equations
edited Dec 27 '18 at 18:58
greedoid
38.2k114795
asked Dec 27 '18 at 18:06
user160040
21
This question already has an answer here:
Find all integer roots of: $x^2(y-1)+y^2(x-1)=1$
5 answers
Find all values of $(x,y)$ where $x,y$ belongs to integers if:
$$x^2(y-1)+y^2(x-1)=1$$
I m a beginner so I do need some help.
This question already has an answer here:
Find all integer roots of: $x^2(y-1)+y^2(x-1)=1$
5 answers
number-theory elementary-number-theory divisibility diophantine-equations
number-theory elementary-number-theory divisibility diophantine-equations
edited Dec 27 '18 at 18:58
greedoid
38.2k114795
asked Dec 27 '18 at 18:06
user160040
21
edited Dec 27 '18 at 18:58
greedoid
38.2k114795
asked Dec 27 '18 at 18:06
user160040
21
edited Dec 27 '18 at 18:58
greedoid
38.2k114795
edited Dec 27 '18 at 18:58
greedoid
38.2k114795
edited Dec 27 '18 at 18:58
greedoid
38.2k114795
38.2k114795
asked Dec 27 '18 at 18:06
user160040
21
asked Dec 27 '18 at 18:06
user160040
21
asked Dec 27 '18 at 18:06
user160040
21
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Dec 27 '18 at 19:11
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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add a comment |
2 Answers
2
active
oldest
votes
Hint:
Write as quadratic on $x$ and parameter $y$: $$(y-1)x^2+y^2x-(1+y^2)=0$$
If $yne 1$ then it discriminat must be perfect square:$$y^4+4y^3-4y^2+4y-4=z^2$$
for some $z$. Now use: $$y^4< z^2<(y+1)^4$$
for $y>1$...and then try something similary for $y<1$.
answered Dec 27 '18 at 18:15
greedoid
38.2k114795
1
appears you meant $y^4 + 4 y^3 ...=z^2$
– Will Jagy
Dec 27 '18 at 18:18
add a comment |
Hint:
Perhaps it would be easier if you write the equation like this
$$ xy(x+y)-(x+y)^2+2xy = 1$$ and now put $a=x+y$ and $b=xy$. Then you get: $$ ba-a^2+2b=1implies b ={a^2+1over a+2}$$
This means, since $b$ is integer, that $$a+2mid a^2+1$$
and now you should continue...
or you could write $c=a+2$ and then $$b = {(c-2)^2+1over c} = {c^2-4c+5over c}= c-4 +{5over c}$$
so $cmid 5$....
answered Dec 27 '18 at 18:22
greedoid
38.2k114795
We can try odd even parity?
– user160040
Dec 27 '18 at 18:36
But how prove when a is odd?
– user160040
Dec 27 '18 at 18:37
Note that $a^2+1=(a+2)(a-2)+5$.
– Servaes
Dec 27 '18 at 18:38
a^2+1 will be even when a is odd but a+2 will be odd ,still will divide it or not ?
– user160040
Dec 27 '18 at 18:43
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint:
Write as quadratic on $x$ and parameter $y$: $$(y-1)x^2+y^2x-(1+y^2)=0$$
If $yne 1$ then it discriminat must be perfect square:$$y^4+4y^3-4y^2+4y-4=z^2$$
for some $z$. Now use: $$y^4< z^2<(y+1)^4$$
for $y>1$...and then try something similary for $y<1$.
answered Dec 27 '18 at 18:15
greedoid
38.2k114795
1
appears you meant $y^4 + 4 y^3 ...=z^2$
– Will Jagy
Dec 27 '18 at 18:18
add a comment |
Hint:
Write as quadratic on $x$ and parameter $y$: $$(y-1)x^2+y^2x-(1+y^2)=0$$
If $yne 1$ then it discriminat must be perfect square:$$y^4+4y^3-4y^2+4y-4=z^2$$
for some $z$. Now use: $$y^4< z^2<(y+1)^4$$
for $y>1$...and then try something similary for $y<1$.
answered Dec 27 '18 at 18:15
greedoid
38.2k114795
1
appears you meant $y^4 + 4 y^3 ...=z^2$
– Will Jagy
Dec 27 '18 at 18:18
add a comment |
Hint:
Write as quadratic on $x$ and parameter $y$: $$(y-1)x^2+y^2x-(1+y^2)=0$$
If $yne 1$ then it discriminat must be perfect square:$$y^4+4y^3-4y^2+4y-4=z^2$$
for some $z$. Now use: $$y^4< z^2<(y+1)^4$$
for $y>1$...and then try something similary for $y<1$.
answered Dec 27 '18 at 18:15
greedoid
38.2k114795
Hint:
Write as quadratic on $x$ and parameter $y$: $$(y-1)x^2+y^2x-(1+y^2)=0$$
If $yne 1$ then it discriminat must be perfect square:$$y^4+4y^3-4y^2+4y-4=z^2$$
for some $z$. Now use: $$y^4< z^2<(y+1)^4$$
for $y>1$...and then try something similary for $y<1$.
answered Dec 27 '18 at 18:15
greedoid
38.2k114795
edited Dec 27 '18 at 18:19
answered Dec 27 '18 at 18:15
greedoid
38.2k114795
answered Dec 27 '18 at 18:15
greedoid
38.2k114795
answered Dec 27 '18 at 18:15
greedoid
38.2k114795
38.2k114795
1
appears you meant $y^4 + 4 y^3 ...=z^2$
– Will Jagy
Dec 27 '18 at 18:18
add a comment |
1
appears you meant $y^4 + 4 y^3 ...=z^2$
– Will Jagy
Dec 27 '18 at 18:18
1
1
appears you meant $y^4 + 4 y^3 ...=z^2$
– Will Jagy
Dec 27 '18 at 18:18
appears you meant $y^4 + 4 y^3 ...=z^2$
– Will Jagy
Dec 27 '18 at 18:18
add a comment |
Hint:
Perhaps it would be easier if you write the equation like this
$$ xy(x+y)-(x+y)^2+2xy = 1$$ and now put $a=x+y$ and $b=xy$. Then you get: $$ ba-a^2+2b=1implies b ={a^2+1over a+2}$$
This means, since $b$ is integer, that $$a+2mid a^2+1$$
and now you should continue...
or you could write $c=a+2$ and then $$b = {(c-2)^2+1over c} = {c^2-4c+5over c}= c-4 +{5over c}$$
so $cmid 5$....
answered Dec 27 '18 at 18:22
greedoid
38.2k114795
We can try odd even parity?
– user160040
Dec 27 '18 at 18:36
But how prove when a is odd?
– user160040
Dec 27 '18 at 18:37
Note that $a^2+1=(a+2)(a-2)+5$.
– Servaes
Dec 27 '18 at 18:38
a^2+1 will be even when a is odd but a+2 will be odd ,still will divide it or not ?
– user160040
Dec 27 '18 at 18:43
add a comment |
Hint:
Perhaps it would be easier if you write the equation like this
$$ xy(x+y)-(x+y)^2+2xy = 1$$ and now put $a=x+y$ and $b=xy$. Then you get: $$ ba-a^2+2b=1implies b ={a^2+1over a+2}$$
This means, since $b$ is integer, that $$a+2mid a^2+1$$
and now you should continue...
or you could write $c=a+2$ and then $$b = {(c-2)^2+1over c} = {c^2-4c+5over c}= c-4 +{5over c}$$
so $cmid 5$....
answered Dec 27 '18 at 18:22
greedoid
38.2k114795
We can try odd even parity?
– user160040
Dec 27 '18 at 18:36
But how prove when a is odd?
– user160040
Dec 27 '18 at 18:37
Note that $a^2+1=(a+2)(a-2)+5$.
– Servaes
Dec 27 '18 at 18:38
a^2+1 will be even when a is odd but a+2 will be odd ,still will divide it or not ?
– user160040
Dec 27 '18 at 18:43
add a comment |
Hint:
Perhaps it would be easier if you write the equation like this
$$ xy(x+y)-(x+y)^2+2xy = 1$$ and now put $a=x+y$ and $b=xy$. Then you get: $$ ba-a^2+2b=1implies b ={a^2+1over a+2}$$
This means, since $b$ is integer, that $$a+2mid a^2+1$$
and now you should continue...
or you could write $c=a+2$ and then $$b = {(c-2)^2+1over c} = {c^2-4c+5over c}= c-4 +{5over c}$$
so $cmid 5$....
answered Dec 27 '18 at 18:22
greedoid
38.2k114795
Hint:
Perhaps it would be easier if you write the equation like this
$$ xy(x+y)-(x+y)^2+2xy = 1$$ and now put $a=x+y$ and $b=xy$. Then you get: $$ ba-a^2+2b=1implies b ={a^2+1over a+2}$$
This means, since $b$ is integer, that $$a+2mid a^2+1$$
and now you should continue...
or you could write $c=a+2$ and then $$b = {(c-2)^2+1over c} = {c^2-4c+5over c}= c-4 +{5over c}$$
so $cmid 5$....
answered Dec 27 '18 at 18:22
greedoid
38.2k114795
edited Dec 27 '18 at 18:40
answered Dec 27 '18 at 18:22
greedoid
38.2k114795
answered Dec 27 '18 at 18:22
greedoid
38.2k114795
answered Dec 27 '18 at 18:22
greedoid
38.2k114795
38.2k114795
We can try odd even parity?
– user160040
Dec 27 '18 at 18:36
But how prove when a is odd?
– user160040
Dec 27 '18 at 18:37
Note that $a^2+1=(a+2)(a-2)+5$.
– Servaes
Dec 27 '18 at 18:38
a^2+1 will be even when a is odd but a+2 will be odd ,still will divide it or not ?
– user160040
Dec 27 '18 at 18:43
add a comment |
We can try odd even parity?
– user160040
Dec 27 '18 at 18:36
But how prove when a is odd?
– user160040
Dec 27 '18 at 18:37
Note that $a^2+1=(a+2)(a-2)+5$.
– Servaes
Dec 27 '18 at 18:38
a^2+1 will be even when a is odd but a+2 will be odd ,still will divide it or not ?
– user160040
Dec 27 '18 at 18:43
We can try odd even parity?
– user160040
Dec 27 '18 at 18:36
We can try odd even parity?
– user160040
Dec 27 '18 at 18:36
But how prove when a is odd?
– user160040
Dec 27 '18 at 18:37
But how prove when a is odd?
– user160040
Dec 27 '18 at 18:37
Note that $a^2+1=(a+2)(a-2)+5$.
– Servaes
Dec 27 '18 at 18:38
Note that $a^2+1=(a+2)(a-2)+5$.
– Servaes
Dec 27 '18 at 18:38
a^2+1 will be even when a is odd but a+2 will be odd ,still will divide it or not ?
– user160040
Dec 27 '18 at 18:43
a^2+1 will be even when a is odd but a+2 will be odd ,still will divide it or not ?
– user160040
Dec 27 '18 at 18:43
add a comment |
