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$fracx^2+y^2x-y|1995$. Find positive integer $x,y$
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$x,y$ are positive integer . Find all $x,y$ such that
$fracx^2+y^2x-y|1995$.
My answer: $x^2+y^2|1995$.
Therefore, $x^2+y^2$ can be $1$, $3$, $5$, $7$, $19$, $15$, $21$, $35$, $57$, $95$, $133$, $105$, $285$, $399$, $665$ or $1,995$.
I observed that only $5$ can be represented as the sum of two square ( I haven't check the last two number because they are very big).
So, $(x,y)=(1,2),(2,1)$.
Am I right??
I believe there is a easier method.I am looking for that. Please help me.
elementary-number-theory proof-verification divisibility
edited Jul 17 at 12:56
José Carlos Santos
114k1698177
asked Jul 17 at 12:47
Sufaid Saleel
1,671625
add a comment |Â
up vote
4
down vote
favorite
$x,y$ are positive integer . Find all $x,y$ such that
$fracx^2+y^2x-y|1995$.
My answer: $x^2+y^2|1995$.
Therefore, $x^2+y^2$ can be $1$, $3$, $5$, $7$, $19$, $15$, $21$, $35$, $57$, $95$, $133$, $105$, $285$, $399$, $665$ or $1,995$.
I observed that only $5$ can be represented as the sum of two square ( I haven't check the last two number because they are very big).
So, $(x,y)=(1,2),(2,1)$.
Am I right??
I believe there is a easier method.I am looking for that. Please help me.
elementary-number-theory proof-verification divisibility
edited Jul 17 at 12:56
José Carlos Santos
114k1698177
asked Jul 17 at 12:47
Sufaid Saleel
1,671625
The sum of two squares theorem and $1995=3 times 5 times 7 times 19$ may get you to $5$ more quickly, since $19 equiv 7 equiv 3 pmod 4$ so none of them can appear once in the divisor.
– Henry
Jul 17 at 12:55
You might also check that $x^2+y^2=1$ does not give a solution in positive integers
– Henry
Jul 17 at 12:59
I am looking for any other method. If anyone have please post!
– Sufaid Saleel
Jul 17 at 13:18
If you insist that the divisor be a positive integer $(1,2)$ doesn't work
– Ross Millikan
Jul 17 at 13:44
2
If $x=1197, y=399$, then $x^2+y^2=1,592,010$, but $dfracx^2+y^2x-y=1995$.
– InterstellarProbe
Jul 17 at 14:13
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
$x,y$ are positive integer . Find all $x,y$ such that
$fracx^2+y^2x-y|1995$.
My answer: $x^2+y^2|1995$.
Therefore, $x^2+y^2$ can be $1$, $3$, $5$, $7$, $19$, $15$, $21$, $35$, $57$, $95$, $133$, $105$, $285$, $399$, $665$ or $1,995$.
I observed that only $5$ can be represented as the sum of two square ( I haven't check the last two number because they are very big).
So, $(x,y)=(1,2),(2,1)$.
Am I right??
I believe there is a easier method.I am looking for that. Please help me.
elementary-number-theory proof-verification divisibility
edited Jul 17 at 12:56
José Carlos Santos
114k1698177
asked Jul 17 at 12:47
Sufaid Saleel
1,671625
$x,y$ are positive integer . Find all $x,y$ such that
$fracx^2+y^2x-y|1995$.
My answer: $x^2+y^2|1995$.
Therefore, $x^2+y^2$ can be $1$, $3$, $5$, $7$, $19$, $15$, $21$, $35$, $57$, $95$, $133$, $105$, $285$, $399$, $665$ or $1,995$.
I observed that only $5$ can be represented as the sum of two square ( I haven't check the last two number because they are very big).
So, $(x,y)=(1,2),(2,1)$.
Am I right??
I believe there is a easier method.I am looking for that. Please help me.
elementary-number-theory proof-verification divisibility
edited Jul 17 at 12:56
José Carlos Santos
114k1698177
asked Jul 17 at 12:47
Sufaid Saleel
1,671625
edited Jul 17 at 12:56
José Carlos Santos
114k1698177
edited Jul 17 at 12:56
José Carlos Santos
114k1698177
edited Jul 17 at 12:56
José Carlos Santos
114k1698177
114k1698177
asked Jul 17 at 12:47
Sufaid Saleel
1,671625
asked Jul 17 at 12:47
Sufaid Saleel
1,671625
asked Jul 17 at 12:47
Sufaid Saleel
1,671625
1,671625
The sum of two squares theorem and $1995=3 times 5 times 7 times 19$ may get you to $5$ more quickly, since $19 equiv 7 equiv 3 pmod 4$ so none of them can appear once in the divisor.
– Henry
Jul 17 at 12:55
You might also check that $x^2+y^2=1$ does not give a solution in positive integers
– Henry
Jul 17 at 12:59
I am looking for any other method. If anyone have please post!
– Sufaid Saleel
Jul 17 at 13:18
If you insist that the divisor be a positive integer $(1,2)$ doesn't work
– Ross Millikan
Jul 17 at 13:44
2
If $x=1197, y=399$, then $x^2+y^2=1,592,010$, but $dfracx^2+y^2x-y=1995$.
– InterstellarProbe
Jul 17 at 14:13
add a comment |Â
The sum of two squares theorem and $1995=3 times 5 times 7 times 19$ may get you to $5$ more quickly, since $19 equiv 7 equiv 3 pmod 4$ so none of them can appear once in the divisor.
– Henry
Jul 17 at 12:55
You might also check that $x^2+y^2=1$ does not give a solution in positive integers
– Henry
Jul 17 at 12:59
I am looking for any other method. If anyone have please post!
– Sufaid Saleel
Jul 17 at 13:18
If you insist that the divisor be a positive integer $(1,2)$ doesn't work
– Ross Millikan
Jul 17 at 13:44
2
If $x=1197, y=399$, then $x^2+y^2=1,592,010$, but $dfracx^2+y^2x-y=1995$.
– InterstellarProbe
Jul 17 at 14:13
The sum of two squares theorem and $1995=3 times 5 times 7 times 19$ may get you to $5$ more quickly, since $19 equiv 7 equiv 3 pmod 4$ so none of them can appear once in the divisor.
– Henry
Jul 17 at 12:55
The sum of two squares theorem and $1995=3 times 5 times 7 times 19$ may get you to $5$ more quickly, since $19 equiv 7 equiv 3 pmod 4$ so none of them can appear once in the divisor.
– Henry
Jul 17 at 12:55
You might also check that $x^2+y^2=1$ does not give a solution in positive integers
– Henry
Jul 17 at 12:59
You might also check that $x^2+y^2=1$ does not give a solution in positive integers
– Henry
Jul 17 at 12:59
I am looking for any other method. If anyone have please post!
– Sufaid Saleel
Jul 17 at 13:18
I am looking for any other method. If anyone have please post!
– Sufaid Saleel
Jul 17 at 13:18
If you insist that the divisor be a positive integer $(1,2)$ doesn't work
– Ross Millikan
Jul 17 at 13:44
If you insist that the divisor be a positive integer $(1,2)$ doesn't work
– Ross Millikan
Jul 17 at 13:44
2
2
If $x=1197, y=399$, then $x^2+y^2=1,592,010$, but $dfracx^2+y^2x-y=1995$.
– InterstellarProbe
Jul 17 at 14:13
If $x=1197, y=399$, then $x^2+y^2=1,592,010$, but $dfracx^2+y^2x-y=1995$.
– InterstellarProbe
Jul 17 at 14:13
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
Your first assertion that $x^2+y^2 mid 1995$ doesn't follow from your hypotheses. Note that $(6,3)$ and $(38,19)$ are also solutions. Certainly $(x^2+y^2)/(x-y) >1995$ for $x>1995$, so there can be only finitely many solutions. I believe that $(399,1197)$ is the largest.
I observe that if $a=(x^2+y^2)/(x-y)$ and that $(x,y)$ is a solution, then so is $(a-x,y)$.
Edit: I also observe that every solution is of the form $(2y,y)$ or $(3y,y)$ and in all cases $a=5y$.
Another Edit. Let $k$ be a divisor of $1995$. Then you have $x^2+y^2=k(x-y).$ Put everything on one side and complete the square:
$$left(x-frack2right)^2+left(y+frack2right)^2 = frack^22.$$
Multiply through by $4$:
$$(2x-k)^2 +(2y+k)^2 = 2k^2.$$
So find, in the usual way, all the ways of writing $2k^2$ as a sum of two squares. Since many of the prime divisors are congruent to $3$ mod $4$, there are not many solutions. Then solve for $x$ and $y$. Do this for each $k$ and you have all solutions. Here's one example. Take $k=35$. So we find all solutions to
$$u^2+v^2 = 2cdot35^2$$
and these are $(7,49)$ and $(35,35)$. The second one leads to $x=0$, which is not positive. The first one gives $2x-35 = 7$ and $2y+35 = 49$ which leads to $(x,y) = (21,7)$. Then from my observation above $y=35-21 = 14$ gives a second solution. Lather, rinse, repeat for all $k$.
answered Jul 17 at 14:15
B. Goddard
16.5k21238
add a comment |Â
up vote
1
down vote
Beautiful number.
$$fracx^2+y^2x-y=a$$
Solutions have the form:
$$a=p^2+s^2$$
$$y=s(p-s)$$
$$x=p(p-s)$$
answered Jul 18 at 9:57
individ
3,1941715
Please include an explanation or derivation for your answer - unjustified claims are much less useful
– Carl Mummert
Jul 28 at 20:55
add a comment |Â
up vote
0
down vote
$(1,2),(1,3),(3,6),(3,9),(7,14),(7,21),(19,38),(19,57),(21,42),(21,63),(57,114),(57,171),(133,266),(133,399),(399,798),(399,1197)$
This list is complete for all pairs in $(i,j) in mathbbZ^2 mid 1le i < j le 30000$
I just wrote a small program to find solutions. I figured, having the list of solutions would be a good way to check any answers people come up with, although this is not technically a complete answer, as it does not explain why these are the only pairs that work.
answered Jul 17 at 14:23
InterstellarProbe
2,207518
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Your first assertion that $x^2+y^2 mid 1995$ doesn't follow from your hypotheses. Note that $(6,3)$ and $(38,19)$ are also solutions. Certainly $(x^2+y^2)/(x-y) >1995$ for $x>1995$, so there can be only finitely many solutions. I believe that $(399,1197)$ is the largest.
I observe that if $a=(x^2+y^2)/(x-y)$ and that $(x,y)$ is a solution, then so is $(a-x,y)$.
Edit: I also observe that every solution is of the form $(2y,y)$ or $(3y,y)$ and in all cases $a=5y$.
Another Edit. Let $k$ be a divisor of $1995$. Then you have $x^2+y^2=k(x-y).$ Put everything on one side and complete the square:
$$left(x-frack2right)^2+left(y+frack2right)^2 = frack^22.$$
Multiply through by $4$:
$$(2x-k)^2 +(2y+k)^2 = 2k^2.$$
So find, in the usual way, all the ways of writing $2k^2$ as a sum of two squares. Since many of the prime divisors are congruent to $3$ mod $4$, there are not many solutions. Then solve for $x$ and $y$. Do this for each $k$ and you have all solutions. Here's one example. Take $k=35$. So we find all solutions to
$$u^2+v^2 = 2cdot35^2$$
and these are $(7,49)$ and $(35,35)$. The second one leads to $x=0$, which is not positive. The first one gives $2x-35 = 7$ and $2y+35 = 49$ which leads to $(x,y) = (21,7)$. Then from my observation above $y=35-21 = 14$ gives a second solution. Lather, rinse, repeat for all $k$.
answered Jul 17 at 14:15
B. Goddard
16.5k21238
add a comment |Â
up vote
2
down vote
Your first assertion that $x^2+y^2 mid 1995$ doesn't follow from your hypotheses. Note that $(6,3)$ and $(38,19)$ are also solutions. Certainly $(x^2+y^2)/(x-y) >1995$ for $x>1995$, so there can be only finitely many solutions. I believe that $(399,1197)$ is the largest.
I observe that if $a=(x^2+y^2)/(x-y)$ and that $(x,y)$ is a solution, then so is $(a-x,y)$.
Edit: I also observe that every solution is of the form $(2y,y)$ or $(3y,y)$ and in all cases $a=5y$.
Another Edit. Let $k$ be a divisor of $1995$. Then you have $x^2+y^2=k(x-y).$ Put everything on one side and complete the square:
$$left(x-frack2right)^2+left(y+frack2right)^2 = frack^22.$$
Multiply through by $4$:
$$(2x-k)^2 +(2y+k)^2 = 2k^2.$$
So find, in the usual way, all the ways of writing $2k^2$ as a sum of two squares. Since many of the prime divisors are congruent to $3$ mod $4$, there are not many solutions. Then solve for $x$ and $y$. Do this for each $k$ and you have all solutions. Here's one example. Take $k=35$. So we find all solutions to
$$u^2+v^2 = 2cdot35^2$$
and these are $(7,49)$ and $(35,35)$. The second one leads to $x=0$, which is not positive. The first one gives $2x-35 = 7$ and $2y+35 = 49$ which leads to $(x,y) = (21,7)$. Then from my observation above $y=35-21 = 14$ gives a second solution. Lather, rinse, repeat for all $k$.
answered Jul 17 at 14:15
B. Goddard
16.5k21238
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your first assertion that $x^2+y^2 mid 1995$ doesn't follow from your hypotheses. Note that $(6,3)$ and $(38,19)$ are also solutions. Certainly $(x^2+y^2)/(x-y) >1995$ for $x>1995$, so there can be only finitely many solutions. I believe that $(399,1197)$ is the largest.
I observe that if $a=(x^2+y^2)/(x-y)$ and that $(x,y)$ is a solution, then so is $(a-x,y)$.
Edit: I also observe that every solution is of the form $(2y,y)$ or $(3y,y)$ and in all cases $a=5y$.
Another Edit. Let $k$ be a divisor of $1995$. Then you have $x^2+y^2=k(x-y).$ Put everything on one side and complete the square:
$$left(x-frack2right)^2+left(y+frack2right)^2 = frack^22.$$
Multiply through by $4$:
$$(2x-k)^2 +(2y+k)^2 = 2k^2.$$
So find, in the usual way, all the ways of writing $2k^2$ as a sum of two squares. Since many of the prime divisors are congruent to $3$ mod $4$, there are not many solutions. Then solve for $x$ and $y$. Do this for each $k$ and you have all solutions. Here's one example. Take $k=35$. So we find all solutions to
$$u^2+v^2 = 2cdot35^2$$
and these are $(7,49)$ and $(35,35)$. The second one leads to $x=0$, which is not positive. The first one gives $2x-35 = 7$ and $2y+35 = 49$ which leads to $(x,y) = (21,7)$. Then from my observation above $y=35-21 = 14$ gives a second solution. Lather, rinse, repeat for all $k$.
answered Jul 17 at 14:15
B. Goddard
16.5k21238
Your first assertion that $x^2+y^2 mid 1995$ doesn't follow from your hypotheses. Note that $(6,3)$ and $(38,19)$ are also solutions. Certainly $(x^2+y^2)/(x-y) >1995$ for $x>1995$, so there can be only finitely many solutions. I believe that $(399,1197)$ is the largest.
I observe that if $a=(x^2+y^2)/(x-y)$ and that $(x,y)$ is a solution, then so is $(a-x,y)$.
Edit: I also observe that every solution is of the form $(2y,y)$ or $(3y,y)$ and in all cases $a=5y$.
Another Edit. Let $k$ be a divisor of $1995$. Then you have $x^2+y^2=k(x-y).$ Put everything on one side and complete the square:
$$left(x-frack2right)^2+left(y+frack2right)^2 = frack^22.$$
Multiply through by $4$:
$$(2x-k)^2 +(2y+k)^2 = 2k^2.$$
So find, in the usual way, all the ways of writing $2k^2$ as a sum of two squares. Since many of the prime divisors are congruent to $3$ mod $4$, there are not many solutions. Then solve for $x$ and $y$. Do this for each $k$ and you have all solutions. Here's one example. Take $k=35$. So we find all solutions to
$$u^2+v^2 = 2cdot35^2$$
and these are $(7,49)$ and $(35,35)$. The second one leads to $x=0$, which is not positive. The first one gives $2x-35 = 7$ and $2y+35 = 49$ which leads to $(x,y) = (21,7)$. Then from my observation above $y=35-21 = 14$ gives a second solution. Lather, rinse, repeat for all $k$.
answered Jul 17 at 14:15
B. Goddard
16.5k21238
edited Jul 17 at 16:39
answered Jul 17 at 14:15
B. Goddard
16.5k21238
answered Jul 17 at 14:15
B. Goddard
16.5k21238
answered Jul 17 at 14:15
B. Goddard
16.5k21238
16.5k21238
add a comment |Â
add a comment |Â
up vote
1
down vote
Beautiful number.
$$fracx^2+y^2x-y=a$$
Solutions have the form:
$$a=p^2+s^2$$
$$y=s(p-s)$$
$$x=p(p-s)$$
answered Jul 18 at 9:57
individ
3,1941715
Please include an explanation or derivation for your answer - unjustified claims are much less useful
– Carl Mummert
Jul 28 at 20:55
add a comment |Â
up vote
1
down vote
Beautiful number.
$$fracx^2+y^2x-y=a$$
Solutions have the form:
$$a=p^2+s^2$$
$$y=s(p-s)$$
$$x=p(p-s)$$
answered Jul 18 at 9:57
individ
3,1941715
Please include an explanation or derivation for your answer - unjustified claims are much less useful
– Carl Mummert
Jul 28 at 20:55
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Beautiful number.
$$fracx^2+y^2x-y=a$$
Solutions have the form:
$$a=p^2+s^2$$
$$y=s(p-s)$$
$$x=p(p-s)$$
answered Jul 18 at 9:57
individ
3,1941715
Beautiful number.
$$fracx^2+y^2x-y=a$$
Solutions have the form:
$$a=p^2+s^2$$
$$y=s(p-s)$$
$$x=p(p-s)$$
answered Jul 18 at 9:57
individ
3,1941715
answered Jul 18 at 9:57
individ
3,1941715
answered Jul 18 at 9:57
individ
3,1941715
answered Jul 18 at 9:57
individ
3,1941715
3,1941715
Please include an explanation or derivation for your answer - unjustified claims are much less useful
– Carl Mummert
Jul 28 at 20:55
add a comment |Â
Please include an explanation or derivation for your answer - unjustified claims are much less useful
– Carl Mummert
Jul 28 at 20:55
Please include an explanation or derivation for your answer - unjustified claims are much less useful
– Carl Mummert
Jul 28 at 20:55
Please include an explanation or derivation for your answer - unjustified claims are much less useful
– Carl Mummert
Jul 28 at 20:55
add a comment |Â
up vote
0
down vote
$(1,2),(1,3),(3,6),(3,9),(7,14),(7,21),(19,38),(19,57),(21,42),(21,63),(57,114),(57,171),(133,266),(133,399),(399,798),(399,1197)$
This list is complete for all pairs in $(i,j) in mathbbZ^2 mid 1le i < j le 30000$
I just wrote a small program to find solutions. I figured, having the list of solutions would be a good way to check any answers people come up with, although this is not technically a complete answer, as it does not explain why these are the only pairs that work.
answered Jul 17 at 14:23
InterstellarProbe
2,207518
add a comment |Â
up vote
0
down vote
$(1,2),(1,3),(3,6),(3,9),(7,14),(7,21),(19,38),(19,57),(21,42),(21,63),(57,114),(57,171),(133,266),(133,399),(399,798),(399,1197)$
This list is complete for all pairs in $(i,j) in mathbbZ^2 mid 1le i < j le 30000$
I just wrote a small program to find solutions. I figured, having the list of solutions would be a good way to check any answers people come up with, although this is not technically a complete answer, as it does not explain why these are the only pairs that work.
answered Jul 17 at 14:23
InterstellarProbe
2,207518
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$(1,2),(1,3),(3,6),(3,9),(7,14),(7,21),(19,38),(19,57),(21,42),(21,63),(57,114),(57,171),(133,266),(133,399),(399,798),(399,1197)$
This list is complete for all pairs in $(i,j) in mathbbZ^2 mid 1le i < j le 30000$
I just wrote a small program to find solutions. I figured, having the list of solutions would be a good way to check any answers people come up with, although this is not technically a complete answer, as it does not explain why these are the only pairs that work.
answered Jul 17 at 14:23
InterstellarProbe
2,207518
$(1,2),(1,3),(3,6),(3,9),(7,14),(7,21),(19,38),(19,57),(21,42),(21,63),(57,114),(57,171),(133,266),(133,399),(399,798),(399,1197)$
This list is complete for all pairs in $(i,j) in mathbbZ^2 mid 1le i < j le 30000$
I just wrote a small program to find solutions. I figured, having the list of solutions would be a good way to check any answers people come up with, although this is not technically a complete answer, as it does not explain why these are the only pairs that work.
answered Jul 17 at 14:23
InterstellarProbe
2,207518
answered Jul 17 at 14:23
InterstellarProbe
2,207518
answered Jul 17 at 14:23
InterstellarProbe
2,207518
answered Jul 17 at 14:23
InterstellarProbe
2,207518
2,207518
add a comment |Â
add a comment |Â
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The sum of two squares theorem and $1995=3 times 5 times 7 times 19$ may get you to $5$ more quickly, since $19 equiv 7 equiv 3 pmod 4$ so none of them can appear once in the divisor.
– Henry
Jul 17 at 12:55
You might also check that $x^2+y^2=1$ does not give a solution in positive integers
– Henry
Jul 17 at 12:59
I am looking for any other method. If anyone have please post!
– Sufaid Saleel
Jul 17 at 13:18
If you insist that the divisor be a positive integer $(1,2)$ doesn't work
– Ross Millikan
Jul 17 at 13:44
2
If $x=1197, y=399$, then $x^2+y^2=1,592,010$, but $dfracx^2+y^2x-y=1995$.
– InterstellarProbe
Jul 17 at 14:13