Mixing
Find all integers $x, y, u, v$ for which holds: $x^2 + y^2 = 3(u^2 + v^2)$.
$begingroup$
Find all integers $x, y, u, v$ for which holds: $x^2 + y^2 = 3(u^2 + v^2)$.
My approach was to say that $u^2+v^2$ is the divisor of $x^2 + y^2$ and $u^2+v^2$ is the divisor of $x^2$ and $y^2$, but I somehow need to use the information about number $3$ that is given. Can someone help me? Thanks in advance!
number-theory elementary-number-theory diophantine-equations
edited Jan 18 at 17:46
J. W. Tanner
2,6171217
asked Jan 18 at 16:28
Wolf M.Wolf M.
1097
$endgroup$
add a comment |
$begingroup$
Find all integers $x, y, u, v$ for which holds: $x^2 + y^2 = 3(u^2 + v^2)$.
My approach was to say that $u^2+v^2$ is the divisor of $x^2 + y^2$ and $u^2+v^2$ is the divisor of $x^2$ and $y^2$, but I somehow need to use the information about number $3$ that is given. Can someone help me? Thanks in advance!
number-theory elementary-number-theory diophantine-equations
edited Jan 18 at 17:46
J. W. Tanner
2,6171217
asked Jan 18 at 16:28
Wolf M.Wolf M.
1097
$endgroup$
3
$begingroup$
$$x^2+y^2notequiv-1pmod4$$
$endgroup$
– lab bhattacharjee
Jan 18 at 16:32
1
$begingroup$
Reduce so there are no common factors. Consider modulo $8$.
$endgroup$
– Mark Bennet
Jan 18 at 16:35
$begingroup$
Wait, I'm confused again. Why is the right side $-1 (mod 4)$? I understand the fact that left side is congruent to 0, or 1 $(mod 4)$ but I don't know why the right side is $-1$.
$endgroup$
– Wolf M.
Jan 18 at 16:36
add a comment |
$begingroup$
Find all integers $x, y, u, v$ for which holds: $x^2 + y^2 = 3(u^2 + v^2)$.
My approach was to say that $u^2+v^2$ is the divisor of $x^2 + y^2$ and $u^2+v^2$ is the divisor of $x^2$ and $y^2$, but I somehow need to use the information about number $3$ that is given. Can someone help me? Thanks in advance!
number-theory elementary-number-theory diophantine-equations
edited Jan 18 at 17:46
J. W. Tanner
2,6171217
asked Jan 18 at 16:28
Wolf M.Wolf M.
1097
$endgroup$
Find all integers $x, y, u, v$ for which holds: $x^2 + y^2 = 3(u^2 + v^2)$.
My approach was to say that $u^2+v^2$ is the divisor of $x^2 + y^2$ and $u^2+v^2$ is the divisor of $x^2$ and $y^2$, but I somehow need to use the information about number $3$ that is given. Can someone help me? Thanks in advance!
number-theory elementary-number-theory diophantine-equations
number-theory elementary-number-theory diophantine-equations
edited Jan 18 at 17:46
J. W. Tanner
2,6171217
asked Jan 18 at 16:28
Wolf M.Wolf M.
1097
edited Jan 18 at 17:46
J. W. Tanner
2,6171217
asked Jan 18 at 16:28
Wolf M.Wolf M.
1097
edited Jan 18 at 17:46
J. W. Tanner
2,6171217
edited Jan 18 at 17:46
J. W. Tanner
2,6171217
edited Jan 18 at 17:46
J. W. Tanner
2,6171217
2,6171217
asked Jan 18 at 16:28
Wolf M.Wolf M.
1097
asked Jan 18 at 16:28
Wolf M.Wolf M.
1097
asked Jan 18 at 16:28
Wolf M.Wolf M.
1097
1097
3
$begingroup$
$$x^2+y^2notequiv-1pmod4$$
$endgroup$
– lab bhattacharjee
Jan 18 at 16:32
1
$begingroup$
Reduce so there are no common factors. Consider modulo $8$.
$endgroup$
– Mark Bennet
Jan 18 at 16:35
$begingroup$
Wait, I'm confused again. Why is the right side $-1 (mod 4)$? I understand the fact that left side is congruent to 0, or 1 $(mod 4)$ but I don't know why the right side is $-1$.
$endgroup$
– Wolf M.
Jan 18 at 16:36
add a comment |
3
$begingroup$
$$x^2+y^2notequiv-1pmod4$$
$endgroup$
– lab bhattacharjee
Jan 18 at 16:32
1
$begingroup$
Reduce so there are no common factors. Consider modulo $8$.
$endgroup$
– Mark Bennet
Jan 18 at 16:35
$begingroup$
Wait, I'm confused again. Why is the right side $-1 (mod 4)$? I understand the fact that left side is congruent to 0, or 1 $(mod 4)$ but I don't know why the right side is $-1$.
$endgroup$
– Wolf M.
Jan 18 at 16:36
3
3
$begingroup$
$$x^2+y^2notequiv-1pmod4$$
$endgroup$
– lab bhattacharjee
Jan 18 at 16:32
$begingroup$
$$x^2+y^2notequiv-1pmod4$$
$endgroup$
– lab bhattacharjee
Jan 18 at 16:32
1
1
$begingroup$
Reduce so there are no common factors. Consider modulo $8$.
$endgroup$
– Mark Bennet
Jan 18 at 16:35
$begingroup$
Reduce so there are no common factors. Consider modulo $8$.
$endgroup$
– Mark Bennet
Jan 18 at 16:35
$begingroup$
Wait, I'm confused again. Why is the right side $-1 (mod 4)$? I understand the fact that left side is congruent to 0, or 1 $(mod 4)$ but I don't know why the right side is $-1$.
$endgroup$
– Wolf M.
Jan 18 at 16:36
$begingroup$
Wait, I'm confused again. Why is the right side $-1 (mod 4)$? I understand the fact that left side is congruent to 0, or 1 $(mod 4)$ but I don't know why the right side is $-1$.
$endgroup$
– Wolf M.
Jan 18 at 16:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are no solutions besides the trivial $x=y=u=v=0$. To show this, suppose for a contradiction that non-trivial solutions exist, and choose one with $|x|+|y|+|u|+|v|$ as small as possible. From $3mid x^2+y^2$ it follows that both $x$ and $y$ are divisible by $3$; say, $x=3x_1$ and $y=3y_1$. Substituting, we get $3(x_1^2+y_1^2)=u^2+v^2$, showing that $(u,v,x_1,y_1)$ is also a solution. Moreover, we have $|u|+|v|+|x_1|+|y_1|<|x|+|y|+|u|+|v|$, contradicting the choice of $(x,y,u,v)$ as the "smallest" non-trivial solution.
answered Jan 18 at 16:45
W-t-PW-t-P
1,287611
$endgroup$
2
$begingroup$
This is a good solution. However, I believe the $x$ and $y$ in "... showing that $left(u, v, x, yright)$ is also a solution" should be $x_1$ and $y_1$ instead. Also, there is no $x_2$ so, in the absolute value inequality, I believe $y_1$ is what you want to use.
$endgroup$
– John Omielan
Jan 18 at 22:42
$begingroup$
@JohnOmielan: Thanks for the remarks, I will fix the things immediately.
$endgroup$
– W-t-P
Jan 19 at 7:41
add a comment |
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$begingroup$
There are no solutions besides the trivial $x=y=u=v=0$. To show this, suppose for a contradiction that non-trivial solutions exist, and choose one with $|x|+|y|+|u|+|v|$ as small as possible. From $3mid x^2+y^2$ it follows that both $x$ and $y$ are divisible by $3$; say, $x=3x_1$ and $y=3y_1$. Substituting, we get $3(x_1^2+y_1^2)=u^2+v^2$, showing that $(u,v,x_1,y_1)$ is also a solution. Moreover, we have $|u|+|v|+|x_1|+|y_1|<|x|+|y|+|u|+|v|$, contradicting the choice of $(x,y,u,v)$ as the "smallest" non-trivial solution.
answered Jan 18 at 16:45
W-t-PW-t-P
1,287611
$endgroup$
2
$begingroup$
This is a good solution. However, I believe the $x$ and $y$ in "... showing that $left(u, v, x, yright)$ is also a solution" should be $x_1$ and $y_1$ instead. Also, there is no $x_2$ so, in the absolute value inequality, I believe $y_1$ is what you want to use.
$endgroup$
– John Omielan
Jan 18 at 22:42
$begingroup$
@JohnOmielan: Thanks for the remarks, I will fix the things immediately.
$endgroup$
– W-t-P
Jan 19 at 7:41
add a comment |
$begingroup$
There are no solutions besides the trivial $x=y=u=v=0$. To show this, suppose for a contradiction that non-trivial solutions exist, and choose one with $|x|+|y|+|u|+|v|$ as small as possible. From $3mid x^2+y^2$ it follows that both $x$ and $y$ are divisible by $3$; say, $x=3x_1$ and $y=3y_1$. Substituting, we get $3(x_1^2+y_1^2)=u^2+v^2$, showing that $(u,v,x_1,y_1)$ is also a solution. Moreover, we have $|u|+|v|+|x_1|+|y_1|<|x|+|y|+|u|+|v|$, contradicting the choice of $(x,y,u,v)$ as the "smallest" non-trivial solution.
answered Jan 18 at 16:45
W-t-PW-t-P
1,287611
$endgroup$
2
$begingroup$
This is a good solution. However, I believe the $x$ and $y$ in "... showing that $left(u, v, x, yright)$ is also a solution" should be $x_1$ and $y_1$ instead. Also, there is no $x_2$ so, in the absolute value inequality, I believe $y_1$ is what you want to use.
$endgroup$
– John Omielan
Jan 18 at 22:42
$begingroup$
@JohnOmielan: Thanks for the remarks, I will fix the things immediately.
$endgroup$
– W-t-P
Jan 19 at 7:41
add a comment |
$begingroup$
There are no solutions besides the trivial $x=y=u=v=0$. To show this, suppose for a contradiction that non-trivial solutions exist, and choose one with $|x|+|y|+|u|+|v|$ as small as possible. From $3mid x^2+y^2$ it follows that both $x$ and $y$ are divisible by $3$; say, $x=3x_1$ and $y=3y_1$. Substituting, we get $3(x_1^2+y_1^2)=u^2+v^2$, showing that $(u,v,x_1,y_1)$ is also a solution. Moreover, we have $|u|+|v|+|x_1|+|y_1|<|x|+|y|+|u|+|v|$, contradicting the choice of $(x,y,u,v)$ as the "smallest" non-trivial solution.
answered Jan 18 at 16:45
W-t-PW-t-P
1,287611
$endgroup$
There are no solutions besides the trivial $x=y=u=v=0$. To show this, suppose for a contradiction that non-trivial solutions exist, and choose one with $|x|+|y|+|u|+|v|$ as small as possible. From $3mid x^2+y^2$ it follows that both $x$ and $y$ are divisible by $3$; say, $x=3x_1$ and $y=3y_1$. Substituting, we get $3(x_1^2+y_1^2)=u^2+v^2$, showing that $(u,v,x_1,y_1)$ is also a solution. Moreover, we have $|u|+|v|+|x_1|+|y_1|<|x|+|y|+|u|+|v|$, contradicting the choice of $(x,y,u,v)$ as the "smallest" non-trivial solution.
answered Jan 18 at 16:45
W-t-PW-t-P
1,287611
edited Jan 19 at 7:42
answered Jan 18 at 16:45
W-t-PW-t-P
1,287611
answered Jan 18 at 16:45
W-t-PW-t-P
1,287611
answered Jan 18 at 16:45
W-t-PW-t-P
1,287611
1,287611
2
$begingroup$
This is a good solution. However, I believe the $x$ and $y$ in "... showing that $left(u, v, x, yright)$ is also a solution" should be $x_1$ and $y_1$ instead. Also, there is no $x_2$ so, in the absolute value inequality, I believe $y_1$ is what you want to use.
$endgroup$
– John Omielan
Jan 18 at 22:42
$begingroup$
@JohnOmielan: Thanks for the remarks, I will fix the things immediately.
$endgroup$
– W-t-P
Jan 19 at 7:41
add a comment |
2
$begingroup$
This is a good solution. However, I believe the $x$ and $y$ in "... showing that $left(u, v, x, yright)$ is also a solution" should be $x_1$ and $y_1$ instead. Also, there is no $x_2$ so, in the absolute value inequality, I believe $y_1$ is what you want to use.
$endgroup$
– John Omielan
Jan 18 at 22:42
$begingroup$
@JohnOmielan: Thanks for the remarks, I will fix the things immediately.
$endgroup$
– W-t-P
Jan 19 at 7:41
2
2
$begingroup$
This is a good solution. However, I believe the $x$ and $y$ in "... showing that $left(u, v, x, yright)$ is also a solution" should be $x_1$ and $y_1$ instead. Also, there is no $x_2$ so, in the absolute value inequality, I believe $y_1$ is what you want to use.
$endgroup$
– John Omielan
Jan 18 at 22:42
$begingroup$
This is a good solution. However, I believe the $x$ and $y$ in "... showing that $left(u, v, x, yright)$ is also a solution" should be $x_1$ and $y_1$ instead. Also, there is no $x_2$ so, in the absolute value inequality, I believe $y_1$ is what you want to use.
$endgroup$
– John Omielan
Jan 18 at 22:42
$begingroup$
@JohnOmielan: Thanks for the remarks, I will fix the things immediately.
$endgroup$
– W-t-P
Jan 19 at 7:41
$begingroup$
@JohnOmielan: Thanks for the remarks, I will fix the things immediately.
$endgroup$
– W-t-P
Jan 19 at 7:41
add a comment |
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3
$begingroup$
$$x^2+y^2notequiv-1pmod4$$
$endgroup$
– lab bhattacharjee
Jan 18 at 16:32
1
$begingroup$
Reduce so there are no common factors. Consider modulo $8$.
$endgroup$
– Mark Bennet
Jan 18 at 16:35
$begingroup$
Wait, I'm confused again. Why is the right side $-1 (mod 4)$? I understand the fact that left side is congruent to 0, or 1 $(mod 4)$ but I don't know why the right side is $-1$.
$endgroup$
– Wolf M.
Jan 18 at 16:36