Mixing
solution of an algebraic equation with integers [closed]
$begingroup$
Prove that the equation $x^2+y^2+z^2=(x-y)(y-z)(z-x)$ has an infinite number of solutions when $x,y,z$ are integers.
I started from specific cases.
elementary-number-theory diophantine-equations
edited Jan 24 at 2:21
Namaste
1
asked Jan 23 at 23:38
Joshua Haim MamouJoshua Haim Mamou
245
$endgroup$
closed as off-topic by Morgan Rodgers, José Carlos Santos, Kemono Chen, clathratus, Trevor Gunn Jan 28 at 16:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, José Carlos Santos, Kemono Chen, clathratus, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Prove that the equation $x^2+y^2+z^2=(x-y)(y-z)(z-x)$ has an infinite number of solutions when $x,y,z$ are integers.
I started from specific cases.
elementary-number-theory diophantine-equations
edited Jan 24 at 2:21
Namaste
1
asked Jan 23 at 23:38
Joshua Haim MamouJoshua Haim Mamou
245
$endgroup$
closed as off-topic by Morgan Rodgers, José Carlos Santos, Kemono Chen, clathratus, Trevor Gunn Jan 28 at 16:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, José Carlos Santos, Kemono Chen, clathratus, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Expanding the right side may help.
$endgroup$
– Math Lover
Jan 23 at 23:47
$begingroup$
Did you find any solutions, Joshua?
$endgroup$
– Gerry Myerson
Jan 24 at 0:25
2
$begingroup$
@AfronPie no, here are some positive xyz 21 7 14 gcd 7 \ 100 55 85 gcd 5 \ 105 60 75 gcd 15 \ 125 75 100 gcd 25 \ 184 115 161 gcd 23 \ 260 175 205 gcd 5 \ 265 155 250 gcd 5 \ 385 275 330 gcd 55 \ 497 259 266 gcd 7 \ 845 650 715 gcd 65 \ 873 679 776 gcd 97 \ 989 736 943 gcd 23 \
$endgroup$
– Will Jagy
Jan 24 at 0:27
3
$begingroup$
If $x^2+y^2+z^2=d(x-y)(y-z)(z-x)$, then $(dx)^2+(dy)^2+(dz)^2=(dx-dy)(dy-dz)(dz-dx)$.
$endgroup$
– Gerry Myerson
Jan 24 at 0:32
$begingroup$
Generalizing Gerry Myerson's comment, if $x^2+y^2+z^2 =d^a(x-y)(y-z)(z-x) $ then $d^ax, d^ay, d^az$ is a solution. Other combinations of exponents don't work.
$endgroup$
– marty cohen
Jan 26 at 5:33
add a comment |
$begingroup$
Prove that the equation $x^2+y^2+z^2=(x-y)(y-z)(z-x)$ has an infinite number of solutions when $x,y,z$ are integers.
I started from specific cases.
elementary-number-theory diophantine-equations
edited Jan 24 at 2:21
Namaste
1
asked Jan 23 at 23:38
Joshua Haim MamouJoshua Haim Mamou
245
$endgroup$
Prove that the equation $x^2+y^2+z^2=(x-y)(y-z)(z-x)$ has an infinite number of solutions when $x,y,z$ are integers.
I started from specific cases.
elementary-number-theory diophantine-equations
elementary-number-theory diophantine-equations
edited Jan 24 at 2:21
Namaste
1
asked Jan 23 at 23:38
Joshua Haim MamouJoshua Haim Mamou
245
edited Jan 24 at 2:21
Namaste
1
asked Jan 23 at 23:38
Joshua Haim MamouJoshua Haim Mamou
245
edited Jan 24 at 2:21
Namaste
1
edited Jan 24 at 2:21
Namaste
1
edited Jan 24 at 2:21
Namaste
1
1
asked Jan 23 at 23:38
Joshua Haim MamouJoshua Haim Mamou
245
asked Jan 23 at 23:38
Joshua Haim MamouJoshua Haim Mamou
245
asked Jan 23 at 23:38
Joshua Haim MamouJoshua Haim Mamou
245
245
closed as off-topic by Morgan Rodgers, José Carlos Santos, Kemono Chen, clathratus, Trevor Gunn Jan 28 at 16:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, José Carlos Santos, Kemono Chen, clathratus, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Morgan Rodgers, José Carlos Santos, Kemono Chen, clathratus, Trevor Gunn Jan 28 at 16:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Morgan Rodgers, José Carlos Santos, Kemono Chen, clathratus, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
Expanding the right side may help.
$endgroup$
– Math Lover
Jan 23 at 23:47
$begingroup$
Did you find any solutions, Joshua?
$endgroup$
– Gerry Myerson
Jan 24 at 0:25
2
$begingroup$
@AfronPie no, here are some positive xyz 21 7 14 gcd 7 \ 100 55 85 gcd 5 \ 105 60 75 gcd 15 \ 125 75 100 gcd 25 \ 184 115 161 gcd 23 \ 260 175 205 gcd 5 \ 265 155 250 gcd 5 \ 385 275 330 gcd 55 \ 497 259 266 gcd 7 \ 845 650 715 gcd 65 \ 873 679 776 gcd 97 \ 989 736 943 gcd 23 \
$endgroup$
– Will Jagy
Jan 24 at 0:27
3
$begingroup$
If $x^2+y^2+z^2=d(x-y)(y-z)(z-x)$, then $(dx)^2+(dy)^2+(dz)^2=(dx-dy)(dy-dz)(dz-dx)$.
$endgroup$
– Gerry Myerson
Jan 24 at 0:32
$begingroup$
Generalizing Gerry Myerson's comment, if $x^2+y^2+z^2 =d^a(x-y)(y-z)(z-x) $ then $d^ax, d^ay, d^az$ is a solution. Other combinations of exponents don't work.
$endgroup$
– marty cohen
Jan 26 at 5:33
add a comment |
2
$begingroup$
Expanding the right side may help.
$endgroup$
– Math Lover
Jan 23 at 23:47
$begingroup$
Did you find any solutions, Joshua?
$endgroup$
– Gerry Myerson
Jan 24 at 0:25
2
$begingroup$
@AfronPie no, here are some positive xyz 21 7 14 gcd 7 \ 100 55 85 gcd 5 \ 105 60 75 gcd 15 \ 125 75 100 gcd 25 \ 184 115 161 gcd 23 \ 260 175 205 gcd 5 \ 265 155 250 gcd 5 \ 385 275 330 gcd 55 \ 497 259 266 gcd 7 \ 845 650 715 gcd 65 \ 873 679 776 gcd 97 \ 989 736 943 gcd 23 \
$endgroup$
– Will Jagy
Jan 24 at 0:27
3
$begingroup$
If $x^2+y^2+z^2=d(x-y)(y-z)(z-x)$, then $(dx)^2+(dy)^2+(dz)^2=(dx-dy)(dy-dz)(dz-dx)$.
$endgroup$
– Gerry Myerson
Jan 24 at 0:32
$begingroup$
Generalizing Gerry Myerson's comment, if $x^2+y^2+z^2 =d^a(x-y)(y-z)(z-x) $ then $d^ax, d^ay, d^az$ is a solution. Other combinations of exponents don't work.
$endgroup$
– marty cohen
Jan 26 at 5:33
2
2
$begingroup$
Expanding the right side may help.
$endgroup$
– Math Lover
Jan 23 at 23:47
$begingroup$
Expanding the right side may help.
$endgroup$
– Math Lover
Jan 23 at 23:47
$begingroup$
Did you find any solutions, Joshua?
$endgroup$
– Gerry Myerson
Jan 24 at 0:25
$begingroup$
Did you find any solutions, Joshua?
$endgroup$
– Gerry Myerson
Jan 24 at 0:25
2
2
$begingroup$
@AfronPie no, here are some positive xyz 21 7 14 gcd 7 \ 100 55 85 gcd 5 \ 105 60 75 gcd 15 \ 125 75 100 gcd 25 \ 184 115 161 gcd 23 \ 260 175 205 gcd 5 \ 265 155 250 gcd 5 \ 385 275 330 gcd 55 \ 497 259 266 gcd 7 \ 845 650 715 gcd 65 \ 873 679 776 gcd 97 \ 989 736 943 gcd 23 \
$endgroup$
– Will Jagy
Jan 24 at 0:27
$begingroup$
@AfronPie no, here are some positive xyz 21 7 14 gcd 7 \ 100 55 85 gcd 5 \ 105 60 75 gcd 15 \ 125 75 100 gcd 25 \ 184 115 161 gcd 23 \ 260 175 205 gcd 5 \ 265 155 250 gcd 5 \ 385 275 330 gcd 55 \ 497 259 266 gcd 7 \ 845 650 715 gcd 65 \ 873 679 776 gcd 97 \ 989 736 943 gcd 23 \
$endgroup$
– Will Jagy
Jan 24 at 0:27
3
3
$begingroup$
If $x^2+y^2+z^2=d(x-y)(y-z)(z-x)$, then $(dx)^2+(dy)^2+(dz)^2=(dx-dy)(dy-dz)(dz-dx)$.
$endgroup$
– Gerry Myerson
Jan 24 at 0:32
$begingroup$
If $x^2+y^2+z^2=d(x-y)(y-z)(z-x)$, then $(dx)^2+(dy)^2+(dz)^2=(dx-dy)(dy-dz)(dz-dx)$.
$endgroup$
– Gerry Myerson
Jan 24 at 0:32
$begingroup$
Generalizing Gerry Myerson's comment, if $x^2+y^2+z^2 =d^a(x-y)(y-z)(z-x) $ then $d^ax, d^ay, d^az$ is a solution. Other combinations of exponents don't work.
$endgroup$
– marty cohen
Jan 26 at 5:33
$begingroup$
Generalizing Gerry Myerson's comment, if $x^2+y^2+z^2 =d^a(x-y)(y-z)(z-x) $ then $d^ax, d^ay, d^az$ is a solution. Other combinations of exponents don't work.
$endgroup$
– marty cohen
Jan 26 at 5:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $r$ be odd, then $(r+2)^2+r^2+(r+1)^2$ is even, so it is a multiple of $((r+2)-r)(r-(r+1))((r+1)-(r+2))=2$. The quotient is $d=(3r^2+6r+5)/2$. Then $x=d(r+2)$, $y=dr$, $z=d(r+1)$ works.
Of course, there are many more. E.g., let $b=a+1$, $c=a+3$. Then $(a-b)(b-c)(c-a)=6$. If $a$ is even, then $a^2+b^2+c^2$ is divisible by 2, and if $a-1$ is divisible by $3$, then so is $a^2+b^2+c^2$, so if $a=6r-2$, then $a^2+b^2+c^2=108r^2-24r+6=(a-b)(b-c)(c-a)s$ where $s=18r^2-4r+1$. So, $x=(6r-2)s$, $y=(6r-1)s$, $z=(6r+1)s$ is a solution for $r=1,2,3,dots$.
answered Jan 24 at 0:44
Gerry MyersonGerry Myerson
147k8150302
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $r$ be odd, then $(r+2)^2+r^2+(r+1)^2$ is even, so it is a multiple of $((r+2)-r)(r-(r+1))((r+1)-(r+2))=2$. The quotient is $d=(3r^2+6r+5)/2$. Then $x=d(r+2)$, $y=dr$, $z=d(r+1)$ works.
Of course, there are many more. E.g., let $b=a+1$, $c=a+3$. Then $(a-b)(b-c)(c-a)=6$. If $a$ is even, then $a^2+b^2+c^2$ is divisible by 2, and if $a-1$ is divisible by $3$, then so is $a^2+b^2+c^2$, so if $a=6r-2$, then $a^2+b^2+c^2=108r^2-24r+6=(a-b)(b-c)(c-a)s$ where $s=18r^2-4r+1$. So, $x=(6r-2)s$, $y=(6r-1)s$, $z=(6r+1)s$ is a solution for $r=1,2,3,dots$.
answered Jan 24 at 0:44
Gerry MyersonGerry Myerson
147k8150302
$endgroup$
add a comment |
$begingroup$
Let $r$ be odd, then $(r+2)^2+r^2+(r+1)^2$ is even, so it is a multiple of $((r+2)-r)(r-(r+1))((r+1)-(r+2))=2$. The quotient is $d=(3r^2+6r+5)/2$. Then $x=d(r+2)$, $y=dr$, $z=d(r+1)$ works.
Of course, there are many more. E.g., let $b=a+1$, $c=a+3$. Then $(a-b)(b-c)(c-a)=6$. If $a$ is even, then $a^2+b^2+c^2$ is divisible by 2, and if $a-1$ is divisible by $3$, then so is $a^2+b^2+c^2$, so if $a=6r-2$, then $a^2+b^2+c^2=108r^2-24r+6=(a-b)(b-c)(c-a)s$ where $s=18r^2-4r+1$. So, $x=(6r-2)s$, $y=(6r-1)s$, $z=(6r+1)s$ is a solution for $r=1,2,3,dots$.
answered Jan 24 at 0:44
Gerry MyersonGerry Myerson
147k8150302
$endgroup$
add a comment |
$begingroup$
Let $r$ be odd, then $(r+2)^2+r^2+(r+1)^2$ is even, so it is a multiple of $((r+2)-r)(r-(r+1))((r+1)-(r+2))=2$. The quotient is $d=(3r^2+6r+5)/2$. Then $x=d(r+2)$, $y=dr$, $z=d(r+1)$ works.
Of course, there are many more. E.g., let $b=a+1$, $c=a+3$. Then $(a-b)(b-c)(c-a)=6$. If $a$ is even, then $a^2+b^2+c^2$ is divisible by 2, and if $a-1$ is divisible by $3$, then so is $a^2+b^2+c^2$, so if $a=6r-2$, then $a^2+b^2+c^2=108r^2-24r+6=(a-b)(b-c)(c-a)s$ where $s=18r^2-4r+1$. So, $x=(6r-2)s$, $y=(6r-1)s$, $z=(6r+1)s$ is a solution for $r=1,2,3,dots$.
answered Jan 24 at 0:44
Gerry MyersonGerry Myerson
147k8150302
$endgroup$
Let $r$ be odd, then $(r+2)^2+r^2+(r+1)^2$ is even, so it is a multiple of $((r+2)-r)(r-(r+1))((r+1)-(r+2))=2$. The quotient is $d=(3r^2+6r+5)/2$. Then $x=d(r+2)$, $y=dr$, $z=d(r+1)$ works.
Of course, there are many more. E.g., let $b=a+1$, $c=a+3$. Then $(a-b)(b-c)(c-a)=6$. If $a$ is even, then $a^2+b^2+c^2$ is divisible by 2, and if $a-1$ is divisible by $3$, then so is $a^2+b^2+c^2$, so if $a=6r-2$, then $a^2+b^2+c^2=108r^2-24r+6=(a-b)(b-c)(c-a)s$ where $s=18r^2-4r+1$. So, $x=(6r-2)s$, $y=(6r-1)s$, $z=(6r+1)s$ is a solution for $r=1,2,3,dots$.
answered Jan 24 at 0:44
Gerry MyersonGerry Myerson
147k8150302
edited Jan 26 at 5:06
answered Jan 24 at 0:44
Gerry MyersonGerry Myerson
147k8150302
answered Jan 24 at 0:44
Gerry MyersonGerry Myerson
147k8150302
answered Jan 24 at 0:44
Gerry MyersonGerry Myerson
147k8150302
147k8150302
add a comment |
add a comment |
2
$begingroup$
Expanding the right side may help.
$endgroup$
– Math Lover
Jan 23 at 23:47
$begingroup$
Did you find any solutions, Joshua?
$endgroup$
– Gerry Myerson
Jan 24 at 0:25
2
$begingroup$
@AfronPie no, here are some positive xyz 21 7 14 gcd 7 \ 100 55 85 gcd 5 \ 105 60 75 gcd 15 \ 125 75 100 gcd 25 \ 184 115 161 gcd 23 \ 260 175 205 gcd 5 \ 265 155 250 gcd 5 \ 385 275 330 gcd 55 \ 497 259 266 gcd 7 \ 845 650 715 gcd 65 \ 873 679 776 gcd 97 \ 989 736 943 gcd 23 \
$endgroup$
– Will Jagy
Jan 24 at 0:27
3
$begingroup$
If $x^2+y^2+z^2=d(x-y)(y-z)(z-x)$, then $(dx)^2+(dy)^2+(dz)^2=(dx-dy)(dy-dz)(dz-dx)$.
$endgroup$
– Gerry Myerson
Jan 24 at 0:32
$begingroup$
Generalizing Gerry Myerson's comment, if $x^2+y^2+z^2 =d^a(x-y)(y-z)(z-x) $ then $d^ax, d^ay, d^az$ is a solution. Other combinations of exponents don't work.
$endgroup$
– marty cohen
Jan 26 at 5:33