MHB Integer Solutions for $4x^2 + 9 y^2 = 72z^2$

[kaliprasad]

solve in integers x, y, z(parametric form)

$4x^2 + 9 y^2 = 72z^2$

 

[Albert1]

Re: integer solutions

solve in integers x, y, z(parametric form)

Spoiler

$4x^2 + 9 y^2 = 72z^2$

Spoiler

let $2x=a,3y=b,6z=c$
we get :$a^2+b^2=2c^2$
the only possible sotions will be:$a=b=c$
$\therefore 2x=3y=6z=t$
or $x=\dfrac{t}{2},y=\dfrac{t}{3},z=\dfrac{t}{6}$
where $t\in Z$

 

[kaliprasad]

Re: integer solutions

Spoiler

let $2x=a,3y=b,6z=c$
we get :$a^2+b^2=2c^2$
the only possible sotions will be:$a=b=c$
$\therefore 2x=3y=6z=t$
or $x=\dfrac{t}{2},y=\dfrac{t}{3},z=\dfrac{t}{6}$
where $t\in Z$

Spoiler

the only possible sotions will be:$a=b=c$

is

Spoiler

wrong

 

[Albert1]

Re: integer solutions

Spoiler

let $2x=a,3y=b,6z=c$
we get :$a^2+b^2=2c^2$
the only possible sotions will be:$a=b=c$
$\therefore 2x=3y=6z=t$
or $x=\dfrac{t}{2},y=\dfrac{t}{3},z=\dfrac{t}{6}$
where $t\in Z$

EDIT:

Spoiler

sorry in a haste, I made a mistake
2x=a,3y=b,6z=c
for $a,b,c,x,y,z \in Z$
$a^2+b^2=2c^2$
we have :$\pm2x=\pm3y=\pm6z=6t(for \,\,a^2=b^2=c^2)$
$x=\pm 3t,y=\pm 2t,z=\pm t$
where $t\in Z$

 

[Opalg]

Re: integer solutions

[sp]As Albert points out, the problem reduces to finding integer solutions of the equation $a^2+b^2 = 2c^2$. Apart from the trivial solutions with $a=b=c$, there will be nontrivial solutions whenever $c$ has any prime factors of the form $4k+1.$ For example, if $c=5$ then $7^2+1^1 = 2*5^2 = 50.$ When $c = 13$ we have $17^2+7^2 = 2*13^2 = 338$. With so many solutions, I am not sure that there will be a manageable parametrisation for them.[/sp]

 

[kaliprasad]

Re: integer solutions

EDIT:

Spoiler

sorry in a haste, I made a mitake
for $a,b,c\in Z$
$a^2+b^2=2c^2$
we have :$2x=\pm3y=\pm6z=6t(for \,\,a^2=b^2=c^2)$
$x=\pm 2t,y=\pm 3t,z=\pm t$
where $t\in Z$

Spoiler

a = b =c is wrong conclusion as a = 1 , b = 5, c = 7 satisfies

 

[Albert1]

Re: integer solutions

[sp]As Albert points out, the problem reduces to finding integer solutions of the equation $a^2+b^2 = 2c^2$. Apart from the trivial solutions with $a=b=c$, there will be nontrivial solutions whenever $c$ has any prime factors of the form $4k+1.$ For example, if $c=5$ then $7^2+1^1 = 2*5^2 = 50.$ When $c = 13$ we have $17^2+7^2 = 2*13^2 = 338$. With so many solutions, I am not sure that there will be a manageable parametrisation for them.[/sp]

$2x=a,3y=b,6c=z$
note:$x,y,z,a,b,c \,\, all \in Z$

see post #4

 

[kaliprasad]

Re: integer solutions

[sp]As Albert points out, the problem reduces to finding integer solutions of the equation $a^2+b^2 = 2c^2$. Apart from the trivial solutions with $a=b=c$, there will be nontrivial solutions whenever $c$ has any prime factors of the form $4k+1.$ For example, if $c=5$ then $7^2+1^1 = 2*5^2 = 50.$ When $c = 13$ we have $17^2+7^2 = 2*13^2 = 338$. With so many solutions, I am not sure that there will be a manageable parametrisation for them.[/sp]

Spoiler

it is manageable and I feel that you can do it.

 

[Opalg]

Re: integer solutions

$2x=a,3y=b,6c=z$
note:$x,y,z,a,b,c \,\, all \in Z$

see post #4

[sp]I see now that the way I reduced the problem to $a^2 + b^2 =2 c^2$ is different from that. Starting from the original equation $4x^2 + 9y^2 = 72z^2$, you can see that $x$ must be a multiple of $3$ (because $9$ and $72$ are both multiples of $9$, hence so is $4x^2$). In the same way, $y$ must be a multiple of $2$. If we put $x=3a$ and $y = 2b$, then the equation becomes $36a^2 + 36b^2 = 72z^2$, or $a^2 + b^2 = 2z^2$. Any set of integers $(a,b,z)$ satisfying that equation will give a solution $(x,y,z) = (3a,2b,z)$ of the original equation.[/sp]

 

[Albert1]

Re: integer solutions

the solution will be:
$x=\pm3t,y=\pm2t, z=\pm t$
where $t\in Z$

 

[kaliprasad]

Re: integer solutions

[sp]I see now that the way I reduced the problem to $a^2 + b^2 =2 c^2$ is different from that. Starting from the original equation $4x^2 + 9y^2 = 72z^2$, you can see that $x$ must be a multiple of $3$ (because $9$ and $72$ are both multiples of $9$, hence so is $4x^2$). In the same way, $y$ must be a multiple of $2$. If we put $x=3a$ and $y = 2b$, then the equation becomes $36a^2 + 36b^2 = 72z^2$, or $a^2 + b^2 = 2z^2$. Any set of integers $(a,b,z)$ satisfying that equation will give a solution $(x,y,z) = (3a,2b,z)$ of the original equation.[/sp]

as mentioned above by Opalg

Spoiler

$(x,y,z) = (3a, 2b , z)$ is a solution when

$a^2 + b^2 = 2z^2$

solution of $a^2 + b^2 = 2z^2$

is given by

$ (a,b,z) = (\pm(m^2-n^2-2mn),\,\pm(m^2-n^2+2mn) \,\pm(m^2+n^2))$

as per http://mathhelpboards.com/challenge-questions-puzzles-28/3-consecutive-terms-ap-perfect-square-8701.htmlso one set of solution
$(x,y,z) = (3a, 2b , z) = (\pm3(m^2-n^2-2mn),\,\pm2 (m^2-n^2+2mn), \,\pm(m^2+n^2)))$

2nd set is
$(x,y,z) = (3b, 2a , z) = (\pm3(m^2-n^2+2mn),\,\pm2 (m^2-n^2-2mn), \,\pm(m^2+n^2)))$

as a and b are interchangeable

edit: one can see that interchanging m and n both the 1st and second sets are same.

so solution is
$(x,y,z) = (3a, 2b , z) = (\pm3(m^2-n^2-2mn),\,\pm2 (m^2-n^2+2mn), \,\pm(m^2+n^2)))$