The equation $4x^2 + 9y^2 = 72z^2$ can be solved in integers by transforming it into the form $a^2 + b^2 = 2c^2$. This transformation reveals that nontrivial integer solutions exist when the parameter $c$ has prime factors of the form $4k+1$. For instance, with $c=5$, the solution $(7, 1, 5)$ satisfies the equation, and for $c=13$, the solution $(17, 7, 13)$ holds. The general solution can be expressed as $x = \pm 3t$, $y = \pm 2t$, and $z = \pm t$, where $t$ is an integer.
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kaliprasad said:solve in integers x, y, z(parametric form)
$4x^2 + 9 y^2 = 72z^2$
Albert said: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:Albert said: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 $t\in Z$
Albert said:EDIT:
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 $t\in Z$
$2x=a,3y=b,6c=z$Opalg said:[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]
Opalg said:[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]
[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]Albert said:$2x=a,3y=b,6c=z$
note:$x,y,z,a,b,c \,\, all \in Z$
see post #4
Opalg said:[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]