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kaliprasad
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solve in integers x, y, z(parametric form)
$4x^2 + 9 y^2 = 72z^2$
$4x^2 + 9 y^2 = 72z^2$
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]
The equation $4x^2 + 9 y^2 = 72z^2$ is used to find integer solutions for a three-variable Diophantine equation. This means that the equation is looking for whole number solutions for the variables x, y, and z.
The process of solving this equation involves using number theory methods and algebraic manipulations. One approach is to use the Pythagorean triple method, where the equation is transformed into the form $a^2 + b^2 = c^2$ and then using Pythagorean triples to find solutions.
Yes, there are restrictions on the values of x, y, and z in this equation. They must be integers and there also may be additional restrictions depending on the specific problem being solved.
This type of equation has many applications in fields such as cryptography, physics, and engineering. It can be used to solve problems involving distance, velocity, and acceleration, as well as in the design of efficient communication systems.
It is possible to solve this equation by hand, but it can be a complex and time-consuming process. Many mathematicians and scientists use computer programs and algorithms to find solutions to these types of equations more efficiently.