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My question is:

Solve for $a$, $b$, and $c$ if: $a^2 - b^2=36,$ $b^2 - c^2=\frac{116}{3},$ $a^2 - c^2=\frac{224}{3}.$

Any solution for this question would be greatly appreciated.

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    hey all sorry yes this is not a diophantine equation...i tagged it wrong2012-07-09

3 Answers 3

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As Adam Rubinson has commented, the equations have a dependence, so you can ignore one. If you don't require integer solutions, the easiest is to ignore the first. Then pick any value for $c$ you want. You can plug it into the others to solve for $a$ and $b$. For example, $a^2=\frac{224}3 + c^2$, so $a = \pm \sqrt{\frac{224}3 + c^2}$. The reason to use $c$ as the choosable value is you avoid any problem of square roots of negative numbers.

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$a^2=b^2+(6)^2$ , so $(b,6,a)$ is a Pythagorean triple. Hence there exist $k,m,n$ such that $m>n$ and $b=k(m^2-n^2),6=k(2mn),a=k(m^2+n^2)$ so $kmn=3$, this implies $m=3,n=1,k=1$( as $m>n$). But then $b=m^2-n^2=8$ and $a=m^2+n^2=10$ Now $c^2=b^2-116/3=64-116/3=(192-116)/3=76/3$, so $c$ is not an integer. So there are no integer soloutions.

If you are asking for real soloutions then there are infinitely many real soloutions. Take any real $r>10$ then $b=r,a=\sqrt{r^2+36},c=\sqrt{r^2-116/3}$ is a soloution.

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If this isn't a set of Diophantine equations, as the unknowns appear only as squares, you can solve the (linear!) system for $a^2$, $b^2$, and $c^2$, and then take square roots. Don't forget to consider the signs.

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    The system of linear equations you refer to doesn't have a unique solution.2013-01-25