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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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Consider any two of the three equations. The third equation can be derived from the other two, and so gives you no more information about the unknowns. So you have two equations with 3 unknowns and so the best you can do is write a in terms of b and c (for example) –  Adam Rubinson Jul 9 '12 at 16:34
    
If this is truly a Diophantine equation, your solution set is empty. –  user17794 Jul 9 '12 at 16:38
    
To see this more directly, let's just look at equations 1 and 2 and ignore the 3rd for a moment. Consider a = 10, Then b and c can both take two values (b can take +-8 and c can take +-(sqrt(73/3)). These values of c will satisfy the 3rd equation (check). Moreover, for ANY value of a >= 6, b and c will be uniquely determined (up to +-) –  Adam Rubinson Jul 9 '12 at 16:41
    
Surely if this is a diophantine equation then the question is silly –  Adam Rubinson Jul 9 '12 at 16:42
    
hey all sorry yes this is not a diophantine equation...i tagged it wrong –  mgh Jul 9 '12 at 16:44

3 Answers 3

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. –  coffeemath Jan 25 '13 at 16:39

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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