Indeterminate Equation #3

Add the two equations sides by sides.

$\begin{aligned} &4c=a^2+3 \\ &4c-4a=a^2-4a+3 \\ &4(c-a)=(a-3)(a-1) \quad\cdots [A] \end{aligned}$

Now, subtract the second equation from the first equation.

$\begin{aligned} &a^2-2a-4b-3=0 \\ &4b=a^2-2a-3 \quad\cdots [B]\\ &a^2-2a-3>0 \\ &(a-3)(a+1)>0 \\ \end{aligned}$

Since $a>0$ , we can say that $3<a<5$ . Then $a-1>0$ and $a-3>0$ . Substitute that into $[A]$ and we get

$\boxed{c>a}$

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Subtract $4a$ from both sides of $[B]$ .

$\begin{aligned} &4b-4a=a^2-6a-3 \\ &4(b-a)=(a-3)^2-12<0\quad(\because3<a<5) \\ \end{aligned}$

The inequality leads us to:

$\boxed{a>b}$

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Therefore, the three numbers satisfy

$\boxed{c>a>b}$