Why do elementary steps of elimination appear to fail for a system of equations with each term of degree 2? I apologise in advance if this is irritating due to its simplicity - I have only done High School Mathematics.
Given:
(1) $(a+1)^2$ = $(b+2)^2$+$(c+3)^2$
(2) $(a+3)^2$ = $(b+1)^2$+$(c+2)^2$
(3) $(a+4)^2$ = $(b+3)^2$+$(c+1)^2$
(4) $(a+2)^2$ = $(b+4)^2$+$(c+4)^2$  
Subtracting pairs of equations:
$(2)-(1)=(5)$-->$4a+8 = -2b-3 + -2c-5$
$(3)-(2)=(6)$-->$2a+7 = 4b+8 + -2c-3$
$(4)-(3)=(7)$-->$-4a-12 = 2b+7 + 6c + 15$  
Upon simplification this is:
$$\begin{align}2a+b+c&= -8 \\  
a-2b+c &= -1 \\ 
-2a-b-3c &= 17 \\ 
\end{align}
$$
Which, when solved, gives:
$a=-0.7$
$b=-2.1$
$c=-4.5$  
Unless I have made a calculation error, none of these values when re-substituted into (1), (2), (3), or (4) results in a correct statement. Yet, apart from a general sense that the existence of degree 2 terms invalidates some part of the reasoning, or the fact that the system is overdetermined, I can't pinpoint which step is false or contains an incorrect assumption. Can anyone shed light on this for me? Alternatively, if someone has a method for solving equations (1),(2),(3) without the overdetermining (4) could you explain that or point me in the direction of an explanation. 

Edit: For anyone interested in the specific worked solution, I followed Ted's advice (after learning about general solutions here) and looked to solve (1), (2), (3) with the result:
$$\begin{align}
a&=-3.4 -0.6t\\  
b&=-1.2 +0.2t \\ 
c&=t\\ 
\end{align}
$$
Substitution of these into equation (1) yielded $0.68t^2 +3.44t + 3.88$  
This quadratic equation appeared twice more after substitution into (2) and (3) respectively (meeting the criteria that the solution(s) for t be common to all three if solutions for a,b,c are to be found).  
Using the ordinary quadratic formula, the two solutions for the parameter t were therefore: $t = \frac{-43}{17}+\frac{25}{34}\sqrt{1.28} \approx -1.698$  or $t = \frac{-43}{17}-\frac{25}{34}\sqrt{1.28} \approx -3.361$ 
Finally using those approximations,
$a \approx -2.381 $ or $a \approx -1.383$
$b \approx -1.540 $ or $b \approx -1.872$ and
$c \approx -1.698$ or $c \approx -3.361$  
For confirmation, I re-substituted corresponding results for a,b,c into the original equations (1), (2), and (3). This generated three correct statements with LHS = RHS.  
Thank you to Ted and to anyone else who considered this.
I hope this post is useful to someone else.    
 A: The fact that the system is overdetermined is the whole problem.  It has nothing to do with the fact that the equations have degree 2.  The same thing can happen with an overdetermined linear system.  Let's take an even simpler example with only 2 variables and 3 equations:
$$x+y=0, x-y = 0, x+2y=1$$
Now proceeding as you did above, if we add the first 2 equations we get $x=0$, and if we subtract the third equation from the second, we get $y=1/3$.  Now plug these values back into the original system.  Nothing works.  The system was overdetermined and there was no reason to expect a solution.
If you just want to solve your original equations (1), (2), (3)  (without the overdetermining (4)), you can proceed as you did above, ending up with a linear system of 2 equations in 3 variables.  You can solve this linear system; the general solution of the linear system will give $a$, $b$, $c$ in terms of a single parameter $t$.
Then you plug in these expressions for $a$, $b$, $c$ in terms of $t$ into the original 3 equations, getting 3 quadratic equations in $t$.  If there is a common value (or values) of $t$ that solves all 3 quadratic equations, then you can get solutions for $a$, $b$, $c$ using the common value(s) of $t$.  If there is no common value of $t$, then there is no solution.
