Mathematical Reflections 2016, Issue 5 - Problem J385

Problem:
 If the equalities $$\begin{array}{rcl} 2(a+b)-6c-3(d+e)&=&6 \\ 3(a+b)-2c+6(d+e)&=&2 \\ 6(a+b)+3c-2(d+e)&=&-3 \end{array}$$ hold simultaneously, evaluate $a^2-b^2+c^2-d^2+e^2$.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Multiplying the second equation by $3$ and subtracting the first equation, we get $$a+b=-3(d+e).$$ Multiplying the third equation by $2$ and adding the first equation, we get $$2(a+b)=d+e.$$ It follows that $a+b=-6(a+b)$, which gives $a+b=0$ and $d+e=0$. So, $c=-1$ and
$$a^2-b^2+c^2-d^2+e^2=(a-b)(a+b)+c^2-(d-e)(d+e)=1.$$