What is the value of $a+b+c+d$ with following conditions? 
$$
\begin{split}
a^2 &+b   &+c   &+d&=10\\
a   &+b^2 &+c   &+d&=12\\
a   &+b   &+c^2 &+d&=16\\
a   &+b   &+c   &+d^2&=22\\
a   &+b   &+c   &+d&=?
\end{split}
$$

By adding these equations I get that $3(a+b+c+d)=60-(a^2+b^2+c^2+d^2)$.
So can I get some hint that how can I get the value of $(a^2+b^2+c^2+d^2)$ (or relate it with the equation) or my approach is wrong and there are some easy way to solve this problem? 

Subtract the first equation from the second to get $b^2 + a - a^2 - b = 2$, which becomes $2 = (b - a)(b + a - 1)$.
Assuming that $a,b,c,d$ are all positive integers, then $b - a = 1$ and $b + a - 1 = 2$ OR $b - a = 2$ and $b + a - 1 = 1$. The restriction that all four numbers are positive integers gives $a = 1$ and $b = 2$.
Substituting these values into the first equation gives $c + d = 7$ and into the fourth equation gives $c + d^2 = 19$. Solving simultaneously gives $c = 3, d = 4$.
$\therefore a + b + c + d = 1 + 2 + 3 + 4 = 10$
