$$ \begin{split} a^2 &+b &+c &+d&=10\\ a &+b^2 &+c &+d&=12\\ a &+b &+c^2 &+d&=16\\ a &+b &+c &+d^2&=22\\ a &+b &+c &+d&=? \end{split} $$
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?
On
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 $c = 3, d = 4$.
$\therefore a + b + c + d = 1 + 2 + 3 + 4 = 10$
From $a^2 + b + c + d = 10$ we get $d = 10 - a^2 - b - c$. Plugging into the other three equations, we get:
We now make a change of variables $A = 2a - 1$, $B = 2b - 1$, $C = 2c - 1$, and get:
$B^2 = A^2 + 8$
$C^2 = A^2 + 24$
$A^4 + 4A^3 + 4A^2B + 4A^2C - 66A^2 + 8AB + 4B^2 + 8AC + 8BC + 4C^2 - 132A - 132B - 132C + 897 = 0$
Let $f(A, B, C)$ be the polynomial on the left hand side of 6. We then have $f(A, B, C)f(A, -B, C)f(A, B, -C)f(A, -B, -C) = 0$. Expanding this, and plugging in 4 and 5, we get a polynomial in $A$:
This polynomial has only one root in $\Bbb Q$, but has all $16$ simple roots in $\Bbb R$. These are:
\begin{eqnarray*}A &=& -11.4538743253946,\\ &&-8.60113697282605,\\ &&-8.34595743762429,\\ &&-7.96349902754117,\\ &&-5.88026229356758,\\ &&-5.34761209324886,\\ &&-4.81524129575445,\\ &&-2.22253720582883,\\ &&1,\\ &&2.95891204052723,\\ &&3.62156418653849,\\ &&4.21059576045945,\\ &&5.60461219532045,\\ &&6.08827822184500,\\ &&6.41671529164311,\\ &&8.72944295545209.\end{eqnarray*}
We may then obtain the corresponding values of $a, b, c, d$. The complete list of solutions is:
\begin{eqnarray*}(a, b, c, d) &= &(-5.2269, -5.3990, -5.7288, -6.1931),\\ &&(-3.8006, 5.0271, -4.4492, -5.0222),\\ &&(-3.6730, -3.9061, 5.3388, -4.9234),\\ &&(-3.4818, -3.7254, -4.1749, 5.7777),\\ &&(-2.4401, 3.7626, 4.3268, -4.0436),\\ &&(-2.1738, 3.5248, -3.1262, 4.8760),\\ &&(-1.9076, -2.2922, 3.9346, 4.7186),\\ &&(-0.6113, 2.2986, 3.1898, 4.1380),\\ &&(1, 2, 3, 4),\\ &&(1.9795, -1.5467, 3.3616, 4.2668),\\ &&(2.3108, 2.7976, -2.5461, 4.4088),\\ &&(2.6053, 3.0362, 3.7299, -3.5537),\\ &&(3.3023, -2.6389, -3.2220, 4.9557),\\ &&(3.5441, -2.8566, 4.4073, -4.1116),\\ &&(3.7084, 4.0062, -3.5365, -4.2216),\\ &&(4.8647, -4.0881, -4.5051, -5.0723).\end{eqnarray*}
And these corresponds to $16$ different possibilities of $a + b + c + d$:
\begin{eqnarray*} a + b + c + d &=& -22.5478092654833,\\ &&-8.24488930632880,\\ &&-7.16375138765904,\\ &&-5.60432919041228,\\ &&1.60562883971184,\\ &&3.10076122503463,\\ &&4.45336281591524,\\ &&9.01508209217664,\\ &&10,\\ &&8.06120988410575,\\ &&6.97106821069546,\\ &&5.81772083550022,\\ &&2.39708053501633,\\ &&0.983217073351971,\\ &&-0.0435587835016298,\\ &&-8.80079357812302.\end{eqnarray*}