System of 4 tedious nonlinear equations: $ (a+k)(b+k)(c+k)(d+k) = $ constant for $1 \le k \le 4$ It is given that
$$(a+1)(b+1)(c+1)(d+1)=15$$$$(a+2)(b+2)(c+2)(d+2)=45$$$$(a+3)(b+3)(c+3)(d+3)=133$$$$(a+4)(b+4)(c+4)(d+4)=339$$ How do I find the value of $(a+5)(b+5)(c+5)(d+5)$. I could think only of opening each expression and then manipulating, and I also thought of integer solutions(none exist). How do I solve it then?

First, by expanding the four equations, we note that the system of equations is equivalent to
$$ \begin{bmatrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
1 & 2 & 2 & 4 & 2 & 4 & 4 & 8 & 2 & 4 & 4 & 8 & 4 & 8 & 8 \\
1 & 3 & 3 & 9 & 3 & 9 & 9 & 27 & 3 & 9 & 9 & 27 & 9 & 27 & 27 \\
1 & 4 & 4 & 16 & 4 & 16 & 16 & 64 & 4 & 16 & 16 & 64 & 16 & 64 & 64
\end{bmatrix}
\begin{bmatrix}
abcd \\
abc \\
abd \\
ab \\
acd \\
ac \\
ad \\
a \\
bcd \\
bc \\
bd \\
b \\
cd \\
c \\
d
\end{bmatrix} = 
\begin{bmatrix}
14 \\
29 \\
52 \\
83
\end{bmatrix}.
$$
Using row reduction, we see that this is equivalent to the matrix equation
$$ \begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 1
\end{bmatrix}
\begin{bmatrix}
abcd \\
abc \\
abd \\
ab \\
acd \\
ac \\
ad \\
a \\
bcd \\
bc \\
bd \\
b \\
cd \\
c \\
d
\end{bmatrix} = 
\begin{bmatrix}
7 \\
3 \\
4 \\
0
\end{bmatrix}.
$$
This resolves into the following system of equations:
$$ \begin{aligned}
abcd = 7 \\
abc + abd + acd + bcd = 3 \\
ab + ac + ad + bc + bd + cd = 4 \\
a + b + c + d = 0.
\end{aligned}
$$
Finally, we note that $(a+5)(b+5)(c+5)(d+5)$ is equal to
$$abcd + 5(abc + abd + acd + bcd) + 25(ab + ac + ad + bc + bd + cd) + 125(a + b + c + d) + 625.$$
So $(a+5)(b+5)(c+5)(d+5) = 7 + (5\times3) + (25\times4) + (125\times0) + (625) = 747$.
