<s>[INST] 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?
 [/INST] Note that 
\begin{align}
f(k)&=(a+k)(b+k)(c+k)(d+k) \\
&=k^4+(a+b+c+d)k^3+(a b+a c+a d+b c+b d+c d) k^2
+(a b c+a b d+a c d+b c d) k+a b cd \\
&=k^4 + k^3 e_1 + k^2 e_2 + k e_3 + e_4.  
\end{align}
Now you have a linear system of 4 equations and 4 unknowns ($e_1,e_2,e_3,e_4$).
Solve it and find $f(5)$.
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