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?

Hint: 
Use finite differences  as was suggested earlier by @user314 and Robert Israel. They behave similar to derivatives. 
With $\Delta [p](x) \colon = p(x+1) - p(x)$, we have for $p(x) = a_n x^n + \cdots $ a polynomial of degree $n$,
$$\Delta^n [p(x)] \equiv n ! \cdot a_n$$
 Therefore, for our monic polynomial $p(x)= (x+a)(x+b)(x+c)(x+d)$ of degree $4$ we have 
$$\Delta^4 [p] (x)=  p(x+4) - \binom{4}{1}p(x+3) + \binom{4}{2} p(x+2) - \binom{4}{1} p(x+1) + p(x) \equiv 4! = 24$$
and so for $x=1$ we get 
$$p(5) = 4\, p(4) - 6\, p(3) + 4 \,p(2) - p(1) + 24 = \,...$$
