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?

According to Maple, there are no real solutions.  There are complex solutions.
They have $a$ as a root of the irreducible quartic $x^4+4 x^2-3 x+7$.
But as for the value of $(a+5)(b+5)(c+5)(d+5)$: note that 
$(a+x)(b+x)(c+x)(d+x)$ is a monic quartic polynomial in $x$. Find a 
monic quartic with values $15,\;45,\;133,\;339$ at $1,2,3,4$.  This part involves linear equations, or you might use a table of differences...
