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?
 A: 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 
$$p(5) = 4\, p(4) - 6\, p(3) + 4 \,p(2) - p(1) + 24 = \,...$$
A: 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 $f(5)$.
A: 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 $(a+5)(b+5)(c+5)(d+5) = 7 + (5\times3) + (25\times4) + (125\times0) + (625) = 747$.
A: I found the solution numerically using a variant of Newton's method. It is,
\begin{align}
a &= 0.6625 + 1.1432i \\
b &=  -0.6625 + 1.8897i \\
c &=   0.6625 - 1.1432i \\
d &=  -0.6625 - 1.8897i,
\end{align}
which can be verified by pluging into the original equations.
The desired product is,
$$(a+5)(b+5)(c+5)(d+5) = 747.00$$
and indeed the zeros after the decimal point continue out to machine precision.
Perhaps these numerical results will be of use in finding an exact (non-numerical) solution through more analytic means.

For completeness, the Matlab script I used to get this answer is included below. It's worth noting that the initial guess must be complex in order to get convergence.
%Solves system of nonlinear equations found here:
%http://math.stackexchange.com/questions/1140178/system-of-4-tedious-nonlinear-equations-akbkckdk-constant-for
%using Newton's method

g = [15; 45; 133; 339];
f_fct = @(v) [prod(v+1); prod(v+2); prod(v+3); prod(v+4)] - g; %want f(v)=0

%Jacobian matrix:
J_fct = @(v) ...
    [prod(v([2,3,4])+1), prod(v([1,3,4])+1), prod(v([1,2,4])+1), prod(v([1,2,3])+1); ...
    prod(v([2,3,4])+2), prod(v([1,3,4])+2), prod(v([1,2,4])+2), prod(v([1,2,3])+2); ...
    prod(v([2,3,4])+3), prod(v([1,3,4])+3), prod(v([1,2,4])+3), prod(v([1,2,3])+3); ...
    prod(v([2,3,4])+4), prod(v([1,3,4])+4), prod(v([1,2,4])+4), prod(v([1,2,3])+4)];

v0 = 1i*[1;2;3;4]; %initial guess must be complex for convergence
v = v0;
aa = 0;
c1 = 1e-4;
for k=1:100
    f = f_fct(v);
    disp(['k= ',num2str(k),', aa= ',num2str(aa,3),', ||f||= ', num2str(norm(f),3)]);
    if (norm(f) < 1e-9)
        break;
    end
    J = J_fct(v);
    p = -J\f; %Newton search direction

    %Find step size 'aa' satisfying first Armijo linesearch condition
    aa = 1;
    if (k < 1)
        aa = 0.01;
    else
        for jj=0:5
            aa = 2^(-jj);
            armijo_lhs = norm(f_fct(v + aa*p));
            armijo_rhs = norm(f_fct(v)) + c1*aa*p'*J_fct(v)'*f_fct(v);
            if (armijo_lhs < armijo_rhs)
                break;
            end
        end
    end
    v = v + aa*p;
end
a = v(1)
b = v(2)
c = v(3)
d = v(4)
answer = prod(v+5)

A: 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...
