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?

5These equations may be tedious, but they're not linear, assuming all variables are unknowns. – Matt Samuel Feb 09 '15 at 04:31

29"Today class, we'll be covering how to solve systems of tedious equations, for the people in engineering that love tedious calculations." – Dair Feb 09 '15 at 04:33

I see someone else who uses clevermath – Faraz Masroor Feb 10 '15 at 02:31
5 Answers
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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4Oh, might as well. So $f(k)=k^4+4k^2+3k+7$ hence $k(5) = 747$. – Tito Piezas III Feb 09 '15 at 07:06
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 $(a+5)(b+5)(c+5)(d+5) = 7 + (5\times3) + (25\times4) + (125\times0) + (625) = 747$.
According to Maple, there are no real solutions. There are complex solutions. They have $a$ as a root of the irreducible quartic $x^4+4 x^23 x+7$.
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...
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Could you please elaborate a little bit. I am unable to understand what you want me to do? – Feb 09 '15 at 05:08

6@GarvilSinghal The problem is actually very trivial and extremely quickly solvable if you use his suggested 'table of differences'. [See here about the easily provable Method of Differences](https://brilliant.org/discussions/thread/methodofdifferences2/). – user26486 Feb 09 '15 at 17:48

10@GarvilSinghal $$\begin{array}{}n&f(x)& D_1& D_2&D_3&D_4\\1& 15& 30& 58& 60& 24\\2& 45&88&118&84\\3&133& 206&202\\4&339& 408\\5&\boxed{747}\end{array}$$ – user26486 Feb 09 '15 at 17:54

@user314: Can you just confirm with me? The given data allows you to fill in everything strictly above the diagonal. Then you use : dominant coefficient $1$ to put $24$ in the top right corner, and then you go down on the diagonal $24$, $84$, $202$, $\ldots$. I think I understand... – orangeskid Mar 06 '15 at 16:52

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
$$p(5) = 4\, p(4)  6\, p(3) + 4 \,p(2)  p(1) + 24 = \,...$$
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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 (nonnumerical) 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/systemof4tediousnonlinearequationsakbkckdkconstantfor
%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 = 1e4;
for k=1:100
f = f_fct(v);
disp(['k= ',num2str(k),', aa= ',num2str(aa,3),', f= ', num2str(norm(f),3)]);
if (norm(f) < 1e9)
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)
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1Err, just plug the values into the formula $(a+5)(b+5)(c+5)(d+5)$. You get 747. – Nick Alger Feb 10 '15 at 11:57