# 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?

• These equations may be tedious, but they're not linear, assuming all variables are unknowns. Feb 9, 2015 at 4:31
• "Today class, we'll be covering how to solve systems of tedious equations, for the people in engineering that love tedious calculations."
– Dair
Feb 9, 2015 at 4:33
• I see someone else who uses clevermath Feb 10, 2015 at 2:31

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 $f(5)$.

• Oh, might as well. So $f(k)=k^4+4k^2+3k+7$ hence $k(5) = 747$. Feb 9, 2015 at 7:06
• Is this newton girard identities Mar 8 at 18:17

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 $(a+5)(b+5)(c+5)(d+5) = 7 + (5\times3) + (25\times4) + (125\times0) + (625) = 747$.

• Amazing! Plus One!!! Feb 10, 2015 at 6:12

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...

• Could you please elaborate a little bit. I am unable to understand what you want me to do?
– user167045
Feb 9, 2015 at 5:08
• @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. Feb 9, 2015 at 17:48
• @GarvilSinghal $$\begin{array}{|||||||}n&f(x)& D_1& D_2&D_3&D_4\\1& 15& 30& 58& 60& 24\\2& 45&88&118&84\\3&133& 206&202\\4&339& 408\\5&\boxed{747}\end{array}$$ Feb 9, 2015 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... Mar 6, 2015 at 16:52
• @orangeskid yes, that is how you do it. The $24$ there is $4!$. Apr 5, 2015 at 15:08

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

$$p(5) = 4\, p(4) - 6\, p(3) + 4 \,p(2) - p(1) + 24 = \,...$$

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)

• Err, just plug the values into the formula $(a+5)(b+5)(c+5)(d+5)$. You get 747. Feb 10, 2015 at 11:57