Find all the integer numbers $a$, $b$, $c$, $d$, $e$, $f$, $k$ to this equation have three integer different solutions?

Question

How to find all integer numbers $a$, $b$, $c$, $d$, $e$, $f$, $k$ that belong to interval $[-10,10]$, so that the equation $$\sqrt[3]{a x + b} +\sqrt[3]{c x + d} + \sqrt[3]{e x + f} =k$$ has three different integer solutions? I tried

Reduce[{Root[#^3 - (a x + b) &, 1] + Root[#^3 - (c x + d) &, 1] + 
    Root[#^3 - (e x + f) &, 1] == k, -10 <= a <= 10, -10 <= b <= 
   10, -10 <= c <= 10, -10 <= d <= 10, -10 <= e <= 10, -10 <= f <= 
   10, -10 <= k <= 10}, {a, b, c, d, e, f, k, x}, Integers]

Update

With $k=0$, we can transform the given equation of the form $$(a+c+e)x + b + d + f = 3\sqrt[3]{(ax + b)(cx+d)(ex+f)}$$ Cubing both sides of the equation:

ClearAll;
f[x_] := (a + c + e) x + b + d + f;
g[x_] := 3 ((a x + b) (c x + d) (e x + f))^(1/3);
Collect[f[x]^3 - g[x]^3, x];
Discriminant[%, x]

Next, we solve for Discriminant > 0 But I can not get the answer.

This is some equations

• $\sqrt[3]{-2 x + 8}+ \sqrt[3]{x + 2} + \sqrt[3]{x - 10}=0$.
• $\sqrt[3]{-2 x + 4}+ \sqrt[3]{x + 6} + \sqrt[3]{x - 10}=0$.
• $\sqrt[3]{-3 x + 9}+ \sqrt[3]{x + 1} + \sqrt[3]{2x - 10}=0$.
• $\sqrt[3]{-6 x + 6}+ \sqrt[3]{4 x + 4} + \sqrt[3]{2x - 10}=0$.

Answer 1

I presume what you have written, is an algebraic identity. Otherwise it just doesn't make sense to me. That being the case, I can show that either k or {a,c,e} must be 0.

Since $$\sqrt[3]{a x + b} + \sqrt[3]{c x + d} + \sqrt[3]{e x + f} = k$$ is an identity, it should be valid for all values of x, including $$ x = 0 $$ which will give: $$ \sqrt[3]{b} + \sqrt[3]{d} + \sqrt[3]{f} = k $$

On the other hand we can take derivatives from both sides of an algebraic identity with respect to its variables, which happens to be x in this case. Taking the derivative: $$\text{Simplify}\left[\frac{\partial \left(\sqrt[3]{a x+b}+\sqrt[3]{c x+d}+\sqrt[3]{e x+f}\right)}{\partial x}\right] = \frac{1}{3} \left(\frac{a}{(a x+b)^{2/3}}+\frac{c}{(c x+d)^{2/3}}+\frac{e}{(e x+f)^{2/3}}\right) = 0$$

It can be verified that this will happen when $$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=1 \And \sqrt[3]{a}+\sqrt[3]{c}+\sqrt[3]{e}=0 \Rightarrow k=0 $$ OR $$ a = c = d =0.$$ You can verify this by using Mathematica:

Simplify[Factor[ (a/(b + a x)^(2/3) + c/(d + c x)^(2/3) + e/(f + e x)^(2/3))], Assumptions -> b^(1/3) + d^(1/3) + f^(1/3) == 0 && a == b && c == d && e == f] // Factor

I noticed this myself by expanding the function around x=0:

Series[1/3 (a/(b + a x)^(2/3) + c/(d + c x)^(2/3) + e/(f + e x)^(2/3)), {x, 0, 10}] // Simplify

Anyway, knowing that either $$ a=c=e=0$$ or $$a=b,c=d,e=f$$, it's pretty simple and straightforward to find all the integer solutions of a,b,c,d,e,f,k.

Answer 2

This is not a final solution, but I hope it is an useful approach: First, define a function which depends on your variable x (for me x is a real number) and your parameters a,b,c,d,e,f (integers between -10 and 10):

k[x_,a_,b_,c_,d_,e_,f_]:=(a*x+b)^(1/3)+(c*x+d)^(1/3)+(e*x+f)^(1/3)

You could use a graphical solution. E.g. use

Manipulate[Plot[k[x,a,b,c,d,e,f],{x,-10,10},PlotRange -> {All,All}, 
AxesLabel -> {"x", "k"}],{a,-10,10,1},{b,-10,10,1},{c,-10,10,1},{d,-10,10,1},
{e,-10,10,1},{f,-10,10,1}]

As you can see, depending on x there are several integer values for k.

If you want to find for which value of x k is an integer, you can try the following (for simplicity I fix here arbitrarily c, d, e, f to 10 and k to 8):

bigMatrix=Quiet[Table[{a,b,If[#∈Reals,#,no]&[FindRoot[k[x,a,b,10,10,10,10]==8
(*adapt it to integers between-10 and 10*),{x,1}][[1,2]]]},
{a,-10,10,1},{b,-10,10,1}]];

With Quiet I suppress some warnings (be careful, due to the suppression there are some wrong solutions produced).

With

solutions = Flatten[Delete[bigMatrix, Position[bigMatrix, no][[All, {1, 2}]]], 1];

I delete all solutions where x is not a real number ("no").

And with

output = Table[k[solutions[[i, 3]], solutions[[i, 1]], solutions[[i, 2]], 10, 10, 
10, 10], {i, 1, Length[solutions]}];
solutions2 = Delete[solutions, Position[output, val_ /; val != 8]];

I delete all wrong solutions.

Let us take a look at one of the solutions, e.g. solutions2[[158]] ({10, -7, 1.59669}).

If you set in this solution in your function with k[solutions2[[158,3]],solutions2[[158,1]],solutions2[[158,2]],10,10,10,10], you can see that the output is indeed 8 again.

You can find a lot of solution by changing the 8 in the FindRoot by some of the other integers. And you can find even more solutions if you do not fix as many parameters as I do – but then calculation times increases much (which can perhaps be improved by some better code).

Appendix 1:

If you are interested in a solution for fixed values of x (x ≠0), you could do the following:

Calculate all values of the function of k for a fixed x (e.g. x=1)

start=-4;
stop=4;
bigMatrixB=ParallelTable[{a,b,c,d,e,f,k[1,a,b,c,d,e,f]},{a,start,stop,1},
{b,start,stop,1},{c,start,stop,1},{d,start,stop,1},{e,start,stop,1},{f,start,stop,1}];

As start and stop value I have chosen -4 and 4 to avoid long calculation times.

Then filter for integer solutions with:

solutionsB=Flatten[bigMatrixB,5];
solutionsB2=Delete[solutionsB,Partition[Union[Position[solutionsB[[All,7]],
val_/;val∉Integers][[All,1]]],1]];

In this special case (a,b,c,d,e and f are going from -4 to 4 and x = 1) you can find 5832 solutions! Among them are solutions such as a=-2, b=3, c=4, d=-3, e=-3, f=4, k=3 (output of solutionsB2[[1587]]) which shows you that Alis answer is not really correct (neither a=c=e=0 or a=b, c=d, e=f is here fulfilled).

Appendix 2:

For the special case that x=0 your equation simplifies and also the function k. In this case we have the more simple function k2 given by:

k2[b_, d_, f_] := b^(1/3) + d^(1/3) + f^(1/3);

It does not cost much calculation time to calculate all possible values of k2 with

start=-10;
stop=10;
bigMatrixC=ParallelTable[{b,d,f,k2[b,d,f]},{b,start,stop,1},{d,start,stop,1},
{f,start,stop,1}];

Filter again for all integer solution with:

solutionsC=Flatten[bigMatrixC,2];
solutionsC2=Delete[solutionsC,Partition[Union[Position[solutionsC[[All,4]],
val_/;val∉Integers][[All,1]]],1]];

And you will find 27 solutions for b,d,f and k which are:

{{0,0,0,0},{0,0,1,1},{0,0,8,2},{0,1,0,1},{0,1,1,2},{0,1,8,3},{0,8,0,2},{0,8,1,3},{0,8,8,4},{1,0,0,1},{1,0,1,2},{1,0,8,3},{1,1,0,2},{1,1,1,3},{1,1,8,4},{1,8,0,3},{1,8,1,4},{1,8,8,5},{8,0,0,2},{8,0,1,3},{8,0,8,4},{8,1,0,3},{8,1,1,4},{8,1,8,5},{8,8,0,4},{8,8,1,5},{8,8,8,6}}

So, both appendices give you in total 5859 integer solutions to your problem. Tell us, if you need more ;-)

Answer 3

There are no solutions. Rewrite your equation as

y[i_] = a[i] x + b[i];
k = Inactive[Sum][y[i]^(1/3), {i, 1, 3}];

For real $k$, $y(i)$ cannot be negative, because $\sqrt[3]{y(i)} = \sqrt[3]{-1}\sqrt[3]{-y(i)}$, which, for negative $y(i)$, is a constant complex number multiplied by a positive function increasing monotonically as $y(i)$ decreases, so no cancellation of imaginary terms in $k$ is possible. Positive $y(i)$ requires

Reduce[{y[i] >= 0, a[i]==0},x,Reals]||
Reduce[{y[i] >= 0, a[i]<0},x,Reals]||
Reduce[{y[i] >= 0, a[i]>0},x,Reals]

$$ (b(i)\geq 0\land a(i)=0)\lor \left(a(i)<0\land x\leq -\frac{b(i)}{a(i)}\right)\lor \left(a(i)>0\land x\geq -\frac{b(i)}{a(i)}\right) $$

Therefore,

$$ \max \left(\left\{\left.-\frac{b(i)}{a(i)}\right|a(i)>0\right\}\right)\leq x\leq \min \left(\left\{\left.-\frac{b(i)}{a(i)}\right|a(i)<0\right\}\right) $$

In this domain, continuity of $\sqrt[3]{y(i)}$ implies continuity of $k$. Also, $\sqrt[3]{y(i)}$ is positive and monotonically increasing with $y(i)$.

If all $a(i)=0$, any $x$ gives the same $k$, which thus has infinitely many $x$ solutions.

Otherwise, if all $a(i)$ have the same sign, $k$ is monotonic in $x$, so each $k$ has no more than 1 $x$ solution.

Otherwise, for some $k$ to have 3 $x$ solutions, continuity requires

D[k, x] /. y[i] -> Defer[y[i]]

$$ \underset{i=1}{\overset{3}{\sum }}\frac{a(i)}{3 y(i)^{2/3}} $$

to change signs twice. Since $\frac{\partial k}{\partial x}$ is continuous for $y(i)\geq0$, changing signs twice requires $\frac{\partial k}{\partial x}=0$ twice. However, $\frac{\partial k}{\partial x}$ can be decomposed into the sum of 2 positive functions, 1 with positive $a(i)$ increasing monotonically with $x$, and the other with negative $a(i)$ decreasing monotonically with $x$. These functions can have no more than 1 intersection, so $\frac{\partial k}{\partial x}=0$ at most once. Hence, $k$ can have no more than 2 $x$ solutions.

Further Comments

1. Since $k$ has no more than 2 $x$ solutions unless all $a(i)=0$, the equation can only be an algebraic identity if all $a(i)=0$.

2. There are integer $k, a(i), b(i)$ with 2 $x$ solutions, e.g.

   Solve[1 == x^(1/3) + (1 - x)^(1/3), x]

   $$ \{\{x\to 0\},\{x\to 1\}\} $$

One question might be to find what $k, a(i), b(i)$ have 1 or 2 $x$ solutions, but I won't work on this question unless it's of interest to anyone.