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

How to find all integer numbers $a$, $b$, $c$, $d$, $e$, $f$, $k$ that belong to interval $[-10,10]$, so that the equation $$\sqrt{a x + b} +\sqrt{c x + d} + \sqrt{e x + f} =k$$ 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 $$(a+c+e)x + b + d + f = 3\sqrt{(ax + b)(cx+d)(ex+f)}$$ 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

• $\sqrt{-2 x + 8}+ \sqrt{x + 2} + \sqrt{x - 10}=0$.
• $\sqrt{-2 x + 4}+ \sqrt{x + 6} + \sqrt{x - 10}=0$.
• $\sqrt{-3 x + 9}+ \sqrt{x + 1} + \sqrt{2x - 10}=0$.
• $\sqrt{-6 x + 6}+ \sqrt{4 x + 4} + \sqrt{2x - 10}=0$.
• Have you tried FindInstance? If so, please place the code so we can see. – bill s May 17 '13 at 6:22
• Here's an answer: a=b=c=d=e=f=g=0. – bill s May 17 '13 at 6:27
• You need to simplify this problem. Think that you have 7 variables, each of which can take 41 different values. This means that if you go the brute force way you will have $41^7 \sim 2 \times 10^11$ different combinations to examine. And you want to find all integers that satisfy this equation. Also, note that in your code you seem to be happy with the values of variables being in the interval $[-10,10]$. – SMeznaric May 17 '13 at 13:54
• I think one has to look at a small number of combinations. If r1=a x + b and similar, then this gives the list of {r1, r2, r2} satisfying the equation : rlist = {Reduce[{r1^(1/3) + r2^(1/3) + r3^(1/3) == #, 0 <= r1 <= 10, 0 <= r2 <= 10, 0 <= r3 <= 10}, {r1, r2, r3}, Integers] /. {r_ == rsol_ -> rsol, And -> List, Or -> List}, #} & /@ Range[0, 20]. The last step is to resolve each of {r1, r2, r3} into triplets of integers with Reduce once more. – b.gates.you.know.what May 17 '13 at 14:23
• There are some ambiguities in your question. 1. $a,b,c,d,e,f,g$ belong to $[−20,20]$ or to $[−10,10]$ ? 2. What about $k$ and $x$ ? – Artes May 17 '13 at 17:54

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

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: $$\text{Simplify}\left[\frac{\partial \left(\sqrt{a x+b}+\sqrt{c x+d}+\sqrt{e x+f}\right)}{\partial x}\right] = \frac{1}{3} \left(\frac{a}{(a x+b)^{2/3}}+\frac{c}{(c x+d)^{2/3}}+\frac{e}{(e x+f)^{2/3}}\right) = 0$$

It can be verified that this will happen when $$\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=1 \And \sqrt{a}+\sqrt{c}+\sqrt{e}=0 \Rightarrow k=0$$ 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.

• Your argument doesn't show $a=c=e=0$ and $a=b,c=d,e=f$ are the only solutions for the equation to be an identity. For example, I could have satisfied your derivative equation with $a=n*b, c=n*d, e=n*f$ for any integer $n$ and $\sqrt{b}+\sqrt{d}+\sqrt{f} = 0$. In reality, $a=c=e=0$ is the only solution, as I show in my answer. – obsolesced Aug 1 '16 at 7:49

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[] ({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[]) 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 ;-)

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 $\sqrt{y(i)} = \sqrt{-1}\sqrt{-y(i)}$, 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 $\sqrt{y(i)}$ implies continuity of $k$. Also, $\sqrt{y(i)}$ 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.

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.