Find the possible values of $a$, $b$ and $c$?

Given $(a,\space b,\space c)\in \mathbb Z^3$ and that $$\sqrt[3]{\sqrt{a}+\sqrt{b}} + \sqrt[3]{\sqrt{a}-\sqrt{b}} = c$$ Find the possible values of $a$, $b$, and $c$.

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Without specifying restrictions on $a, b, c$, you could find countless number of values for positive pairs of $a,b$ that results in a value for $c$. For example, (a,b,c) value such as (0,0,0), (1,0,2), ... could be generated. –  Emmad Kareem Oct 21 '12 at 14:08
True, but all the examples you gave have $b = 0$, so it seems possible to classify all the examples. –  only Oct 21 '12 at 14:19
Actually, I found a more troublesome case: $(a,b,c) = ((m^3+3mn)^2,n(3m^2+n)^2,2m)$ –  only Oct 21 '12 at 14:22
@only It would be cool if you show how you found the case. –  user31280 Oct 21 '12 at 14:47
If $u,v$ have the same parity, then $(a,b,c)=(\frac{u^6+2u^3v^3+v^6}4, \frac{u^6-2u^3v^3+v^6}4, u+v)$ is a solution. In fact, for these solutions all intermediate results are integer, which need not be the case for all solutions. –  Hagen von Eitzen Oct 21 '12 at 14:50

We try to find all solutions in integers $a,b,c$ for which the square and cube root are unambiguously defined, that is we require $a\ge0$ and $b\ge0$.

Let $u=\sqrt[3]{\sqrt a+\sqrt b}$, $v=\sqrt[3]{\sqrt a-\sqrt b}$, $w=uv=\sqrt[3]{a-b}$. (Note that $u,v,w$ need not be integers).

Note that $u^3+v^3=2\sqrt a\ge0$ implies $u^3\ge -v^3$ and hence $u\ge-v$ and finally $c\ge0$. The case $c=0$ leads to $u=-v$ and hence $a=0$. Then we find the solutions $$\tag1(0,b,0)\quad\text{with arbitrary }b\ge0.$$ For the rest of the argument we may assume that $c>0$.

Moreover $u^3+v^3=2\sqrt a$ implies $$\tag22\sqrt a = u^3+v^3=(u+v)^3-3(u+v)uv=c^3-3cw,$$ hence by isolating $-3cw$ and cubing $$-c^{27}+6c^9\sqrt a-12c^3a+8a\sqrt a=-27c^3(a-b),$$ i.e. $$\tag33(3c^9+4a)\sqrt a \in\mathbb Z.$$ Since $c>0$ and $a\ge0$, we have $3c^9+4a\ne0$ and conclude that $a=d^2$ is a perfect square with $d\in\mathbb Z$.

Next observe that $$4b=(u^3-v^3)^2=u^6-3u^3v^3+v^6\\=(u+v)^6-6uv(u+v)^4+9(uv)^2(u+v)^2-4(uv)^3 \\=c^6-6c^4w+9c^2w^2-4(a-b).$$ Thus $w$ is root of a quadratic and of a cubic rational polynomial, hence is rational, i.e. $a-b$ is a perfect cube, say $a=b+e^3$ with $e\in \mathbb Z$. With this, $(2)$ becomes $$\tag42d=c^3-3ce.$$

Note that $u,v$ are roots of $$x^2-cx+e = x^2-(u+v)x+uv= 0,$$ i.e. $$\tag5u=\frac{c+\sqrt{c^2-4e}}2\quad v=\frac{c-\sqrt{c^2-4e}}2$$ and we require $c^2\ge4e$.

Now let us go backwards: Select integers $c>0$ and $e\le\frac{c^2}4$ such that $c\equiv0\pmod 2$ or $e\equiv 1\pmod 2$. Then $a:=\frac{c^2(c^2-3e)^2}4$ is a nonnegative integer. Set $b:=a-e^3$. Then $b\ge0$ because either $e\le 0$ and then $b\ge a\ge0$; or $e>0$ and then $c^2-3e\ge e>0$, i.e. $b = \frac{c^2(c^2-3e)^2-4e^3}4\ge \frac{c^2e^2-4e^3}4=\frac{(c^2-4e)e^2}4\ge0$. With these values, $(a,b,c)$ is a solution. With nice parametrizations depending on the parity of $c$ we thus find for even $c=2m$ and $e=m^2-n$ ($m>0$ ,$n\ge0$): $$\tag6\begin{matrix} a&=&m^2(m^2+3n)^2,\\ b&=&m^2(m^2+3n)^2-(m^2-n)^3=n(3m^2+n)^2,\\ c&=&2m.\end{matrix}$$ And for odd $c$ and odd $e$ ($c=2m+1$, $e=m(m+1)-2n-1$ with $m,n\ge0$): $$\tag7\begin{matrix}a&=&(2m+1)^2\left(\frac{m(m+1)}2+2+3n\right)^2,\\ b&=&(2m+1)^2\left(\frac{m(m+1)}2+2+3n\right)^2-\left(m(m+1)-2n-1\right)^3,\\ c&=&2m+1.\end{matrix}$$

The solutions given by $(1),(6),(7)$ are complete.

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solution 6 is giving me a complex result according to wolframalpha here –  user31280 Oct 24 '12 at 10:13
@F'OlaYinka: Oops, my last edit has introduced two typos in the (simplified) formula for $b$ (cf. comment by only above, which corresponds to (6)). Now I get with $m=5, n=3$: $a=28900$, $b=18252$, $c=10$ and compute $\approx\sqrt[3]{305.1}+\sqrt[3]{34.9}\approx 6.732+2.268=10$. –  Hagen von Eitzen Oct 25 '12 at 13:13

There is a property that says if $$a +b+c=0,$$ then $$a^3 +b^3+c^3 =3abc$$ so since $$\sqrt[3]{\sqrt{a}+\sqrt{b}} + \sqrt[3]{\sqrt{a}-\sqrt{b}} +(-c) =0,$$ then $$({\sqrt{a}+\sqrt{b}}) + ({\sqrt{a}-\sqrt{b}}) +(-c) ^3= 3(\sqrt[3]{(\sqrt{a}+\sqrt{b})({\sqrt{a}-\sqrt{b}})} .(-c)$$ which implies that; $$c^3-3\sqrt[3]{a-b}.c - 2\sqrt a =0$$ Now one can solve for $c$ following the values of $a$ and $b$

Let's we know that $a$, $b$ and $c$ are integers so we substitute as follows $p= -3\sqrt[3]{a-b}$ and $q= -2\sqrt a$ which turns our equation into $$c^3 + pc+q=0$$. We just have to find the roots of this polynomial and then analyze it. The discriminant of this equation is: $$\Delta = -4p^3 - 27q^2 = -4(-27(a-b))-27(4a) = -108b$$. Since we only need one real integer value of $c$ then $b\ge 0$.

• $(0,\space 0,\space 0)$ is an obvious solution.

• If $b=0$ then $c=2a^{1/6}$ which gives us the triplet $(k^6,\space 0,\space 2|k|), \forall k\in \mathbb Z$

• if $b>0$, then we get only one real root and two imaginary roots. Without wasting time, we can as well write that: $$c=\left ({-{q\over 2}+ \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}}\right )^{1/3} +\left ({-{q\over 2}- \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}}\right )^{1/3}$$ $$\sqrt a = \cfrac {-q}{2} \quad \&\ \quad \sqrt b = \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}$$ Since $a$ is an integer then $$q = 2k_1 \quad \&\ \quad a = k_1^2,\quad k_1\in \mathbb Z$$ and then $\displaystyle b = {q^{2}\over 4}+{p^{3}\over 27} = k_1^2+{p^3 \over 27}.\space$ Since $b$ and $k_1$ are both integers then $$27|p^3 \quad \Rightarrow \quad 3|p \quad \Rightarrow \quad p=3k_2 \quad \&\ \quad b = k_1^2 + k_2^3\quad k_2\in \mathbb Z$$ then $$\sqrt[3]{k_1+\sqrt{k_1^2 + k_2^3}} + \sqrt[3]{k_1-\sqrt{k_1^2 + k_2^3}} = c$$ $a-b$ is a perfect cube and it equals to $-k_2^3$. Using the Rational Root theorem, if $c$ is rational, then $c=\cfrac {k_3}{k_4}$ where $k_3$ is a factor of $q$ and $k_4$ is a factor of the coefficient of $c^3$. In our case $c$ is an integer so $q=k_3k_5, \space k_5 \in \mathbb Z$ and $k_4 = 1$ which implies $c = k_3 = \cfrac q{k_5} = \cfrac {2k_1}{k_5}, \space k_5 \in \mathbb Z$ and $k_1 = \cfrac {k_5k_3}{2}$. We have $$k_3^3 + 3k_2k_3 +k_5k_3 = 0$$
• If $c=k_3=0$, $k_1 = 0$, $a = k_1^2 = 0$ and $b>0$ giving another triplet $(0,\space k,\space 0), \forall k\in \mathbb N \cup\{0\}$
• Else if $k_3^2 + 3k_2 +k_5 = 0$ then working on the parity of $c$
• If $c = 2m$, $m\in \mathbb Z$, then $k_3 = 2m$ and we have $$4m^2 + 3k_2 + k_5 = 0 \quad \&\ \quad k_1 = mk_5$$which implies $$k_1 = -m(4m^2+3k_2)$$ Since $b>0$, then $b =a+k_2^3 =m^2(4m^2+3k_2)^2 + k_2^3>0$ implies $(k_2/4+m^2)^2 (k_2+m^2)>0$ which gives us the following cases $$\left\{ \begin{array}{l l} k_2>0, \space m>0\\ k_2\le 0,\space m>\sqrt{-k_2}\\ k_2\le 0, \space m<-\sqrt{-k_2} \\ \end{array} \right.$$ For any couple $(m,k_2) \in \mathbb Z^2$ that satisfy either of the above conditions then we have the following triplet $(a, \space b,\space c)$ $$\boxed { \left\{ \begin{array}{l l} a=k_1^2=m^2(4m^2+3k_2)^2 \\ b = m^2(4m^2+3k_2)^2 + k_2^3 \\ c= 2m\\ \end{array} \right.}$$
• If $c = 2m+1$, $m\in \mathbb Z$, then $k_3 = 2m$ and we have $$(2m+1)^2+ 3k_2 + k_5 = 0\quad \&\ \quad k_5 = 2k_6 \quad \&\ \quad k_1 = k_6(2m+1)$$which implies $$k_1 = -\cfrac{(2m+1)((2m+1)^2+3k_2)}{2}$$ At this point, we conclude that $k_2 = 2n+1, \space \forall n \in \mathbb Z$ since $k_2$ has got to be odd to give an integer value of $k_1$ and $$k_1 = -\cfrac{(2m+1)[(2m+1)^2+3(2n+1)]}{2}$$ Since $b>0$, then $b =a+k_2^3 =\cfrac{(2m+1)^2[(2m+1)^2+3(2n+1)]^2}{4} + (2n+1)^3>0$ $(2 m^2+2 m+n+1)^2 (4 m^2+4 m+8 n+5)>0\quad \Rightarrow \quad (4 m^2+4 m+8 n+5)>0$ which leaves us with only one condition $$n > -\cfrac 58 - \cfrac m2 - \cfrac {m^2}2$$ For any couple $(m,n) \in \mathbb Z^2$ that satisfy the above condition then we have the following triplet $(a, \space b,\space c)$ $$\boxed { \left\{ \begin{array}{l l} a=k_1^2=\cfrac{(2m+1)^2[(2m+1)^2+3(2n+1)]^2}{4} \\ b =\cfrac{(2m+1)^2[(2m+1)^2+3(2n+1)]^2}{4} + (2n+1)^3\\ c= 2m+1\\ \end{array} \right.}$$

$c$ is always positive but I can't seem to find where that property lies in my solution or maybe that is just a calculator problem.

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Start: $(\sqrt{a}+\sqrt{b})^{1/3}+(\sqrt{a}-\sqrt{b})^{1/3}=c$

Cube the equation:
$(\sqrt{a}+\sqrt{b})+3((a-b)(\sqrt{a}+\sqrt{b}))^{1/3}+3((a-b)(\sqrt{a}-\sqrt{b})^{1/3}+(\sqrt{a}-\sqrt{b})=c^3$

Simplify:
$2\sqrt{a}+3(a-b)^{1/3}((\sqrt{a}+\sqrt{b})^{1/3}+(\sqrt{a}-\sqrt{b})^{1/3})=c^3$

Notice that the cuberoot terms can be substituted back to $c$:
$2\sqrt{a}+3(a-b)^{1/3}c=c^3$

Rearranging (treat $c$ as the indeterminate):
$c^3-3(a-b)^{1/3}c-2\sqrt{a}=0$

Since you require $c\in\mathbb{Z}$, hence $a$ must be a square.
Let $a=k^2$:
$f(x)=c^3-3(k^2-b)^{1/3}c-2k=0$

From here, you can refer to cube root formula to get all the root.
Then with the restrictions of $a,b,k\in\mathbb{Z}$ you should get the solutions.

As you can see, the goal is to get a polynomial of 1 variable first.
The reason is once this is done, you can base the information that $c\in\mathbb{Z}$ to get several deductions.
For example, you can narrow down the possibilities of the roots using rational root theorem.

You might also choose to "guess" a solution of $f(x)=(c-x_1)(c-x_2)(c-x_3)=0$ and try to compare the coefficients, which must be equal.
A question that you might want to ask is: Must $x_1,x_2,x_3\in\mathbb{Z}$?
Or can two of them be complex numbers?

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since the value of $c$ is unique for any couple $(a,b)$ then I think $c$ can only have one correct value i.e two of $x_i$ can be complex. –  user31280 Oct 22 '12 at 17:10
Sorry for the late reply, was busy. Wanted to point you towards the fact that discriminant can help to differentiate the cases. Seems like it is applicable, as shown in your solution. –  Yong Hao Ng Oct 25 '12 at 18:51