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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. – NoChance 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
up vote 8 down vote accepted

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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