Determine the positive numbers a

[anemone]

Determine the positive numbers $a$ such that $$\sqrt[3]{3+\sqrt{a}}+\sqrt[3]{3-\sqrt{a}}$$ is an integer.

 

[kaliprasad]

Determine the positive numbers $a$ such that $$\sqrt[3]{3+\sqrt{a}}+\sqrt[3]{3-\sqrt{a}}$$ is an integer.

(3 + a^(1/2))^(1/3) + (3 – a^(1/2))^(1/3) is integer

Let (3 + a^(1/2))^(1/3) + (3 – a^(1/2))^(1/3) = -c

So (3 + a^(1/2))^(1/3) + (3 – a^(1/2))^(1/3) +c = 0

As x + y +z = 0 => x^3 + y^3 + z^3 = 3xyz

So (3 + a^(1/2))+ (3 – a^(1/2)) +c^3 = 3c(9-a)

Or c^3 + 6 = 3c(9-a)

C cannot be zero

9-a = (c^3 + 6)/3c or a = 9 – (c^3+6)/(3c) …(1)

Because a is positive we can choose c^3 + 6 > 3c if c is negative and c^3 + 6 < 3c if c is positive

So c^3 + 6 > 3c and c < 0 => c = -2 or – 1

c^3 + 6 < 3c and c > 0 no solutionso c = -2 or =-1 from (1) you can get a = 26/3 or 20/3
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[anemone]

so c = -2 or =-1 from (1) you can get a = 26/3 or 20/3
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Hi kaliprasad,

Thanks for participating but I see that when $$a=\frac{26}{3}$$, then $$\sqrt[3]{3+\sqrt{\frac{26}{3}}}+\sqrt[3]{3-\sqrt{\frac{26}{3}}}\ne \text{integer}$$...

And the same applies to the case when $$a=\frac{20}{3}$$.

 

[kaliprasad]

I am sorry. Unable to find the flaw in the logic second solution seems to be 22/3 a calcluation mistake

 

[anemone]

...
So (3 + a^(1/2))^(1/3) + (3 – a^(1/2))^(1/3) +c = 0

As x + y +z = 0 => x^3 + y^3 + z^3 = 3xyz

So (3 + a^(1/2))+ (3 – a^(1/2)) +c^3 = 3c(9-a) I believe this is where it went wrong...

I think you should get $$3 + a^{\frac{1}{2}}+3 – a^{\frac{1}{2}}+c^3 = 3c(9-a)^{\frac{1}{3}} $$, i.e. the factor $(9-a)$ is raised to the one-third power.

 

[Opalg]

Determine the positive numbers $a$ such that $$\sqrt[3]{3+\sqrt{a}}+\sqrt[3]{3-\sqrt{a}}$$ is an integer.

I very much like kaliprasad's idea of using the implication $x + y +z = 0 \ \Longrightarrow \ x^3 + y^3 + z^3 = 3xyz.$ So suppose that $$\sqrt[3]{3+\sqrt{a}}+\sqrt[3]{3-\sqrt{a}} = n$$. Then $$(3+\sqrt{a})+ (3-\sqrt{a}) + (-n)^3 = 3(-n)\sqrt[3]{9-a}.$$ Therefore $\sqrt[3]{9-a} = \dfrac{n^3-6}{3n}$, and $a = 9 - \dfrac{(n^3-6)^3}{27n^3}.$

We are told that $a>0$, so we must have $\dfrac{(n^3-6)^3}{27n^3} < 9$. The function $f(x) = \dfrac{(x-6)^3}{27x}$ is always positive when $x<0$, and has a minimum value $9$ when $x=-3$. When $x>0$, $f(x)$ is an increasing function, and $f(24) = 9$. So the condition $f(n^3) < 9$ implies that $n^3>0$ and $n^3 < 24$. Thus the only integer solutions for $n$ are $n=1$ and $n=2.$

If $n=1$ then $a = 9 - \dfrac{(-5)^3}{27} = \dfrac{368}{27}$. If $n=2$ then $a = 9 - \dfrac1{27} = \dfrac{242}{27}.$ Those are the only two solutions for $a$.

 

[anemone]

I very much like kaliprasad's idea of using the implication $x + y +z = 0 \ \Longrightarrow \ x^3 + y^3 + z^3 = 3xyz.$ So suppose that $$\sqrt[3]{3+\sqrt{a}}+\sqrt[3]{3-\sqrt{a}} = n$$. Then $$(3+\sqrt{a})+ (3-\sqrt{a}) + (-n)^3 = 3(-n)\sqrt[3]{9-a}.$$ Therefore $\sqrt[3]{9-a} = \dfrac{n^3-6}{3n}$, and $a = 9 - \dfrac{(n^3-6)^3}{27n^3}.$

We are told that $a>0$, so we must have $\dfrac{(n^3-6)^3}{27n^3} < 9$. The function $f(x) = \dfrac{(x-6)^3}{27x}$ is always positive when $x<0$, and has a minimum value $9$ when $x=-3$. When $x>0$, $f(x)$ is an increasing function, and $f(24) = 9$. So the condition $f(n^3) < 9$ implies that $n^3>0$ and $n^3 < 24$. Thus the only integer solutions for $n$ are $n=1$ and $n=2.$

If $n=1$ then $a = 9 - \dfrac{(-5)^3}{27} = \dfrac{368}{27}$. If $n=2$ then $a = 9 - \dfrac1{27} = \dfrac{242}{27}.$ Those are the only two solutions for $a$.

Thanks Opalg for participating and yes, the two values of $a$ that you found are correct, of course!(Sun)

My solution:

Let $$y=\sqrt[3]{3+\sqrt{a}}+\sqrt[3]{3-\sqrt{a}}$$ where $y$ is an integer and $a>0$.

Then we have

$$y^3=\left(\sqrt[3]{3+\sqrt{a}}+\sqrt[3]{3-\sqrt{a}}\right)^3$$

$$y^3=\left(3+\sqrt{a}\right)+3(3+\sqrt{a})^{\frac{1}{3}}(3-\sqrt{a})^{\frac{1}{3}}\left(\sqrt[3]{3+\sqrt{a}}+\sqrt[3]{3-\sqrt{a}}\right)+(3-\sqrt{a})$$

$$y^3=\left(3+\sqrt{a}\right)+3y\left((3+\sqrt{a})(3-\sqrt{a})\right)^{\frac{1}{3}}+(3-\sqrt{a})$$

$$y^3=\left(3+\sqrt{a}\right)+3y(9-a)^{\frac{1}{3}}+(3-\sqrt{a})$$

Rearrange the equation a bit and we get

$$y(y^2-3(9-a)^{\frac{1}{3}})=6$$

Since $y$ is an integer, we see that

$$y(y^2-3(9-a)^{\frac{1}{3}})
\stackrel{\text{or}}{=}1(6)\stackrel{\text{or}}{=}2(3)$$

If $y=1$, then $$1^2-3(9-a)^{\frac{1}{3}}=6\;\;\rightarrow a=\frac{368}{27}$$.

If $y=6$, then $$6^2-3(9-a)^{\frac{1}{3}}=6\;\;\rightarrow a=-\frac{42632}{27}$$.

If $y=2$, then $$2^2-3(9-a)^{\frac{1}{3}}=3\;\;\rightarrow a=\frac{242}{27}$$.

If $y=3$, then $$3^2-3(9-a)^{\frac{1}{3}}=2\;\;\rightarrow a=-\frac{100}{27}$$.

Since $a>0$, the only possible values of $a$ in this problem are $$a=\frac{368}{27}$$ and $$a=\frac{242}{27}$$.