จำนวนเต็ม

[Thamma]

พิสูจน์ว่าเป็นจำนวนเต็มหรือไม่

$ \left( 2\root3\of{2} +1 - \root\of{12\root3\of{2} - 15}\right)^3 $

[Anton]

Problem. Justify whether the expression
$$\bigg(2\sqrt[3]{2}+1-\sqrt{12\sqrt[3]{2}-15}\bigg)^3$$
is an integer.

Let $t:=\sqrt[3]{2}$. We note that $t^3=2$, so
$$12\sqrt[3]{2}-15=12t-15=4t^4+4t-15\,.$$
We make an Ansatz that
$$4t^4+4t-15=(2t^2+at+b)^2$$
for some rational numbers $a$ and $b$. Thus, the coefficient of $t^2$ in $(2t^2+at+b)^2$, after reduction by the relation $t^3=2$, should become $0$. That is,
$$a^2+4b=0\,.$$
The constant term of $(2t^2+at+b)^2$, after reduction by the relation $t^3=2$, should become $-15$. Therefore,
$$8a+b^2=-15\,.$$
From the relations above, we have
$$(b^2+15)^2=(-8a)^2=8^2a^2=8^2(-4b)=-256b\,.$$
Thus,
$$b^4+30b^2+256b+225=0\,.$$
This means
$$(b+1)(b^3-b^2+31b+225)=0\,.$$
We try $b=-1$, which means $a=-\dfrac{b^2+15}{8}=-2$. Then, we check whether $12t-15=(2t^2+at+b)^2=(2t^2-2t+1)^2$. Now,
$$(2t^2-2t-1)^2=4t^4+4t^2+1-8t^3-4t^2+4t=4(2t)+4t^2+1-8(2)-4t^2+4t=12t-15\,.$$
That is,
$$\sqrt{12\sqrt[3]{2}-15}=\sqrt{12t-15}=\sqrt{(2t^2-2t-1)^2}=|2t^2-2t-1|=1+2t-2t^2=1+2\sqrt[3]{2}-2\sqrt[3]{4}\,.$$
Therefore,
$$\left(2\sqrt[3]{2}+1-\sqrt{12\sqrt[3]{2}-15}\right)^3=\Big(1+2\sqrt[3]{2}-\big(1+2\sqrt[3]{2}-2\sqrt[3]{4}\big)\Big)^3=(2\sqrt[3]{4})^3=2^3\cdot 4=32$$
is an integer.