##p(\sqrt 3+\sqrt 2)=(\sqrt 3)-1## find minimal ##p(x)## polynomial

[littlemathquark]

Homework Statement

$p(\sqrt 3+\sqrt 2)=(\sqrt 3)-1$ find minimal $p(x)$ polynomial

Relevant Equations

$p(\sqrt 3+\sqrt 2)=(\sqrt 3)-1$ find minimal $p(x)$ polynomial with rational coefficient

I have tried $p(x)=ax^3+bx^2+cx+d$ but this is a long way. How can I solve easily that question?

 

[fresh_42]

Homework Statement: $p(\sqrt 3+\sqrt 2)=(\sqrt 3)-1$ find minimal $p(x)$ polynomial
Relevant Equations: $p(\sqrt 3+\sqrt 2)=(\sqrt 3)-1$ find minimal $p(x)$ polynomial with rational coefficient

I have tried $p(x)=ax^3+bx^2+cx+d$ but this is a long way. How can I solve easily that question?

I'm not quite sure what you are looking for, the minimal polynomial of $\sqrt{2}+\sqrt{3}$ or a polynomial of minimal degree that satisfies $p\left(\sqrt{2}+\sqrt{3}\right)=\sqrt{3}-1.$

If it is the former, then you will probably need a polynomial of degree four.
If it is the latter, then a standard method should be to multiply the equation by $\sqrt{3}+1=p\left(\sqrt{2}+\sqrt{3}\right)+2.$ Maybe we can get away with $\deg p(x)=2$ since we can use $\sqrt{3}$ as coefficients.

 

[littlemathquark]

I'm looking for polynomial of minimal degree that satisfies $p\left(\sqrt{2}+\sqrt{3}\right)=\sqrt{3}-1$

 

[littlemathquark]

But $p(x)=\dfrac{-x^3+11x}{2}+1$ satisfies $p(\sqrt{2}+\sqrt{3})=\sqrt{3}-1$

 

[fresh_42]

But $p(x)=\dfrac{-x^3+11x}{2}+1$ satisfies $p(\sqrt{2}+\sqrt{3})=\sqrt{3}-1$

I haven't solved it for you. Now I changed that. Calculate $\left( \sqrt{2}+\sqrt{3} \right)^3$ and express it in terms of $p\left( \sqrt{2}+\sqrt{3} \right)$ and $\left( \sqrt{2}+\sqrt{3} \right).$ That makes seven lines total.

 

[littlemathquark]

I found that solution:
$x=\sqrt2+\sqrt3$
$x^3=9\sqrt3+11\sqrt2$
$11x-x^3=2\sqrt3$
$\sqrt3=\dfrac{11x-x^3}{2}$
Can you share detaile your solution?
Bu the way the correct $p(x)=\dfrac{11x-x^3}{2}-1$ not plus 1,sorry.

 

[fresh_42]

I found that solution:
$x=\sqrt2+\sqrt3$
$x^3$=9\sqrt3+11\sqrt2
$11x-x^3=2\sqrt3$
$\sqrt3=\dfrac{11x-x^3}{2}$
Can you share detaile your solution?

I first calculated the low powers of $\sqrt{3}+\sqrt{2}$ since they are crucial for any approach to this question. It is something you can do without wasting too much time. It is like equipping your toolbox with additional screw drivers. So I got
\begin{align*}
\left( \sqrt{2}+\sqrt{3} \right)^2&= 2 \sqrt{6}+5\\
\left( \sqrt{2}+\sqrt{3} \right)^3&= \left( 2\sqrt{6}+5 \right) \left(\sqrt{2}+\sqrt{3}\right)\\
&=4\sqrt{3}+ 5\sqrt{2}+6\sqrt{2}+5\sqrt{3}\\
&=9\sqrt{3}+11\sqrt{2}\\
&=11\left( \sqrt{2}+\sqrt{3} \right)-2\sqrt{3}\\
&=11\left( \sqrt{2}+\sqrt{3} \right)-2p\left( \sqrt{2}+\sqrt{3} \right)+2
\end{align*}
and re-arranging the last equation yields the desired result
$$
p\left( \sqrt{2}+\sqrt{3} \right)=\dfrac{11 \left( \sqrt{2}+\sqrt{3} \right)-\left( \sqrt{2}+\sqrt{3} \right)^3}{2}+1
$$

 

[fresh_42]

I forgot that we also have to show that $\deg p(x)=3$ is the minimal possible degree.

If $\deg p(x)=1$ then
\begin{align*}
p(x)&=ax+b\\
p\left(\sqrt{2}+\sqrt{3}\right)&=a\sqrt{2}+a\sqrt{3}+ b=\sqrt{3}-1
\end{align*}
and thus $a=0$ contradicting $b\in \mathbb{Q}$ since $\sqrt{2}$ and $\sqrt{3}$ are lineraly independent over $\mathbb{Q}.$

The case $\deg p(x)=2$ is a bit longer but along the same lines: $\{\sqrt{2},\sqrt{3},\sqrt{6}\}$ is $\mathbb{Q}$-linearly independent.