Let $(\sqrt{3} + \sqrt{2})^5 = a\sqrt{3} + b\sqrt{2}, a,b \in \mathbb Z$ Find $a+b$.

Question

Let

$$(\sqrt{3} + \sqrt{2})^{\color{red}{5}} = a\sqrt{3} + b\sqrt{2}, a,b \in \mathbb Z$$

Find $a+b$.

I don't know if that's supposed to be $\color{red}{5}$ or $\color{red}{3}$.

By binomial theorem, we have

$$(\sqrt{3} + \sqrt{2})^{\color{red}{5}} = \sum_{k=0}^{5} \binom{5}{k} (\sqrt{3})^k (\sqrt{2})^{5-k} $$

Also,

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

That's all I got. Idk what to do. Please suggest

Answer 1

The minimal polynomial of $b=\sqrt{3}+\sqrt{2}$ over the rationals can be computed by squaring $b-\sqrt{2}=\sqrt{3}$, getting $b^2-2b\sqrt{2}+2=3$, so $2b\sqrt{2}=b^2-1$ and, squaring again, $b^4-2b^2+1=8b^2$. Thus the minimal polynomial is $X^4-10X^2+1$ (it is known that $b$ has degree $4$).

This implies that $b^5=10b^3-b=b(10b^2-1)$. Thus \begin{align} (\sqrt{3}+\sqrt{2})^5 &=(\sqrt{3}+\sqrt{2})(10(3+2\sqrt{6}+2)-1)\\ &=(\sqrt{3}+\sqrt{2})(49+20\sqrt{6})\\ &=49\sqrt{3}+60\sqrt{2}+49\sqrt{2}+40\sqrt{3}\\ &=89\sqrt{3}+109\sqrt{2} \end{align}

On the other hand, the binomial theorem gives \begin{align} (\sqrt{3}+\sqrt{2})^5 &=\binom{5}{0}(\sqrt{3})^5 +\binom{5}{1}(\sqrt{3})^4\,\sqrt{2} +\binom{5}{2}(\sqrt{3})^3\,(\sqrt{2})^2\\ &\qquad+\binom{5}{3}(\sqrt{3})^2\,(\sqrt{2})^3 +\binom{5}{4}\sqrt{3}\,(\sqrt{2})^4 +\binom{5}{5}(\sqrt{2})^5\\ &=9\sqrt{3}+45\sqrt{2}+60\sqrt{3}+60\sqrt{2}+20\sqrt{3}+4\sqrt{2}\\ &=89\sqrt{3}+109\sqrt{2} \end{align}

Take your pick.

Answer 2

$(\sqrt{3} + \sqrt{2})^3=\sqrt{3}^3+3\sqrt{3}^2\sqrt{2}+3\sqrt{3}\sqrt{2}^2+\sqrt{2}^3=3\sqrt{3}+9\sqrt{2}++6\sqrt{3}+2\sqrt{2}=11\sqrt{2}+9\sqrt{3}$ so $a=11, b=9$.

Also note $(a\sqrt{2}+b\sqrt{3})(5+2\sqrt{2}\sqrt{3})=(5a+6b)\sqrt{2}+(5b+4a)\sqrt3$

Answer 3

Solution without Binomial Theorem

let $a_n, b_n, c_n, d_n$ be integers, and $(\sqrt3+\sqrt2)^n=a_n\sqrt3+b_n\sqrt2+c_n\sqrt6+d_n$

By multiplying $(\sqrt3+\sqrt2)^2=5+2\sqrt6$, we get $a_{n+2}=5a_n+4b_n, b_{n+2}=5b_n+6a_n$

You can now calculate by hand, or get the general formula.