The number $(3+\sqrt{5})^n+(3-\sqrt{5})^n$ is an integer

Prove by induction that this number is an integer:

$$u_n=(3+\sqrt{5})^n+(3-\sqrt{5})^n$$

Progress

I assumed that it holds for $n$ and I tried to do it for $n+1$ but the algebra gets quite messy and I'm unable to prove that the following term is an integer: $\sqrt{5}((3+\sqrt{5})^n-(3-\sqrt{5})^n)$

Solution 1:

Outline: For the (strong) induction step, we can use the fact that $$(3+\sqrt{5})^{n+1}+(3-\sqrt{5})^{n+1}=\left[(3+\sqrt{5})^{n}+(3-\sqrt{5})^{n}\right]\left[(3+\sqrt{5})+(3-\sqrt{5})\right]-(3+\sqrt{5})(3-\sqrt{5})\left[(3+\sqrt{5})^{n-1}+(3-\sqrt{5})^{n-1}\right].$$

Note that $(3+\sqrt{5})+(3-\sqrt{5})$ and $(3+\sqrt{5})(3-\sqrt{5})$ are integers.

Remark: There are better "non-induction" ways. For example, imagine expanding each of $(3+\sqrt{5})^n$ and $(3-\sqrt{5})^n$, using the Binomial Theorem. Now add. The terms in odd powers of $\sqrt{5}$ cancel.

Solution 2:

From the theory of sequences defined by a linear recurrence relation with constant coefficients, the sequence $u_n$ satisfies $u_{n+2}=6u_{n+1}-4u_n$ and $u_0=2$ and $u_1=6$.

Then, if you assume that $u_n$ and $u_{n+1}$ are integers, it follows immediately that $u_{n+2}$ is also an integer. You can write a strong induction from this to have a complete proof.