Prove that $(1 + \sqrt2)^{2n} + (1 - \sqrt{2})^{2n}$ is an even integer.

Question

Prove that $(1+\sqrt2)^{2n} + (1-\sqrt2)^{2n}$ is an even integer.

I'm not sure how to prove that it is an even integer. What would I do for the Inductive Step? And for the basic step, can I plug in zero and prove something from that?

Answer 1

Base case: $P(0)$ $$ (1 + \sqrt{2})^0 + (1 - \sqrt{2})^0 = 1 + 1 = 2 $$ which is even since $2 = 2\cdot 1$ and of course $1 \in \mathbb{Z}$.

Inductive step: Assume true for $P(k)$, i.e. $$ (1 + \sqrt{2})^{2k} + (1 - \sqrt{2})^{2k} $$ is true. Show that $P(k+1)$ is true.

Answer 2

Hint: Prove by induction on $k$ that $(1+\sqrt2)^k = a+b\sqrt2$ for some integers $a$ and $b$ such that $(1-\sqrt2)^k = a-b\sqrt2$.

Then set $k=2n$.

You can choose either $k=0$ or $k=1$ to be the base case.

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Alternatively: Use the binomial theorem on each of $(1+\sqrt2)^{2n}$ and $(1-\sqrt2)^{2n}$. Note that the terms that involve an odd power of $\sqrt2$ cancel out each other between the two sums, and that terms with an even power of $\sqrt2$ are (a) integers and (b) are the same in each of the two sums.

Answer 3

For the basic step, yes, you can plug $n=0$ (and also $n=1$, if you want) and do some computations.

On the other hand, for the inductive step you can always use the following equality: $$x^{2n} + y^{2n} = (x^2+y^2)\cdot (x^{2(n-1)} + y^{2(n-1)}) - x^2y^2\cdot (x^{2(n-2)} + y^{2(n-2)})$$ and use complete induction (after you have observed that $(1+\sqrt{2})^2(1-\sqrt{2})^2 =1$).

Answer 4

Note that $(1+\sqrt2)(1-\sqrt2) =-1 $ and $(1+\sqrt2)^2 =3+2\sqrt{2} $.

Therefore, if $a =3+2\sqrt{2} $, then $1/a = 3-2\sqrt{2} $, so that $a+1/a =6 $ and $(1+\sqrt2)^{2n} + (1-\sqrt2)^{2n} =a^n+1/a^n $.

We now use the identity true for any $a$ that $a^{n+1}+1/a^{n+1} =(a+1/a)(a^n+1/a^n)-(a^{n-1}+a^{n-1}) $.

Therefore, for this particular $a$, $a^{n+1}+1/a^{n+1} =6(a^n+1/a^n)-(a^{n-1}+a^{n-1}) $.

Since $a^n+1/a^n$ is an integer for $n=0$ and $n=1$, it is an integer for all $n$.

Explicitly, if $u_n = a^n+1/a^n =(1+\sqrt2)^{2n} + (1-\sqrt2)^{2n} $, $u_{n+1} =6u_n-u_{n-1} $ with $u_0 = 1$ and $u_1 = 6$.