\begin{eqnarray} (\sqrt{2}-1)^1 &=& \sqrt{2} - 1 = \sqrt{2} - \sqrt{1} \\ (\sqrt{2}-1)^2 &=& -2\sqrt{2} +3 = \sqrt{9} - \sqrt{8} \\ (\sqrt{2}-1)^3 &=& 5\sqrt{2} - 7 = \sqrt{50} - \sqrt{49} \\ (\sqrt{2}-1)^4 &=& -12\sqrt{2} + 17 = \sqrt{289} - \sqrt{288} \\ (\sqrt{2}-1)^5 &=& 29\sqrt{2} - 41 = \sqrt{1682} - \sqrt{1681} \\ \end{eqnarray}

If we turn this around we get a nicer pattern:

\begin{eqnarray} (1-\sqrt{2})^1 &=& 1 - \sqrt{2} = \sqrt{1} - \sqrt{2} \\ (1-\sqrt{2})^2 &=& 3 -2\sqrt{2} = \sqrt{9} - \sqrt{8} \\ (1-\sqrt{2})^3 &=& 7 - 5\sqrt{2} = \sqrt{49} - \sqrt{50} \\ (1-\sqrt{2})^4 &=& 17 -12\sqrt{2} = \sqrt{289} - \sqrt{288} \\ (1-\sqrt{2})^5 &=& 41 - 29\sqrt{2} = \sqrt{1681} - \sqrt{1682} \\ \end{eqnarray}

Based on these examples, it appears that the result is correct, and in fact the first square root is the square root of a perfect square. Do you see any relationships between the numbers for different powers?

Need another hint? Click here.

Click here for the complete solution.

Hint 3

Let's write

(1-\sqrt{2})^n = A_n - B_n\sqrt{2}.

From the table it appears that each A value is about twice the previous value, and in fact it appears that the difference is

A_{n+1} - 2A_n = A_{n-1}.

Also it appears that

B_{n+1} = B_n + A_n \mathrm{\ and \ } A_{n+1} = A_n + 2B_n.

This table swarms with recursions! See if you can prove some of them.

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Click here for the complete solution.

Hint 4

Since the thing we are interested in (a power) has a recursive definition, we can use that to prove the recursions:

\begin{eqnarray} A_{n+1} - B_{n+1} \sqrt{2} &=& (1 - \sqrt{2})^{n+1} \\ &=& (A_n - B_n \sqrt{2})(1 - \sqrt{2}) \\ &=& (A_n + 2B_n) + (B_n + A_n)\sqrt{2}, \end{eqnarray}

and therefore

B_{n+1} = B_n + A_n \mathrm{\ and \ } A_{n+1} = A_n + 2B_n.

Now see if you can prove the difference in the squares is 1 or -1.

Click here for rest of the solution.

The Rest of the Solution

Again we proceed by recursion: If it is true for n, then

\begin{eqnarray} (A_{n+1}^2 - 2B_{n+1}^2) &=& (A_n + 2B_n)^2 - 2(B_n+A_n)^2 \\ &=& (A_n^2 + 4A_n B_n + 4B_n^2) - 2(B_n^2 + 2A_n B_n + A_n^2) \\ &=& -A_n^2 + 2B_n^2 \\ &=& -(A_n^2 - 2B_n^2) \\ &=& \pm 1. \end{eqnarray}

Is This a Surprising Result?

What special property does \sqrt{2}-1 have, that its powers are always the difference of the squares roots of two consecutive integers? It's easy to see that any power of \sqrt{2}-1 has the same property; are there other kinds of numbers with this property? Yes! Here are some even more unlikely numbers with this property:

\begin{gather} (\sqrt{3} - 2)^n \\ (8 - 3\sqrt{7})^n \\ (1766319049 - 226153980 \sqrt{61})^n \end{gather}

What's going on here? We can get a better understanding by approaching the problem from the other end. In our proof we dealt with an equation of the form

A^2 - NB^2 = 1

(for N=2). We can factor this as

(A - B\sqrt{N})(A + B\sqrt{N}) = 1

and raise each term to a power

(A - B\sqrt{N})^n (A + B\sqrt{N})^n = 1

When you multiply these powers out, you will see that the terms have the form

(C - D\sqrt{N})(C + D\sqrt{N}) = 1

that is,

C^2 - ND^2 = 1

In other words, any solution of this equation leads to additional solutions, by raising a certain expression to a power, and all powers of this expression are the difference of the square roots of two consecutive integers.

The equation

A^2 - NB^2 = 1

is called Pell's equation, and there is a great deal of Pell lore in Number Theory. We've seen just a little bit of this lore here; there is much more in connection with recursions, rational approximations to square roots, and continued fractions. See Shanks's book for more.

References

Edmund Landau, Elementary Number Theory, AMS Chelsea, 1966, pp. 76-84. Many books on number theory cover Pell's equation; this book gives an especially clear and simple development.
Daniel Shanks, Solved and Unsolved Problems in Number Theory, AMS Chelsea, 1993. Pell's equation is spread all through Chapter III, but see especially Theorem 77 and the surrounding pages.
Click here to view the original problem submitted by Billy.
Wikipedia has an article on Pell's Equation.