Buri
Lv 7

# CMO Problem: Just for fun :-)?

Show that every positive integral power of √2 - 1 is of the form √m - √(m - 1) for some positive integer m. (e.g. (√2 - 1)² = 3 - 2√2 = √9 - √8).

If you've seen the solution, please don't post it. And no cheating! :-)

Have fun! :-)

Update:

To my surprise (because I even starred the question) one of my contacts, DaNNiX, asked this exact same question 3 months ago. Here's the link and contains a nice solution by ☮ Vašek (the best answer):

Relevance

Using the binomial theorem

(sqrt(2)-1)^k = C sqrt(2) - D

where C and D are constants with the same sign. This implies that

(sqrt(2)-1)^k = sqrt(A) - sqrt(B)

where if both C and D are positive, A = 2C^2, B = D^2

and if both C and D are negative, A = D^2, B = 2C^2

The binomial theorem also shows that

(sqrt(2)+1)^k = |C|sqrt(2) + |D| = sqrt(A) + sqrt(B)

Multiply the two, we get

[(sqrt(2)+1)(sqrt(2)-1)]^k = A - B

[2-1]^k = A-B

A = B+1

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The binomial theorem assertion can be found by either inspection or by induction:

suppose

(sqrt(2)-1)^(k-1) = Csqrt(2) - D

then

(sqrt(2)-1)^k = -(C+D)sqrt(2) + (2C+D)

If C and D has the same sign, so would C+D and 2C+D.

Similarly for the other claim.

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Edit @ +3 hours:

Half-Blood prince brought up a good point with symmetricity and analogues to hyperbolic trigonometric functions.

In fact, this problem can be generalized to the following statement: let p be some positive integer, then every positive integral power of

sqrt(p^2+1) - p

is of the form sqrt(m+1) - sqrt(m) for some positive integer m. The hyperbolicity that Half-Blood Prince remarked on is (I think) equivalent to the following fact: that over the field Q of rational numbers, the polynomial

x^2 - p^2 -1

is irreducible for p a positive integer. (Basically saying that the only consecutive squares are 0 and 1.) So we can define the field extension Q[sqrt(p^2+1)] = F. We endow it with a simple pseudo-norm given by the following: for any element z in F, we can write

z = a + b sqrt(p^2+1)

where a, b are in Q. We can then define the conjugate of z to be

z* = -a + b sqrt(p^2+1)

and define the norm

|z| = z.z* = b^2(p^2+1) - a^2

Notice that |z| is non-degenerate: if |z| = 0, we have that

p^2+1 = (a/b)^2

and so z = 0 since p^2+1 has no square roots over Q. But |z| is allowed to be negative. Now here's the cool bit: this is "almost" like the Gaussian integers, except that if you draw the level curves of the norm on the a-b plane, you find that instead of circles, you get hyperbolas! (This is exactly why Half-Blood Prince can use the hyperbolic angle formulae in this problem.) (In physics terminology, you can think of this as a "Wick rotation" of the Gaussian integers.)

In any case, the point is that the pseudo-norm is multiplicative:

|w.z| = |w||z|

|z| = b^2(p^2+1) - a^2 = 1

Then z^k will still have |z^k| = 1, which implies that if we write

z^k = c + d.sqrt(p^2+1)

we get

|z^k| = d^2(p^2+1) - c^2 = 1

and hence the claim.

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Edit: @ a little bit later:

aww...shucks. Half-Blood Prince's really neat answer was removed. I hope you don't mind me bludgeoning the question to death.

• Anonymous

I killed the previous answer. Since jaz_will mentioned that solution in his edited version, I would like to paste it again so that people can have an idea about the hyperbolic analogue. I should say this proof is essential the same as jaz_will's. So, Buri, please choose jaz_will's answer as the best answer.

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Denote x = √2 + 1, 1/x = √2 − 1

Now introduce

A_n = (xⁿ +1/xⁿ )/2,

B_n = (xⁿ −1/xⁿ )/2,

particularly, A_1 = √2, B_1 = 1.

A_n and B_n look like hyperbolic cosine and sine functions. Use the angle addition formulas (for hyperbolic functions), we have

A_{n+1} = A_n A_1 + B_n B_1 = A_n √2 + B_n

B_{n+1} = A_n B_1 + B_n A_1 = A_n + B_n √2

We also have an identity corresponds to cosh²(x) − sinh²(x) = 1

(A_{n+1})² − (B_{n+1})² = (A_n)² − (B_n)² = ... = 1...............(a)

Since 1/xⁿ = A_n − B_n, we identify (A_n)² = m, and Eq. (a) asserts that (B_n)² = m − 1.

To be more rigorous, we need to show that (A_n)² is an integer in the last step. That can be done by a simple induction:

Suppose (A_n)², (B_n)² and (A_n)(B_n) √2 are integers (as they are initially), so are (A_{n+1})², (B_{n+1})² and (A_{n+1})(B_{n+1}) √2. This can be easily shown:

(A_{n+1})² = 2 (A_n)² + (B_n)² + 2 [√2(A_n)(B_n) ]

(A_{n+1})(B_{n+1}) = 2 [(A_n)² + (B_n)²] + 2 [√2(A_n)(B_n) ]

as well as (a).

Since every terms on the r.h.s are integers, their sums l.h.s. are also integers.

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PS. I still don't like my solution, and I still have an impluse to delete the answer. I like the question though.