The curious powers of 1 + sqrt 2: conjecture

In my previous post I related the following puzzle from Colin Wright:

What’s the 99th digit to the right of the decimal point in the decimal expansion of (1 + \sqrt 2)^{500}?

Let’s play around with this a bit and see if we notice any patterns. First, 1 + \sqrt 2 itself is approximately

1 + \sqrt 2 \approx 2.414213562373095\dots

so its powers are going to get large. Let’s use a computer to find the first ten or so:

\displaystyle \begin{array}{cc} n & (1 + \sqrt 2)^n \\ \hline 1 & 2.414213562373095 \\ 2 & 5.82842712474619 \\ 3 & 14.071067811865474 \\ 4 & 33.970562748477136 \\ 5 & 82.01219330881975 \\ 6 & 197.99494936611663 \\ 7 & 478.002092041053 \\ 8 & 1153.9991334482227 \\ 9 & 2786.0003589374983 \\ 10 & 6725.9998513232185 \end{array}

Sure enough, these are getting big (the tenth power is already bigger than 6000), but look what’s happening to the part after the decimal: curiously it seems that the powers of (1 + \sqrt 2) are getting rather close to being integers! For example, (1 + \sqrt 2)^{10} is just under 6726, only about 0.0002 away.

At this point, I had seen enough to notice and conjecture the following patterns (and I hope you have too):

  • The powers of (1 + \sqrt 2) seem to be getting closer and closer to integers.
  • In particular, they seem to alternate between being just under an integer (for even powers) and just over an integer (for odd powers).

If this is true, the decimal expansion of (1 + \sqrt 2)^{500} must be of the form n.99999999\dots for some big integer n and some number of 9s after the decimal point. And it seems reasonable that if Colin is posing this question, it must have more than 99 nines, which means the answer would be 9.

But why does this happen? Do the powers really keep alternating being just over and under an integer? And how close do they get—how do we know for sure that (1 + \sqrt 2)^{500} is close enough to an integer that the 99th digit will be a 9? This is what I want to explore in a series of future posts—and as should come as no surprise it will take us on a tour of some fascinating mathematics!

About Brent

Associate Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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3 Responses to The curious powers of 1 + sqrt 2: conjecture

  1. llambda says:

    Easy! This is obvious to anyone familiar with the Binet formula, which exhibits similar behaviour. It’s easy to see that f(n) = (1 – √2)^n + (1 + √2)^n follows a recurrent relation f(n+1) = 2f(n) + f(n-1) and hence is an integer for any n. This means that the difference between the target integer f(n) and (1 + √2)^n is (1 – √2)^n, which is less than one in absolute value (hence “convergence” to integers) and negative (hence oscillating)

  2. Pingback: The curious powers of 1 + sqrt 2: a clever solution | The Math Less Traveled

  3. jess tauber says:

    The generalized Pascal Triangle with 2’s down one side and 1’s down the other generates upshifted Fibonacci numbers on one side and Lucas numbers on the other when you sum terms from shallow diagonals. Interestingly, those along the shallow diagonal leading to the Lucas numbers are also the coefficients of terms in the equations for determining the powers of the metallic means (1+sqrt2 being the ‘silver mean’). And the variables in these terms are raised to powers identical to the dimensions associated with the normal edge-parallel diagonals of the (2,1) sided generalized Pascal Triangle that each term in the shallow diagonal is found in. I discovered this by accident some years ago, and don’t know whether I was the first (though I doubt it).

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