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This question comes out of a mathematics calendar problem that asked for the tenths digit of the expression $(17 + \sqrt{280})^{17}$. The calendar implied the digit should be 9, but after playing with the expression for while, factoring it in various ways, binomial expansions, trying series approximations, etc., I made little progress.

I tried a more brute-force approach and entered the expression into a high-precision calculator, which showed the value to be very close to an integer, specifically with a decimal that begins $.999999999824...$ After further trial-and-error, I saw the problem had a number of red herrings (e.g., the fact that the first time and the exponent are equal, or that the number under the radical is different from the square of the first time by a perfect square).

Someone can correct me if I'm wrong, but in general it seems expressions of this form $$(a + \sqrt {a^2 \pm x})^n$$ where $a$ is a positive integer and $x$ is an integer relatively small compared to $a^2$, tend to rapidly approach integer values as $n$ grows.

I feel like I'm missing something rather obvious here, but could someone enlighten me (or at least give me a hint) about why this expression should give values increasingly close to integers when raised to large powers?

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    $\begingroup$ Hint: show that $a_n=(17+\sqrt{280})^n+(17-\sqrt{280})^n$ is always an integer. $\endgroup$ – lulu Apr 6 at 16:10
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Let $\alpha=17+\sqrt{280}$ and $\beta=17-\sqrt{280}$. Then $\alpha$ and $\beta$ are the roots of the quadratic $$X^2-34X+9=0.$$ Now let $$c_n=\alpha^n+\beta^n.$$ Then $c_0=2$, $c_1=34$ and in general $$c_n=34c_{n-1}-9c_{n-2}.$$ Then the $c_n$ are all integers, by induction. But $0<\beta<1$ and for large $n$, $\beta^n$ is positive and close to zero. This means that $\alpha^n$ is just a little below the integer $c_n$. This is the phenomenon you are observing.

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  • $\begingroup$ Ah, thank you! I did create that quadratic and stare at it for a while, but couldn't quite sort out how to make it do what I wanted. $\endgroup$ – Athanasius Apr 6 at 16:20
  • $\begingroup$ Another way to see that $c_n$ is always an interger is to look at the binomial expansion of the terms in $\alpha^n+\beta^n$ and realize that corresponding terms with a given odd power of $\sqrt{280}$ have opposing signs in $\alpha^n$ and $\beta^n$, so cancel each other. The remaining terms only have even powers of $\sqrt{280}$ (creating an integer as the power), thus $c_n$ is the sum of many integers and thus itself an integer. $\endgroup$ – Ingix Apr 6 at 17:46

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