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I'm really not sure how to do this. I factored out a power of $3$ and squared so that I have $2^3 (6+\sqrt{35})^3 \gt n$ , and I know that if I can prove that $12^3-1 \le (6+\sqrt{35})^3 \lt 12^3$ I am basically done, but I don't know how to do that. Any help is appreciated. Thanks!

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  • $\begingroup$ An (algebra-precalculus) tag? $\endgroup$ – Shuhao Cao Jun 22 '13 at 2:18
  • $\begingroup$ I thought this might involve some heavy algebraic manipulation $\endgroup$ – Ovi Jun 22 '13 at 2:19
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    $\begingroup$ What prevents you from just using a calculator? $\endgroup$ – Potato Jun 22 '13 at 2:19
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    $\begingroup$ $6768+1144*\sqrt{35}$,I think you can try numbers below $6768+1144*6$ one by one. $\endgroup$ – eccstartup Jun 22 '13 at 2:39
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    $\begingroup$ @Ovi When presenting such problems, it would be good to add some context and note that this is a contest problem and you would like a solution that does not involved using a calculator. Otherwise we don't know to look for cheap shots. $\endgroup$ – Potato Jun 22 '13 at 2:45
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Let
$$N=(\sqrt{7}+\sqrt{5})^6 +(\sqrt{7}-\sqrt{5})^6.$$ If we imagine expanding via the Binomial Theorem we see that $N$ is an integer. The second term is pretty small, since already $\sqrt{7}-\sqrt{5}$ is well below $1$. So our $n$ is equal to $N-1$.

Now expand. There is a fair bit of cancellation. We get $$N=2\left(7^3 + \binom{6}{2}(7^2)(5)+\binom{6}{4}(7)(5^2)+5^3\right).\tag{1}$$ Finding $N$ explicitly is fairly easy, it is exact arithmetic .

Remarks: $1.$ Much more efficiently, we can use exactly the same idea, but make use of your calculation of the square. Consider $$(6+\sqrt{35})^3+(6-\sqrt{35})^3.$$ This is equal to $2\left(6^3+3(6)(35)\right)$.

$2.$ What we did is not a trick. The general idea is of wide applicability, it is a method.

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  • $\begingroup$ Could you expand on your comment that this is a method? A method for what? $\endgroup$ – Potato Jun 22 '13 at 2:46
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    $\begingroup$ It is used in a large number of problems involving $(a+b\sqrt{d})^n$. And I was making an obscure reference to an old joke about a trick used twice being a method. And this trick is used very much more than twice, Example: Fibonacci numbers. $\endgroup$ – André Nicolas Jun 22 '13 at 2:47
  • $\begingroup$ Thanks, and is there a way of algebraically proving the inequality $12^3-1 \le (6+\sqrt{35})^3 \lt 12^3$ ? Let's say you didn't know this actually came from $(\sqrt5 +\sqrt7)^6$ $\endgroup$ – Ovi Jun 22 '13 at 2:52
  • $\begingroup$ Well, we can use the "trick" on $(6+\sqrt{35})^3$. Your preliminary work enables us to cut down on the arithmetic of my formula (1). Maybe I will add that. $\endgroup$ – André Nicolas Jun 22 '13 at 3:02
  • $\begingroup$ @Ovi: Note that the inequality about $12^3$ is not right. Reasonably close, almost $1692$, not $12^3$. $\endgroup$ – André Nicolas Jun 22 '13 at 3:20
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First, a comment on technique. Heuristically, what you're doing is "wrong" for a competition problem in the sense that you're ignoring and throwing away a lot of the structure. (You have $\sqrt{n+1}-\sqrt{n-1}$ raised to a power. This is a contest problem. That's not a coincidence.) Your way might be fine to find some weak bounds to in order to check your final answer, but before I even found the solution myself, I could tell it probably wasn't going to work.

Whenever you see a problem involving sums of radicals, you should immediately think about taking the conjugate and somehow playing tricks with it. This idea quickly leads to a solution here. The conjugate is $(\sqrt 7 - \sqrt 5)^6$, and it's not hard to notice this is a number less than one. Then

$$(\sqrt 7 + \sqrt 5)^6 + (\sqrt 7 - \sqrt 5)^6$$

is an integer, and the rest of the solution is clear.

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  • $\begingroup$ Conjugate in a «let me wave my hands here» sense, of course :-) $\endgroup$ – Mariano Suárez-Álvarez Jun 22 '13 at 2:50
  • $\begingroup$ @MarianoSuárez-Alvarez It's a Galois conjugate! There is no hand-waving here! $\endgroup$ – Potato Jun 22 '13 at 2:51
  • $\begingroup$ Well, that number has several Galois conjugates... and you picked one of them. $\endgroup$ – Mariano Suárez-Álvarez Jun 22 '13 at 2:52
  • $\begingroup$ @MarianoSuárez-Alvarez Indeed, but using when high school math context people say "the conjugate" in this situation, they mean flipping the signs of exactly one root. And you could get away with $\sqrt 5 - \sqrt 7$ too. $\endgroup$ – Potato Jun 22 '13 at 2:58
  • $\begingroup$ That's the hand-waving I had in mind :-) $\endgroup$ – Mariano Suárez-Álvarez Jun 22 '13 at 3:01
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We want a lower bound on $\sqrt{1-x}$. From $\sqrt{1-x} = (1-x)^{1/2} =1-x/2-x^2/8-x^3/16 -5x^4/128 ... $ I will try $1-x/2-x^2/4$.

$\begin{align} (1-x/2-x^2/4)^2 &=1-x-x^2(1/4+1/2)+x^3/4+x^4/16\\ &=1-x-3x^2/4+x^3/4+x^4/16\\ &= 1-x-x^2(3/4-x/4-x^2/16)\\ &< 1-x\\ \end{align} $

for $0 < x < 1/2$, so $\sqrt{1-x} > 1-x/2-x^2/4$ for $0 < x < 1/2$.

$\sqrt{35} = \sqrt{36-1} = 6\sqrt{1-1/36} > 6(1-1/12-1/(4*36)) =6-1/2-1/24 $ so $6+\sqrt{35} > 12-1/2-1/24 > 12-1$ as you wanted.

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Take the binomial expansion. At $n=6$ pascals is $1$ $6$ $15$ $20$ $15$ $6$ $1$. So $(\sqrt{5}+\sqrt{7})^{6}=$ (grouping $\Bbb N$ seperately from $\Bbb R$) $5^3+7^3+15(5^2 7+7^2 5) + \sqrt{35}(6(5^2+7^2)+20*5*7) = 6768+\sqrt{35}(1144)$. In order to estimate our n, $\sqrt{35}$ must be estimated to 3 places giving us $5.916*1144\approx6767$ rounding down, and $6767+6768=13535$ as the greatest natural number less than $(\sqrt{5}+\sqrt{7})^{6}$.

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