Greatest integer $n$ where $n \lt (\sqrt5 +\sqrt7)^6$ 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!
 A: 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. 
A: 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.
A: 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 $(\sqrt{5}+\sqrt{7})^{6}$.
A: 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. 
