Question

show that $[(\sqrt{3}+1)^{2n}]+1$ is divisible by $2^{n+1}$ where n is a natural number and [] denotes greatest integer function.

Answer 1

[(sqrt3+1)^2n]+1 is divisible by 2^(n+1)

Answer 2

Let $X = (\sqrt{3}+1)^{2n}$. We need to show that $[X]+1$ is divisible by $2^{n+1}$.

First, let's simplify the expression for $X$: $X = ((\sqrt{3}+1)^2)^n = (3+1+2\sqrt{3})^n = (4+2\sqrt{3})^n$.

Now, let's define a conjugate term $Y$: $Y = (\sqrt{3}-1)^{2n} = ((\sqrt{3}-1)^2)^n = (3+1-2\sqrt{3})^n = (4-2\sqrt{3})^n$.

Consider the sum $X+Y$: $X+Y = (4+2\sqrt{3})^n + (4-2\sqrt{3})^n$. Let $a=4$ and $b=2\sqrt{3}$. Then $X+Y = (a+b)^n + (a-b)^n$. Using the binomial expansion: $(a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + \binom{n}{n}b^n$ $(a-b)^n = \binom{n}{0}a^n - \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 - \dots + (-1)^n \binom{n}{n}b^n$ Adding these two expansions, the terms with odd powers of $b$ cancel out: $X+Y = 2 \left[ \binom{n}{0}a^n + \binom{n}{2}a^{n-2}b^2 + \binom{n}{4}a^{n-4}b^4 + \dots \right]$ Substitute $a=4$ and $b^2 = (2\sqrt{3})^2 = 4 \times 3 = 12$: $X+Y = 2 \left[ \binom{n}{0}4^n + \binom{n}{2}4^{n-2}(12) + \binom{n}{4}4^{n-4}(12)^2 + \dots \right]$ Since all terms inside the square bracket are integers, $X+Y$ is an integer. Furthermore, since the entire expression is multiplied by 2, $X+Y$ is an even integer. Let $X+Y = K$, where $K$ is an even integer.

Now, let's evaluate the value of $Y = (4-2\sqrt{3})^n$. Consider the base $4-2\sqrt{3}$. We know that $3^2=9$ and $4^2=16$. So, $3 < \sqrt{12} < 4$. Multiplying by 2, we get $6 < 2\sqrt{3} < 8$. Wait, $2\sqrt{3} = \sqrt{4 \times 3} = \sqrt{12}$. So $3 < \sqrt{12} < 4$. Therefore, $3 < 2\sqrt{3} < 4$. Now, subtract this from 4: $4-4 < 4-2\sqrt{3} < 4-3$ $0 < 4-2\sqrt{3} < 1$. Since the base $4-2\sqrt{3}$ is between 0 and 1, any positive integer power of it will also be between 0 and 1. Thus, $0 < (4-2\sqrt{3})^n < 1$ for any natural number $n$. So, $0 < Y < 1$.

We have $X+Y = K$, where $K$ is an even integer, and $0 < Y < 1$. From $X+Y=K$, we can write $X = K-Y$. Since $0 < Y < 1$, we have $-1 < -Y < 0$. Adding $K$ to all parts of the inequality: $K-1 < K-Y < K$. So, $K-1 < X < K$.

By the definition of the greatest integer function, $[X]$ is the greatest integer less than or equal to $X$. Since $K-1 < X < K$, it implies that $[X] = K-1$.

We need to show that $[X]+1$ is divisible by $2^{n+1}$. Substituting $[X]=K-1$: $[X]+1 = (K-1)+1 = K$. Since $K$ is an even integer, let $K = 2m$ for some integer $m$. We need to show that $K$ is divisible by $2^{n+1}$.

Let's re-examine $K = (4+2\sqrt{3})^n + (4-2\sqrt{3})^n$. $K = 2^n (2+\sqrt{3})^n + 2^n (2-\sqrt{3})^n = 2^n \left[ (2+\sqrt{3})^n + (2-\sqrt{3})^n \right]$. Let $P_n = (2+\sqrt{3})^n + (2-\sqrt{3})^n$. Similar to the argument for $X+Y$, $P_n$ is an integer. $P_n = \sum_{k=0}^n \binom{n}{k} 2^{n-k} (\sqrt{3})^k + \sum_{k=0}^n \binom{n}{k} 2^{n-k} (-\sqrt{3})^k$ $P_n = 2 \left[ \binom{n}{0} 2^n + \binom{n}{2} 2^{n-2} \cdot 3 + \binom{n}{4} 2^{n-4} \cdot 3^2 + \dots \right]$. All terms inside the bracket are integers. The first term is $\binom{n}{0} 2^n = 1 \cdot 2^n = 2^n$. The second term is $\binom{n}{2} 2^{n-2} \cdot 3$. The third term is $\binom{n}{4} 2^{n-4} \cdot 3^2$. So, $P_n = 2 \cdot (\text{integer})$. This means $P_n$ is an even integer.

Therefore, $K = 2^n P_n$. Since $P_n$ is an even integer, let $P_n = 2M$ for some integer $M$. Then $K = 2^n (2M) = 2^{n+1}M$. This shows that $K$ is a multiple of $2^{n+1}$. Thus, $[X]+1 = K$ is divisible by $2^{n+1}$.

This completes the proof.