Question

The integer next above (sqrt3+1)^2m contains

Answer 1

is divisible by 2^(m+1)

Answer 2

To solve this problem, we will use the concept of binomial expansion and properties of conjugate surds, similar to the provided example.

Let the given expression be $X$: $X = (\sqrt{3}+1)^{2m}$

First, simplify the base of the expression: $(\sqrt{3}+1)^2 = (\sqrt{3})^2 + 1^2 + 2 \cdot \sqrt{3} \cdot 1 = 3 + 1 + 2\sqrt{3} = 4+2\sqrt{3}$. So, $X = (4+2\sqrt{3})^m$.

Now, consider its conjugate term, $Y$: $Y = (4-2\sqrt{3})^m$.

Let's analyze the value of $Y$. We know that $3 < \sqrt{12} < 4$, which means $3 < 2\sqrt{3} < 4$. Therefore, $4-4 < 4-2\sqrt{3} < 4-3$, which simplifies to $0 < 4-2\sqrt{3} < 1$. Since $0 < 4-2\sqrt{3} < 1$, any positive integer power of it will also be between 0 and 1. Thus, $0 < Y < 1$.

Next, consider the sum $X+Y$: $X+Y = (4+2\sqrt{3})^m + (4-2\sqrt{3})^m$. Using the binomial expansion for $(a+b)^m + (a-b)^m = 2 \left[ \binom{m}{0}a^m + \binom{m}{2}a^{m-2}b^2 + \dots \right]$, where $a=4$ and $b=2\sqrt{3}$. $b^2 = (2\sqrt{3})^2 = 4 \times 3 = 12$. So, $X+Y = 2 \left[ \binom{m}{0}4^m + \binom{m}{2}4^{m-2}(12) + \binom{m}{4}4^{m-4}(12)^2 + \dots \right]$. All terms inside the square bracket are integers. Therefore, $X+Y$ is an integer. Let $K = X+Y$. So $K$ is an integer.

We have $X+Y=K$ 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, $K-1 < X < K$.

Since $K-1 < X < K$, $X$ is not an integer. The integer next above $X$ is $\lceil X \rceil = K$. Thus, the integer next above $(\sqrt{3}+1)^{2m}$ is $K$.

Now, let's determine the properties of $K$: $K = (4+2\sqrt{3})^m + (4-2\sqrt{3})^m$ $K = (2(2+\sqrt{3}))^m + (2(2-\sqrt{3}))^m$ $K = 2^m (2+\sqrt{3})^m + 2^m (2-\sqrt{3})^m$ $K = 2^m \left[ (2+\sqrt{3})^m + (2-\sqrt{3})^m \right]$.

Let $P_m = (2+\sqrt{3})^m + (2-\sqrt{3})^m$. Using the binomial expansion again: $P_m = 2 \left[ \binom{m}{0}2^m + \binom{m}{2}2^{m-2}(\sqrt{3})^2 + \binom{m}{4}2^{m-4}(\sqrt{3})^4 + \dots \right]$ $P_m = 2 \left[ \binom{m}{0}2^m + \binom{m}{2}2^{m-2}(3) + \binom{m}{4}2^{m-4}(3^2) + \dots \right]$. All terms inside the square bracket are integers, so $P_m$ is an even integer. Let $P_m = 2J$ for some integer $J$.

Substitute $P_m = 2J$ back into the expression for $K$: $K = 2^m (2J) = 2^{m+1}J$. This shows that $K$ is divisible by $2^{m+1}$.

Therefore, the integer next above $(\sqrt{3}+1)^{2m}$ is divisible by $2^{m+1}$.