How do I prove that $\dfrac{(7+\sqrt{48})^n+(7-\sqrt{48})^n}{2}\equiv3\pmod4$ if and only if $n$ is odd? How do I prove that $\dfrac{(7+\sqrt{48})^n+(7-\sqrt{48})^n}{2}\equiv3\pmod4$ if and only if $n$ is odd?
I know that it will always be whole, and that it will never be $\equiv3$ if $n$ is even. But I want to know how to prove that it will always be $\equiv3$ if $n$ is odd...
 A: As mentioned in the comments. Consider $a_n = \frac{(7+\sqrt{48})^n + (7-\sqrt{48})^n}{2}$. Then the characteristic equation of the reccurence relations is the one whose roots are $7+\sqrt{48}$ and $7 - \sqrt{48}$. It's not very difficult to notice that it's: $x^2 - 14x + 1 = 0$. Therefore we have: $a_n = 14a_{n-1} - a_{n-2}$
A: Expand both the terms using binomial expansion. Following are the steps
$$\frac{ (7+\sqrt{48})^n + (7-\sqrt{48})^n}{2}$$
$$\frac{(7^n+7^{n-1}*\sqrt{48}+...\sqrt{48}^n)+(7^n-7^{n-1}*\sqrt{48}+...-\sqrt{48}^n)}{2}$$ 
It is seen   terms having odd power of $\sqrt{48}$ get cancelled. Left are even powers ( which get doubled ). 
$$\frac{2*(7^n+7^{n-2}*\sqrt{48}^2+...+7*\sqrt{48}^{n-1})}{2}$$
The two gets cancelled. All even powers of $\sqrt{48}^{2k}$ make it $48^k$. Now $48$ is divisible by 4. Thus only term left is $7^n$, which is $3 \mod{4}$ when $n$ is odd ( as $7 \equiv 3 \mod{4}$, $7^2 \equiv 1 \mod{ 4}$, $7^3 \equiv 3 \mod {4}$ and so on ).
A: $(a \pm b)^n = a^n \pm n*a^{n-1}b + ...... + (-1)^i*{n \choose i}a^ib^{n-1}+ ....$.
So $(a -b)^n + (a+ b)^n = 2a^n + (n - n)a^{n-1}b + ...... (1 + (-1)^i){n \choose i}a^ib^{n-1}+....$.
$1 + (-1)^i = 0$ if $i$ is odd; $2$ if $i$ is even.
so $(a -b)^n + (a+ b)^n = 2(a^n + {n \choose 2}a^{n-2}b^2 + ...... {n \choose 2k}a^{n-2k}b^{2k}+....)$
So $N= \frac{(7 + \sqrt{48})^n + (7-\sqrt{48})}{2}= \sum_{k = 0;k \le n/2} {n \choose 2k}7^{n-2k}\sqrt{48}^{2k} = \sum_{k = 0;k \le n/2} {n \choose 2k}7^{n-2k}48^k$.
${n \choose 2k}7^{n-2k}48^k \equiv 0 \mod 4$ if $k \ne 0$. (Because $48\equiv 0 \mod 4$).
So $N= \sum_{k = 0;k \le n/2} {n \choose 2k}7^{n-2k}48^k \equiv {n \choose 0}7^n*48^0 = 7^n \equiv (-1)^n \mod 4$.
If $n$ is even $N \equiv (-1)^n \equiv 1 \not \equiv 3 \mod 4$.
If $n$ is odd $N \equiv (-1)^n \equiv -1 \equiv 3 \mod 4$
