Statement 1: If `p` is a prime number `(p!=2),` then `[(2+sqrt(5))^p]-2^(p+1)` is always divisible by `p(w h e r e[dot]` denotes the greatest integer function). Statement 2: if `n` prime, then `^n C_1,^n C_2,^n C_2 ,^n C_(n-1)` must be divisible by `ndot`

Explanation

We have,
`(2+sqrt(5))^(n) + (2-sqrt(5))^(n)`
`= 2{2^(p) + .^(p)C_(2)(2^(p-2))(5) + .^(p)C_(4)(2^(p-4))(5^(2))+"…."+.^(P)C_(p-1)(2)(5^((p-1)//2))}"….."(1)`
From (1), at is clear that `(2+sqrt(5))^(p) + (2-sqrt(5))^(p)` is an integer.
Also, `-1 lt [(2-sqrt(5))^(p)]` , (as p is odd)
`= (2+sqrt(5))^(p) + (2-sqrt(5))^(p)`
So, ` [(2+sqrt(5)y^(2)] = (2+sqrt(5))^(p) + (2-sqrt(5))^(p)`
` = 2^(p+1)+.^(p)C_(2)(2^(p-1)) (5) + "......."`
`+.^(p)C_(p-1)(2)^(2) (5^((p-1)//2)))`
`:. [(2-sqrt(5))^(p)] - 2^(p+1) = 1 {.^(p)C_(2)(2^(p-2))(5) + .^(p)C_(4) (2^(p-4))(5^(2))`
`+ "......." + .^(p)C_(p-1)(2)(5^((p-1)//2))]`
Now, all the binomial coefficients `.^(p)C_(2), .^(p)C_(4),"......"..^(p)C_(p-1)` are divisible by the prime p.
Thus, R.H.S. is divisible by p.