Question

If $p$ is a prime number greater than $2$, then the difference $\left[{(2+\sqrt{5})}^p\right]-2^{p+1}$, where $[.]$ denotes greatest integer. is divisible by

• A. $p+1$
• B. $p-1$
• C. $p+\sqrt{5}$
• D. $p$

Answer 1

Answer: (D)

$p$

Since $p$ is a prime number greater than $2$, it is an odd number.

$(\sqrt{5}+2{)}^p={}^p{C}_0(\sqrt{5}{)}^0{2}^p+{}^p{C}_1(\sqrt{5}{)}^1{2}^{p-1}+...+{}^p{C}_p(\sqrt{5}{)}^p{2}^0$

$(\sqrt{5}-2{)}^p=-{}^p{C}_0(\sqrt{5}{)}^0{2}^p{+}^p{C}_1(\sqrt{5}{)}^1{2}^{p-1}+...{+}^p{C}_p(\sqrt{5}{)}^p{2}^0$

$(\sqrt{5}+2{)}^p-(\sqrt{5}-2{)}^p=2\left[{}^p{C}_0\right(\sqrt{5}{)}^{0}2{}^p{+}^p{C}_2(\sqrt{5}{)}^2{2}^{p-2}+...{+}^p{C}_{p-1}(\sqrt{5}{)}^{p-1}{2}^1]$

Since $(\sqrt{5}-2)$ always lies between 0 and 1, any power of it will also lie in between 0 and 1 and so the greatest integer function of it will always be zero.

$\therefore \left[\right(\sqrt{5}+2{)}^p]=$$2\left[{}^p{C}_0\right(\sqrt{5}{)}^{0}2{}^p{+}^p{C}_2(\sqrt{5}{)}^2{2}^{p-2}+...{+}^p{C}_{p-1}(\sqrt{5}{)}^{p-1}{2}^1]$

$\therefore \left[\right(\sqrt{5}+2{)}^p]=2^{p+1}+2{[}^p{C}_2(\sqrt{5}{)}^2{2}^{p-2}+...+p\left(\sqrt{5}{)}^{p-1}{2}^1\right]$

Since every term on the right hand side will have p as its factor, $\left[\right(\sqrt{5}+2{)}^p]-2^{p+1}$ becomes divisible by $p$