Let $p$ be an odd prime. Prove that the number $$\left\lfloor \left(\sqrt{5}+2\right)^{p}-2^{p+1}\right\rfloor$$is divisible by $20p$.
Problem
Source: Moldova 2017 TSTST, B10
Tags: number theory
Tintarn
21.03.2017 01:10
Snakes wrote: Let $p$ be an odd prime. Prove that the number $$\left\lfloor \left(\sqrt{5}+2\right)^{p}-2^{p+1}\right\rfloor$$is divisible by $20p$.
First, note that the number in question is $a_p=(2-\sqrt{5})^p+(2+\sqrt{5})^p-2^{p+1}$. Now, this is a sequence with $a_0=a_1=0, a_2=10$ and $a_{n+3}=6a_{n+2}-7a_{n+1}-2a_n$ and hence we easily see that $20 \mid a_n$ for all odd $n$, in particular $20 \mid a_p$.
Now, checking the case $p=5$ separately, it suffices to prove that $p \mid a_p$.
But note that $a_p=2\sum_{k=1}^{(p-1)/2} \binom{p}{2k} 5^k2^{p-2k}$ is clearly divisible by $p$ as each of the binomial coefficients is.
Hence the result.
hmida99
21.03.2017 01:37
Tintarn wrote: Snakes wrote: Let $p$ be an odd prime. Prove that the number $$\left\lfloor \left(\sqrt{5}+2\right)^{p}-2^{p+1}\right\rfloor$$is divisible by $20p$.
First, note that the number in question is $a_p=(2-\sqrt{5})^p+(2+\sqrt{5})^p-2^{p+1}$. Now, this is a sequence with $a_0=a_1=0, a_2=10$ and $a_{n+3}=6a_{n+2}-7a_{n+1}-2a_n$ and hence we easily see that $20 \mid a_n$ for all odd $n$, in particular $20 \mid a_p$.
Now, checking the case $p=5$ separately, it suffices to prove that $p \mid a_p$.
But note that $a_p=2\sum_{k=1}^{(p-1)/2} \binom{p}{2k} 5^k2^{p-2k}$ is clearly divisible by $p$ as each of the binomial coefficients is.
Hence the result.
can someone explain me how did #Tintarn get $a_p=(2-\sqrt{5})^p+(2+\sqrt{5})^p-2^{p+1}$ thank you
Tintarn
21.03.2017 01:44
hmida99 wrote: can someone explain me how did #Tintarn get $a_p=(2-\sqrt{5})^p+(2+\sqrt{5})^p-2^{p+1}$ thank you Well, first of all this clearly is an integer as can be seen e.g. by the binomial expansion (the terms with $\sqrt{5}$ cancel). On the other hand, $(2-\sqrt{5})$ is a negative number of absolute value less than $1$. Therefore $(2-\sqrt{5})^p$ will also be of this form and hence the result.
hmida99
21.03.2017 03:20
Tintarn wrote: hmida99 wrote: can someone explain me how did #Tintarn get $a_p=(2-\sqrt{5})^p+(2+\sqrt{5})^p-2^{p+1}$ thank you Well, first of all this clearly is an integer as can be seen e.g. by the binomial expansion (the terms with $\sqrt{5}$ cancel). On the other hand, $(2-\sqrt{5})$ is a negative number of absolute value less than $1$. Therefore $(2-\sqrt{5})^p$ will also be of this form and hence the result. thank you how can we generelize this ?
Snakes
22.03.2017 01:00
^One possible generalization is to show that \[\left\lfloor \left(\sqrt{n^2+1}+n\right)^{p}-2n^p\right\rfloor\]is divisible by $2np(n^2+1).$ The proof is similar