MHB Show that 5 does not divide L_n

[evinda]

Hello! (Wave)

I want to show that $L_{n+1}+L_{n-1}=5 F_n$ for $n \geq 2$ and conclude that $5 \nmid L_n$ for $n \geq 1$.

I have tried the following:

$L_{n+1}+L_{n-1}=F_n+F_{n+2}+F_{n-2}+F_n=F_n+F_{n+1}+F_n+F_n-F_{n-1}+F_n=4F_n+F_{n+1}-F_{n-1}=4F_n+F_n+F_{n-1}-F_{n-1}=5F_n$.

But how do we deduce that $5 \nmid L_n$ for $n \geq 1$ ? (Thinking)

 

[I like Serena]

Hello! (Wave)

I want to show that $L_{n+1}+L_{n-1}=5 F_n$ for $n \geq 2$ and conclude that $5 \nmid L_n$ for $n \geq 1$.

I have tried the following:

$L_{n+1}+L_{n-1}=F_n+F_{n+2}+F_{n-2}+F_n=F_n+F_{n+1}+F_n+F_n-F_{n-1}+F_n=4F_n+F_{n+1}-F_{n-1}=4F_n+F_n+F_{n-1}-F_{n-1}=5F_n$.

But how do we deduce that $5 \nmid L_n$ for $n \geq 1$ ? (Thinking)

Hey evinda!

How about using induction?
Suppose 5 does not divide $L_k$ for k up to n, then... (Thinking)

 

[evinda]

Hey evinda!

How about using induction?
Suppose 5 does not divide $L_k$ for k up to n, then... (Thinking)

So is it right as follows?

Base case: $L_1=1$ and $5 \nmid 1$.

Induction hypothesis: We suppose that $5 \nmid L_k$, $\forall 1 \leq k \leq n$.

Inductive step:$L_{n+1}=5F_n-L_{n-1}$.

Suppose that $5 \mid L_{n+1}$. Then $5 \mid 5F_n-L_{n-1}$ and $5 \mid 5F_n$. So $5 \mid L_{n-1}$, contradiction.
Thus, $5 \nmid L_{n+1}$.

 

[I like Serena]

So is it right as follows?

Base case: $L_1=1$ and $5 \nmid 1$.

Induction hypothesis: We suppose that $5 \nmid L_k$, $\forall 1 \leq k \leq n$.

Inductive step:$L_{n+1}=5F_n-L_{n-1}$.

Suppose that $5 \mid L_{n+1}$. Then $5 \mid 5F_n-L_{n-1}$ and $5 \mid 5F_n$. So $5 \mid L_{n-1}$, contradiction.
Thus, $5 \nmid L_{n+1}$.

Suppose we ckeck the inductive step for n=1, aren't we referring to $L_0$ then?
But that is not defined is it? (Worried)6

 

[evinda]

Suppose we ckeck the inductive step for n=1, aren't we referring to $L_0$ then?
But that is not defined is it? (Worried)6

So we have to pick $n \geq 2$ ? (Thinking)

 

[I like Serena]

So we have to pick $n \geq 2$ ? (Thinking)

Yes. However we will not have proven the statement for $L_2$ will we?
The induction step will cover only $L_3$ and up, and the base case covers only $L_1$. (Sweating)

 

[evinda]

Yes. However we will not have proven the statement for $L_2$ will we?
The induction step will cover only $L_3$ and up, and the base case covers only $L_1$. (Sweating)

Do we include at the base case also the case $n=2$ ?

We get that $L_2=3$ and $5 \mid 5$.

We suppose that $5 \nmid L_k$, $\forall 2 \leq k \leq n$.

Then $L_{n+1}=5F_n-L_{n-1}$.

Let $5 \mid L_{n+1}$. Then we get that $5 \mid L_{n-1}$, contradiction.

So $5 \nmid L_{n+1}$.

 

[I like Serena]

Do we include at the base case also the case $n=2$ ?

Yes.

We get that $L_2=3$ and $5 \mid 5$.

Shouldn't that be $5 \nmid 3$? (Wondering)

We suppose that $5 \nmid L_k$, $\forall 2 \leq k \leq n$.

Shouldn't that be $\forall 1 \leq k \leq n$ with $n \ge 2$? (Wondering)

Then $L_{n+1}=5F_n-L_{n-1}$.

Let $5 \mid L_{n+1}$. Then we get that $5 \mid L_{n-1}$, contradiction.

So $5 \nmid L_{n+1}$.

 

[evinda]

Yes.
Shouldn't that be $5 \nmid 3$? (Wondering)
Shouldn't that be $\forall 1 \leq k \leq n$ with $n \ge 2$? (Wondering)

Nice... (Nod) Thank you! (Happy)