Show that there are no a,b such that a^n-b^n | a^n + b^n

[evinda]

Hey again! (Mmm)

I want to show that there are no $a,b \geq 1, n \geq 2$ such that $a^n-b^n \mid a^n+b^n$.

That's what I have tried so far:

$$\exists k \in \mathbb{Z} \text{ such that } a^n+b^n=k(a^n-b^n)$$

Let $d=(a,b)$, $a_1=\frac{a}{d} \ , b_1=\frac{b}{d}$ , then $(a_1,b_1)=1$

So,we have :

$$d^n \cdot a_1^n+d^n \cdot b_1^n=k(d^n \cdot a_1^n-d^n \cdot b_1^n) \\ \Rightarrow a_1^n+b_1^n=k(a_1^n-b_1^n) \\ \Rightarrow (k-1) a^n=(k+1) b_1^n$$

But...how could I continue? (Thinking) (Thinking)

 

[I like Serena]

Hmm... this is SOLVED?
But... but... I have just figured it out! (Crying)

 

[evinda]

Hmm... this is SOLVED?
But... but... I have just figured it out! (Crying)

It doesn't matter that it is solved.. I would be glad to hear also your idea! (Happy)

 

[I like Serena]

$$(k-1) a_1^n=(k+1) b_1^n$$

Well... we already know that $(a_1, b_1) = 1$.
And since $k-1$ and $k+1$ are only 2 points apart, that must mean that they either have only the factor 2 in common, or they are co-prime.

So let's assume $k-1$ and $k+1$ are co-prime for now.
Then that can only mean that $(k-1) =b_1^n$ and $a_1^n=(k+1)$, since a prime factorization is guaranteed to be unique.

In turn that means that $a_1^n$ and $b_1^n$ are 2 points apart.
But... that is not possible if $(a_1, b_1)=1$ and $n>1$.

Contradiction! (Bandit)(Thinking)

We can apply the same line of reasoning if $(k-1,k+1)=2$.

It doesn't matter that it is solved.. I would be glad to hear also your idea! (Happy)

So what is your solution? (Wondering)

 

[evinda]

Well... we already know that $(a_1, b_1) = 1$.
And since $k-1$ and $k+1$ are only 2 points apart, that must mean that they either have only the factor 2 in common, or they are co-prime.

Is it known,that if we have two numbers that are two points apart,that they either have only the factor 2 in common, or they are co-prime? (Thinking) (Thinking)

So let's assume $k-1$ and $k+1$ are co-prime for now.
Then that can only mean that $(k-1) =b_1^n$ and $a_1^n=(k+1)$, since a prime factorization is guaranteed to be unique.

In turn that means that $a_1^n$ and $b_1^n$ are 2 points apart.
But... that is not possible if $(a_1, b_1)=1$ and $n>1$.

Contradiction! (Bandit)(Thinking)

We can apply the same line of reasoning if $(k-1,k+1)=2$.

Why if $(a_1,b_1)=1$,isn't it possible that $a_1^n \text{ and }b_1^n$ are 2 points apart??

So what is your solution? (Wondering)

We suppose that:

$$a^n-b^n \mid a^n + b^n (1)$$

Let $d=(a,b), a_1=\frac{a}{d}, b_1=\frac{b}{d} \Rightarrow a=a_1 \cdot d , b=b_1 \cdot d$

$$(1) \Rightarrow a_1^n d^n-b_1^n d^n \mid a_1^n d^n+b_1^n d^n \Rightarrow a_1^n-b_1^n \mid a_1^n+b_1^n$$

Now,we have that:

$$a_1^n-b_1^n \mid a_1^n+b_1^n \text{ and } a_1^n-b_1^n \mid a_1^n-b_1^n \text{ , so we conclude that } a_1^n-b_1^n \mid 2 b_1^n$$

As $(a_1,b_1)=1 \Rightarrow (a_1^n-b_1^n,b_1^n)=1$

So, $a_1^n-b_1^n \mid 2 b_1^n \Rightarrow a_1^n-b_1^n \mid 2$,but this cannot be,because:

We suppose that $a_1>b_1$,Then:

$$a_1 \geq b_1+1 \Rightarrow a_1^n \geq (b_1+1)^n \Rightarrow a_1^n-b_1^n \geq (b_1+1)^n-b_1^n=2^n-1>2, \text{ so it is a contradiction.}$$

 

[I like Serena]

Is it known,that if we have two numbers that are two points apart,that they either have only the factor 2 in common, or they are co-prime? (Thinking) (Thinking)

Suppose we have two numbers $u$ and $v$ with $(u,v)=d$.
Then there must be some distinct $m$ and $n$ such that $u=md$ and $v=nd$.
This implies that $|u-v| = |m-n| d \ge d$.
So two numbers are always at least their gcd apart.

If the gcd would be greater than 2 than the numbers have to be more than 2 points apart. (Thinking)

Why if $(a_1,b_1)=1$,isn't it possible that $a_1^n \text{ and }b_1^n$ are 2 points apart??

The smallest possible difference between $a_1^n$ and $b_1^n$ occurs when $a_1$ and $b_1$ are 1 point apart.
So let's assume $b_1=a_1+1$.
Then $b_1^n = (a_1+1)^n = a_1^n + n a_1 + ... + 1$.
So the difference between $a_1^n$ and $b_1^n$ is always greater than $na_1$, which is at least 2 when $n>1$ and $a_1 \ge 1$.

So the difference between $a_1^n$ and $b_1^n$ is greater than 2.

We suppose that:

$$a^n-b^n \mid a^n + b^n (1)$$

Let $d=(a,b), a_1=\frac{a}{d}, b_1=\frac{b}{d} \Rightarrow a=a_1 \cdot d , b=b_1 \cdot d$

$$(1) \Rightarrow a_1^n d^n-b_1^n d^n \mid a_1^n d^n+b_1^n d^n \Rightarrow a_1^n-b_1^n \mid a_1^n+b_1^n$$

Now,we have that:

$$a_1^n-b_1^n \mid a_1^n+b_1^n \text{ and } a_1^n-b_1^n \mid a_1^n-b_1^n \text{ , so we conclude that } a_1^n-b_1^n \mid 2 b_1^n$$

As $(a_1,b_1)=1 \Rightarrow (a_1^n-b_1^n,b_1^n)=1$

So, $a_1^n-b_1^n \mid 2 b_1^n \Rightarrow a_1^n-b_1^n \mid 2$,but this cannot be,because:

We suppose that $a_1>b_1$,Then:

$$a_1 \geq b_1+1 \Rightarrow a_1^n \geq (b_1+1)^n \Rightarrow a_1^n-b_1^n \geq (b_1+1)^n-b_1^n=2^n-1>2, \text{ so it is a contradiction.}$$

Nice! (Cool)

$$a_1^n-b_1^n \mid a_1^n+b_1^n \text{ and } a_1^n-b_1^n \mid a_1^n-b_1^n \text{ , so we conclude that } a_1^n-b_1^n \mid 2 b_1^n$$

Why can we conclude that? (Wondering)

 

[evinda]

Suppose we have two numbers $u$ and $v$ with $(u,v)=d$.
Then there must be some distinct $m$ and $n$ such that $u=md$ and $v=nd$.
This implies that $|u-v| = |m-n| d \ge d$.
So two numbers are always at least their gcd apart.

If the gcd would be greater than 2 than the numbers have to be more than 2 points apart. (Thinking)

Interesting! (Nerd)

The smallest possible difference between $a_1^n$ and $b_1^n$ occurs when $a_1$ and $b_1$ are 1 point apart.
So let's assume $b_1=a_1+1$.
Then $b_1^n = (a_1+1)^n = a_1^n + n a_1 + ... + 1$.
So the difference between $a_1^n$ and $b_1^n$ is always greater than $na_1$, which is at least 2 when $n>1$ and $a_1 \ge 1$.

So the difference between $a_1^n$ and $b_1^n$ is greater than 2.

I understand! (Smile)

Why can we conclude that? (Wondering)

$$a_1^n-b_1^n \mid a_1^n-b_1^n \text{ and } a_1^n-b_1^n \mid a_1^n+b_1^n, \text{ so } a_1^n-b_1^n \text{ divides also the difference of these numbers : } \\ a_1^n-b_1^n \mid a_1^n+b_1^n-a_1^n+b_1^n \Rightarrow a_1^n-b_1^n \mid 2b_1^n$$

 

[I like Serena]

$$a_1^n-b_1^n \mid a_1^n-b_1^n \text{ and } a_1^n-b_1^n \mid a_1^n+b_1^n, \text{ so } a_1^n-b_1^n \text{ divides also the difference of these numbers } $$

Why is that? (Sweating)

 

[evinda]

Why is that? (Sweating)

If a number divides two numbers,it divides also any linear combination of them..Or am I wrong?? (Thinking)

 

[I like Serena]

If a number divides two numbers,it divides also any linear combination of them..Or am I wrong?? (Thinking)

Hmm... maybe... if so, why would that be? (Wondering)

 

[evinda]

Hmm... maybe... if so, why would that be? (Wondering)

$$a_1^n-b_1^n \mid a_1^n-b_1^n \text{ and } a_1^n-b_1^n \mid a_1^n+b_1^n , \text{ so } a_1^n-b_1^n \text{ divides any linear combination of } a_1^n-b_1^n \text{ and } \\ a_1^n+b_1^n \text{ so it divides also their difference: } \\ a_1^n-b_1^n \mid (a_1^n+b_1^n)-(a_1^n-b_1^n) \Rightarrow a_1^n-b_1^n \mid a_1^n+b_1^n-a_1^n+b_1^n \Rightarrow a_1^n-b_1^n \mid 2b_1^n $$ (Nerd) (Smirk)

 

[I like Serena]

$$a_1^n-b_1^n \mid a_1^n-b_1^n \text{ and } a_1^n-b_1^n \mid a_1^n+b_1^n , \text{ so } a_1^n-b_1^n \text{ divides any linear combination of } a_1^n-b_1^n \text{ and } \\ a_1^n+b_1^n \text{ so it divides also their difference: } \\ a_1^n-b_1^n \mid (a_1^n+b_1^n)-(a_1^n-b_1^n) \Rightarrow a_1^n-b_1^n \mid a_1^n+b_1^n-a_1^n+b_1^n \Rightarrow a_1^n-b_1^n \mid 2b_1^n $$ (Nerd) (Smirk)

Okay! (Mmm)