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Number Theory There are no a,b such that a^n-b^n|a^n+b^n

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Hello again!!! (Blush)

How can I prove that there are no $a,b \geq 1,n \geq 2$ such that: $a^n-b^n|a^n+b^n$ ?
I have tried the following..Could you tell me if it is right so far and how I could continue? :rolleyes:

We suppose that $(a,b)=d$.We know that $(\frac{a}{d},\frac{b}{d})=1$.
$d|a,d|b \Rightarrow a=k \cdot d,b=l \cdot d ,k,l \in \mathbb{Z}$
Then we suppose that $a^n-b^n|a^n+b^n$.Then, $\frac{a^n+b^n}{a^n-b^n} \in \mathbb{Z}$.
$\frac{a^n+b^n}{a^n-b^n}=1+\frac{2l^n}{k^n-l^n}$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Hello again!!! (Blush)

How can I prove that there are no $a,b \geq 1,n \geq 2$ such that: $a^n-b^n|a^n+b^n$ ?
I have tried the following..Could you tell me if it is right so far and how I could continue? :rolleyes:

We suppose that $(a,b)=d$.We know that $(\frac{a}{d},\frac{b}{d})=1$.
$d|a,d|b \Rightarrow a=k \cdot d,b=l \cdot d ,k,l \in \mathbb{Z}$
Then we suppose that $a^n-b^n|a^n+b^n$.Then, $\frac{a^n+b^n}{a^n-b^n} \in \mathbb{Z}$.
$\frac{a^n+b^n}{a^n-b^n}=1+\frac{2l^n}{k^n-l^n}$
Hullo!! (Mmm)

Let's take this in a different direction.

Suppose $a^n-b^n|a^n+b^n$, then there is a $k$ such that $a^n+b^n=k(a^n-b^n)$.

Let $(a,b)=d$, then we can divide the expression by $d$, leaving us with an identical expression.
So without loss of generality we can assume that $(a,b)=1$. (Why? :confused:)

Now rearrange the expression to:
$$(k-1)a^n = (k+1)b^n$$
What can you say about $(k-1)$ and $(k+1)$ knowing that $a$ and $b$ have no factors in common?
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Hullo!! (Mmm)

Let's take this in a different direction.

Suppose $a^n-b^n|a^n+b^n$, then there is a $k$ such that $a^n+b^n=k(a^n-b^n)$.

Let $(a,b)=d$, then we can divide the expression by $d$, leaving us with an identical expression.
So without loss of generality we can assume that $(a,b)=1$. (Why? :confused:)

Now rearrange the expression to:
$$(k-1)a^n = (k+1)b^n$$
What can you say about $(k-1)$ and $(k+1)$ knowing that $a$ and $b$ have no factors in common?
I don't really know how to show that $(a,b)=1$.That's what I have tried to do it:
Let $(a,b)=d>1$,so it has a prime divisor,$p$.
$p|d,d|a,d|b \Rightarrow p|a,p|b$
But how from these relations can I conclude that $(a,b)=1$ ?

Supposing that $(a,b)=1 \Rightarrow (a^n,b^n)=1$.
$(k-1)a^n=(k+1)b^n \Rightarrow a^n| (k+1)b^n$ and since $(a^n,b^n)=1$ we get that $a^n|k+1$.Also,$b^n|(k-1)a^n \Rightarrow b^n|k-1$..
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I don't really know how to show that $(a,b)=1$.That's what I have tried to do it:
Let $(a,b)=d>1$,so it has a prime divisor,$p$.
$p|d,d|a,d|b \Rightarrow p|a,p|b$
But how from these relations can I conclude that $(a,b)=1$ ?
Let $(a,b)=d$.
Define a'=(a/d) and b'=(b/d). Then (a',b')=1.

Then, since we are assuming that $a^n+b^n=k(a^n-b^n)$, we have that:
$$(a/d)^n+(b/d)^n=k((a/d)^n-(b/d)^n)$$
$$(a')^n+(b')^n=k((a')^n-(b')^n)$$

So if we can prove the assumption does not hold for a' and b' with (a',b')=1, it will also not hold for a and b with (a,b)=d.
In other words, it suffices to prove it does not hold for a' and b' with (a',b')=1.

For ease of notation, we can continue the proof with (a,b)=1 with the understanding that we actually mean a' and b'.


Supposing that $(a,b)=1 \Rightarrow (a^n,b^n)=1$.
$(k-1)a^n=(k+1)b^n \Rightarrow a^n| (k+1)b^n$ and since $(a^n,b^n)=1$ we get that $a^n|k+1$.Also,$b^n|(k-1)a^n \Rightarrow b^n|k-1$..
We're getting there...

Slightly sharper: a and b have no prime factors in common.
Therefore all prime factors in b have to be in (k-1).
Moreover, the powers of each of those prime factors must also be in (k-1).
It gets even better, we are guaranteed to have that:
$$(k-1)=b^n$$
Has to be, because the (powers of the) prime factors in $b^n$ can not go anywhere else.

Similarly:
$$(k+1)=a^n$$
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Let $(a,b)=d$.
Define a'=(a/d) and b'=(b/d). Then (a',b')=1.

Then, since we are assuming that $a^n+b^n=k(a^n-b^n)$, we have that:
$$(a/d)^n+(b/d)^n=k((a/d)^n-(b/d)^n)$$
$$(a')^n+(b')^n=k((a')^n-(b')^n)$$

So if we can prove the assumption does not hold for a' and b' with (a',b')=1, it will also not hold for a and b with (a,b)=d.
I I found $(a')^n=\frac{k+1}{k-1}(b')^n$,replaced it at the relation $a^n-b^n | a^n+b^n$
and found $k|1$.But that does not give us a contradiction,right?
But,could,we also do it in that way?? :confused:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I I found $(a')^n=\frac{k+1}{k-1}(b')^n$,replaced it at the relation $a^n-b^n | a^n+b^n$
and found $k|1$.But that does not give us a contradiction,right?
But,could,we also do it in that way?? :confused:
I've lost you.
But if $k|1$ that means that $k=1$, meaning your expression $(a')^n=\frac{k+1}{k-1}(b')^n$ is ill defined (which is pretty close to a contradiction).
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
I haven't followed this all the way through, but there is one other possibility for $k|1$, which is:

$k = -1$

contradicting $a,b \geq 1$.