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

  • MHB
  • Thread starter evinda
  • Start date
In summary, the conversation discusses a mathematical problem where it is asked to prove that there are no values for $a,b \geq 1$ and $n \geq 2$ that satisfy the equation $a^n-b^n|a^n+b^n$. The conversation goes through various attempts at solving the problem, including supposing $(a,b)=d$ and proving that $(a,b)=1$, as well as rearranging the equation to find a contradiction.
  • #1
evinda
Gold Member
MHB
3,836
0
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, $\frac{a^n+b^n}{a^n-b^n} \in \mathbb{Z}$.
$\frac{a^n+b^n}{a^n-b^n}=1+\frac{2l^n}{k^n-l^n}$
 
Mathematics news on Phys.org
  • #2
evinda said:
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?
 
  • #3
I like Serena said:
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,$b^n|(k-1)a^n \Rightarrow b^n|k-1$..
 
  • #4
evinda said:
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:
$$(k+1)=a^n$$
 
  • #5
I like Serena said:
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:
 
  • #6
evinda said:
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).
 
  • #7
I haven't followed this all the way through, but there is one other possibility for $k|1$, which is:

$k = -1$

contradicting $a,b \geq 1$.
 

FAQ: There are no a,b such that a^n-b^n|a^n+b^n

What does the statement "a^n-b^n|a^n+b^n" mean?

The statement means that for any positive integer n, there are no values of a and b that can satisfy the equation a^n-b^n|a^n+b^n, or in other words, there are no values of a and b that can evenly divide the sum of their nth powers.

Is this statement always true?

Yes, this statement is always true. It is a mathematical theorem known as the Fermat's Last Theorem, which was proven by Andrew Wiles in 1994.

Can you provide an example of values for a and b that do not satisfy the statement?

One example is a=2 and b=3. When n=2, we have 2^2-3^2=4-9=-5, and 2^2+3^2=4+9=13. Since -5 does not evenly divide 13, this example does not satisfy the statement.

Are there any exceptions to this statement?

No, there are no exceptions to this statement. It has been proven to hold true for all positive integers n.

What implications does this statement have in mathematics?

This statement has significant implications in number theory and algebra. It has also led to the development of new mathematical techniques and theorems, and has inspired further research in this area.

Similar threads

Replies
1
Views
1K
Replies
4
Views
2K
Replies
21
Views
3K
Replies
11
Views
3K
Replies
10
Views
1K
Replies
3
Views
1K
Replies
1
Views
760
Back
Top