If $n$ is an odd prime, and $a$ and $b$ are co-prime, and $a+b$ is not a multiple of $n$, prove that $\frac{a^n+b^n}{a+b}$ is co-prime to $a+b$.

Question

How does even even begin to prove this? I have proved that if $a+b$ is a multiple of $n$, then so is $a^n+b^n$, using FLT, but how to progress next eludes me.

If $n$ is odd, there is an explicit expression for $(a^n + b^n)/(a + b)$. Then you can argue modulo $a + b$. — user49640, Apr 19 '17 at 04:37

score 2 · Accepted Answer · answered Apr 19 '17 at 04:41

2

Note that the problem does not make sense if $n=2$, as $\dfrac{a^2+b^2}{a+b}$ is not generally an integer. So we assume $n\ne2$, and hence $n$ is odd.

If $\dfrac{a^n+b^n}{a+b}$ and $a+b$ have a common factor then they have a common prime factor $p$. So $$p\mid a+b\quad\hbox{and}\quad p\mid a^{n-1}-a^{n-2}b+a^{n-3}b^2-\cdots+b^{n-1}\ .$$ This gives $b\equiv-a\pmod p$ and so the second expression above can be written $$p\mid na^{n-1}\ .$$ Now $p\mid a+b$ and $n\not\mid a+b$, so $p\ne n$, so $p\mid a$ and hence $p\mid b$. Since $a,b$ are coprime this is a contradiction.

Hence $\dfrac{a^n+b^n}{a+b}$ and $a+b$ are coprime.

answered Apr 19 '17 at 04:41

David

82,662

by the way, is my method correct? I doubt its working! – vidyarthi Apr 19 '17 at 04:53
@vidyarthi I think it's correct. A little hard to follow, perhaps... – David Apr 19 '17 at 05:09
oh, thanks. My doubt was regarding whether $\frac{kn}{a+b}$ was an integer, but I now came to its evidence using symmetric polynomials. – vidyarthi Apr 19 '17 at 05:19

score 1 · Answer 2 · answered Apr 19 '17 at 04:41

1

When $n=2$ you have $\frac{a^n+b^n}{a+b}=a-b+\frac{2b}{a+b}$ which is not an integer in general.

So I assume $n=2k+1$ is an odd prime should be part of the statement.

In this case $$\frac{a^n+b^n}{a+b}=a^{n-1}-a^{n-2}b+...+b^{n-1}\\= a^{n-1}+a^{n-2}b-2a^{n-2}b-2a^{n-3}b^2+3a^{n-3}b^2+3a^{n-4}b^3-...\\ =(a+b)(a^{n-2}-2a^{n-3}b+3a^{n-4}b^-...)+nb^{n-1}$$

Therefore $$gcd(\frac{a^n+b^n}{a+b}, a+b)=gcd(nb^{n-1}, a+b)=gcd(n, a+b)$$ with the last equality following from $gcd(a+b, b^{n-1})=1$.

answered Apr 19 '17 at 04:41

N. S.

132,525

what identity did you use for $a^n+b^n$? thanks. – miniparser Apr 20 '17 at 18:14
@miniparser $$a^{2k+1}+b^{2k+1}=(a+b)(a^{2k}-a^{2k-1}b+a^{2k-2}b^2-a^{2k-3}b^3+....+b^{2k})$$ – N. S. Apr 20 '17 at 19:42
what is it called? – miniparser Apr 20 '17 at 20:38
@miniparser It probably doesn't have a name, it is just a well known identity. – N. S. Apr 20 '17 at 20:58
@miniparser thanks for adding the comment! I think my answer was incorrect – vidyarthi Apr 22 '17 at 12:04

If $n$ is an odd prime, and $a$ and $b$ are co-prime, and $a+b$ is not a multiple of $n$, prove that $\frac{a^n+b^n}{a+b}$ is co-prime to $a+b$.

2 Answers2