Show that $(a, b) =1$ and $a> b$ then $(a-b,\frac{a^m-b^m}{a-b})=(a-b,m)$

Question

Exercise complicated Number theory Show that $\left(a, b\right) =1$ and $a> b$ then $$\left(a-b,\frac{a^m-b^m}{a-b}\right)=(a-b,m)$$I tried some, but what I got was.$\;\;\;\;$I tried some, but what I got was $$\left(a-b,\frac{a^m-b^m}{a-b}\right)=(a-b,a^{m-1}+a^{m-2}b+\;...+ab^{m-2}+b^{m-1})$$

Can this help?

@StefanH. Comes from the fact that $(a,b)$ is the ideal generated by the GCD, as far as I know. Out of curiosity, what notation are you used to (hopefully not ggT)? — Daniel Fischer, Aug 01 '13 at 21:33
@DanielFischer Before coming to Math.SE, I had only seen $\gcd(a, b)$, but I've gotten used to $(a, b)$ since then... — apnorton, Aug 01 '13 at 21:38
@DanielFischer: Exactly, we used ggT, and kgV for the least common multiple :-) I like how $(a,b)$ comes from the ideal $(a,b)$ being generated by the gcd. But it's only true in PID, as far as I remember. — Stefan Hamcke, Aug 01 '13 at 21:39
Related: http://math.stackexchange.com/questions/452510/we-need-to-show-gcd-is-1 — lab bhattacharjee, Aug 02 '13 at 07:40

Pedro · Answer 1 · 2013-08-01T21:38:36.753

4

Let's consider the seemingly simpler case. We want to show that for any $x$, $$\left(x-1,\frac{x^{m}-1}{x-1}\right)=(x-1,m)$$

Hint Modulo $x-1$, $$x\equiv 1\implies x^j\equiv 1$$

You can do the very same for your problem, since $\mod a-b$, $a\equiv b\implies a^j\equiv b^j$ for any $j$. Thus, $\mod a-b$ $${a^{m - 1}} + {a^{m - 2}}b + \; \cdots + a{b^{m - 2}} + {b^{m - 1}} \equiv {a^{m - 1}} + {a^{m - 1}} + \cdots + {a^{m - 1}} + {a^{m - 1}} = m{a^{m - 1}}$$

whence the $\gcd$ equals $(a-b,ma^{m-1})$. Can you finish?

If you want to avoid modular arithmetic, recall that $(a,b)=(b,r)$ where $r$ is the remainder in $a=qb+r$. Now, for each $1\leq i\leq m-1$ we have $${a^{m - i}}{b^i} - {a^m} = {a^{m - i}}\left( {{b^i} - {a^i}} \right)$$ and the RHS is divisible by $a-b$, so we can use the division algorithm to make each $a^{m-i}b^i$ into $a^m$ in the RHS, which is what I did above.

edited Aug 01 '13 at 21:38

answered Aug 01 '13 at 21:24

Pedro

122,002

Sorry, I have not seen modular arithmetic has otherwise?$$$$I know taste $$(a-1,\frac{a^{m}-1}{a-1})=(a-1,m)$$ – benjamin_ee Aug 01 '13 at 21:30
Please ... No modular arithmetic :'( – benjamin_ee Aug 01 '13 at 21:36
@marcelolpjunior I have added something using only the division algorithm. Eventually you will have to use congruences... we are the Integers $\mod n$. – Pedro Aug 01 '13 at 21:39

Show that $(a, b) =1$ and $a> b$ then $(a-b,\frac{a^m-b^m}{a-b})=(a-b,m)$

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