Induction on GCD problem

Question

This is a two part question

Given $\gcd(a,b) = 1$

consider $$\gcd \left( \frac{a^n - b^n }{a-b}, a- b\right) $$

It appears that the value of this is always equal to $n$ or $1$. How to prove it?

And with that show

$$\gcd \left( \frac{a^n - b^n }{a-b}, a- b\right) = \gcd \left( n \frac{a^{n-1} - b^{n-1}}{a-b}, a- b \right) $$

I began with tackling the first part

I trie to use an Inductive Proof at first by noting:

$$\frac{a^{n+1} - b^{n+1}}{a-b} = a\frac{a^{n} - b^{n}}{a-b} + b^{n} = b\frac{a^{n} - b^{n}}{a-b} + a^{n} $$

But this didn't pan out any sort of solution since it doesn't become immediately apparent that $n+1$ divides this new quantity or some linear combination of this quantity with $a - b$

I could attempt a combinatorial approach and ty to use the binomial theorem I was thinking. But no apparent way to match terms become apparent.

What should I do for the first part?

http://math.stackexchange.com/questions/247146/how-to-show-that-gcd-left-an-bn-over-a-b-a-b-right-gcdn-dn-1-a-b — lab bhattacharjee, Aug 05 '14 at 17:13
Is this true? Take a= 6, b= 3 and n=2. Then gcd (9,3) = 3 which is neither 2 nor 1! — wannadeleteacct, Aug 05 '14 at 17:19
@Manasi not sure how you came to that conclusion. What is your choice of exponent $n$? — Sidharth Ghoshal, Aug 05 '14 at 17:22
@Manasi It's true if $,d = (a,b) = 1.,$ Otherwise it involves $,d,$ also, see the above link, or follow my hint. This has been asked many times before. — Bill Dubuque, Aug 05 '14 at 17:22
@BillDubuque : Yeah, your statement seems like standard matter which I need to verify but the OP has not mentioned anything about the gcd being only 1! — wannadeleteacct, Aug 05 '14 at 17:25
@BillDubuque True if $d=(a,b)=1$, then that link shows it is $\gcd(n,a-b)$, not "$1$ or $n$." That's only true when $n$ is prime. — Thomas Andrews, Aug 05 '14 at 17:26
@Manasi Yes, but the statement just isn't true in general, so what the OP wants is wrong. :) — Thomas Andrews, Aug 05 '14 at 17:26
@Thomas Yes, that's correct. I misremembered what I read in the question. — Bill Dubuque, Aug 05 '14 at 17:28
Alright People... You got me! They were supposed to be coprime! :) — Sidharth Ghoshal, Aug 05 '14 at 17:29
@ThomasAndrews: yeah,agreed! I referred to BD's statement and not OP's. — wannadeleteacct, Aug 05 '14 at 17:29
@frogeyedpeas : Hey, I am NOT* a gentleman (period) and will not be for the rest of my life... — wannadeleteacct, Aug 05 '14 at 17:30
@ThomasAndrews Sometimes it's annoying when your memory "auto-corrects" errors. Have you ever had that happen? It reminds me of the classical studies on chess and cognition (DeGroot?) where they presented positions to grandmasters for memorization. They were able to reconstruct meaningful positions with high accuracy, but did very poorly with random boards (their memory often "corrected" the random boards into something more meaningful), iirc (w/o auto-error-correction!) — Bill Dubuque, Aug 05 '14 at 17:40

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

1

Hint $\,\ \big(x\!-\!a,\frac{f(x)-f(a)}{x-a}\!\big) = (x\!-\!a,\:f'(a))\,\ $ by $\ \frac{f(x)-f(a)}{x-a} \equiv f'(a)\pmod{\!x\!-\!a}\ $ for $\ f(x)\in \mathbb Z[x]$

For further details see my post here, which elaborates on how this result is a number-theoretical analog of a well-known result about functions (polynomials), viz. about multiplicity of roots.

edited Apr 13 '17 at 12:21

Community

1

answered Aug 05 '14 at 17:19

Bill Dubuque

272,048

I think this is probably far more advanced than what I was expected to use, but damn... thats a very interesting identity there. I don't think I've seen calculus and number theory intersect that curiously before – Sidharth Ghoshal Aug 05 '14 at 17:25
@frogeyedpeas It uses only an infinitesimal amount of calculus, viz. differentiation of polynomials, which can actually be defined purely algebraically by the above congruence. – Bill Dubuque Aug 05 '14 at 17:26

score 0 · Answer 2 · answered Aug 05 '14 at 18:17

If $D$ is the GCD, then $a\equiv b\pmod D$, and therefore:

$$0\equiv \frac{a^n-b^n}{a-b} = \sum_{k=0}^{n-1} a^kb^{n-1-k} \equiv \sum_{k=0}^{n-1} a^{n-1} = na^{n-1}\pmod D$$

So $D\mid na^{n-1}$. Since $D\mid a-b$ and $(a-b,a)=1$, you have: $$D\mid n$$

On the other hand, it is easy to show the converse, that if $d\mid a-b$ and $d\mid n$ then $d\mid \frac{a^n-b^n}{a-b}$.

So $D=\gcd(n,a-b)$.

Induction on GCD problem

2 Answers2