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If [tex]p[/tex] is a prime that divides [tex]a - b[/tex], then show that [tex]p^2[/tex] divides [tex]a^p - b^p[/tex], where [tex]a, b \in \mathbb{Z}[/tex].

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- Thread starter math_grl
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- #1

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If [tex]p[/tex] is a prime that divides [tex]a - b[/tex], then show that [tex]p^2[/tex] divides [tex]a^p - b^p[/tex], where [tex]a, b \in \mathbb{Z}[/tex].

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- #3

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Then

[itex]a^p - b^p = (b + np)^p - b^p = \sum_{k=1}^p \binom{p}{k} n^k p^k b^{p-k}[/itex]

Let's now consider each of the terms. For k = p, then it's p

Thus, all the terms are divisible by p

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That's not the case. That is because binomial(n,k) for 1 <= k <= n-1 is always evenly divisible by n if n is a prime. For n composite, that need not be the case.

Counterexample: binomial(4,2) = 6, which is not divisible by 4.

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In general for an odd prime: (a^p-b^P) =(a-b)(a^(p-1)+(a^p-2)b +a^(p-3)(b^2)+++(a^0)b^(p-1). Since there are p terms in the second expression and a==b Mod p was given, the second term goes to p(a^(p-1))==0 Mod p.

In fact, for the above situation, we need only an odd integer, not a prime.

In fact, for the above situation, we need only an odd integer, not a prime.

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- #6

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a

4 = 2

ETA: robert Ihnot's argument will work for p being

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ETA: robert Ihnot's argument will work for p being any integer >= 2.

Right that's what I thought, which is why I got confused when he said that we only need odd integers.

- #9

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Forgive me if I am off here, but since we know that p divides a - b, then isn't this question essentially equivalent to stating, by Fermat's Little Theorem:

Can...

p (mod 2)*divide* a (mod p) - b (mod p)?

p divides n*mod (p) for any n

p divides a - b

and p (mod 2) , which equals (or is congruent to..) 1, clearly divides a (mod p) - b (mod p)

?

Can...

p (mod 2)

p divides n*mod (p) for any n

p divides a - b

and p (mod 2) , which equals (or is congruent to..) 1, clearly divides a (mod p) - b (mod p)

?

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I don't know exactly what you mean, but I think you're saying that if gcd(a,p)=gcd(b,p)=1, then by fermatForgive me if I am off here, but since we know that p divides a - b, then isn't this question essentially equivalent to stating, by Fermat's Little Theorem:

Can...

p (mod 2)dividea (mod p) - b (mod p)?

p divides n*mod (p) for any n

p divides a - b

and p (mod 2) , which equals 1, clearly divides a (mod p) - b (mod p)

?

[tex]a^p\equiv\ a\ mod\ p\text{, and }b^p\equiv\ b\ mod\ p\ ==>\ a^p-b^p\equiv\ a-b\ mod\ p[/tex]

it is always true only if p is prime or a carmichael number

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it is always true only if p is prime or a carmichael number

I believe the assumption here is that p is prime. Honestly, however, I was not necessarily thinking in terms of gcd, but, rather, in the case of p being prime, then the above statements ought to cover all cases (i.e. permutations) of how a, b, and p can combine.

If I am not off-base here, one need only think about how p can "interact" with a - b, (a - b)(mod p) = n (mod p) & p(mod 2) can "interact" with a(mod p) - b(mod p).

- RF

P.S. Sorry, but the Latex editor does not work with my computer system.

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- #12

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of coursely if [tex]p|a-b\ ==>\ a^p-b^p\equiv\ 0\equiv\ (a-b)^P\ mod\ p[/tex]

if [tex]p \geq 2,\ p|a-b ==> p^2|p^p|(a-b)^p[/tex]

now you can conclude that [tex]p^2|a^p-b^p[/tex] with some observation

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if [tex]p \geq 2,\ p|a-b ==> p^2|p^p|(a-b)^p[/tex]

p > or = to 2 is unnecessary to state since all p, by common definition, are > or = to 2.

Thank you, al mahed, for presenting in clear manner that which I was trying to get across in a less clear manner.

- RF

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We know that (a-b)^p ==a^p-b^p Mod p. That is not the same as showing that (a-b)^p ==a^p-b^p Mod p^p, as you are suggesting above.

It is not the case that 10^7-3^7 is divisble by 7^7, in fact, 49 is the highest 7 divisor.

It is not the case that 10^7-3^7 is divisble by 7^7, in fact, 49 is the highest 7 divisor.

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Hi Robert (see below post),

I realize I completely reversed terms here in rather dyslexic manner and so, in response to your post, and in the interests of not virally spreading misinformation, I am simply replacing what I posted previously with my initial surmise, which I believe was of value.

I realize I completely reversed terms here in rather dyslexic manner and so, in response to your post, and in the interests of not virally spreading misinformation, I am simply replacing what I posted previously with my initial surmise, which I believe was of value.

Forgive me if I am off here, but since we know that p divides a - b, then isn't this question essentially equivalent to stating, by Fermat's Little Theorem:

Can...

p (mod 2)dividea (mod p) - b (mod p)?

p divides n*mod (p) for any n

p divides a - b

and p (mod 2) , which equals (or is congruent to..) 1, clearly divides a (mod p) - b (mod p)

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By Fermat's Litle Theorm, Modulus p. Fermat's Little Theorem says nothing about modulus P^2.

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Prove that if a ^ p + b ^p = 0 (mod p) then a ^ p + b ^p = 0 (mod p ^ 2)

The logic is exactly similar i guess except that p as to be odd & greater than 2. Can somebody please look into & give me clues to solving the problem i posted a couple of days back with the subject line similar to "reduced residues" involving the Euler phi function? I would really appreciate it.

Thanks in advance,

--

Sachin

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Edit

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Fermat don't say nothing about p^2 modulo, but euler do, where did you see modulus p^p in my post??

generally, if I'm not mistaken, [tex]a^{p^e}-b^{p^e}\equiv\ (a-b)^{p^e}\ mod\ p^e[/tex]

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of coursely if [tex]p|a-b\ ==>\ a^p-b^p\equiv\ 0\equiv\ (a-b)^P\ mod\ p[/tex]

if [tex]p \geq 2,\ p|a-b ==> p^2|p^p|(a-b)^p[/tex]

now you can conclude that [tex]p^2|a^p-b^p[/tex] with some observation

Right above!

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Fermat don't say nothing about p^2 modulo, but euler do, where did you see modulus p^p in my post??

generally, if I'm not mistaken, [tex]a^{p^e}-b^{p^e}\equiv\ (a-b)^{p^e}\ mod\ p^e[/tex]

I would be helpful to give a reference or show a proof.

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[tex]2^3+5^3 \equiv133\equiv7 Mod 9.[/tex]

But [tex] 7^3\equiv 1 Mod 9.[/tex]

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my source is

http://planetmath.org/encyclopedia/FrobeniusAutomorphism4.html [Broken]

http://planetmath.org/encyclopedia/FrobeniusAutomorphism4.html [Broken]

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- #24

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Fermat's Little Theorm can be shown by induction: 1^p ==1, (x+1)^p ==x^p+1, but this is because all the intermediate terms in the binomial expansion contain p, a prime, which does not divide out. BUT THIS DOES NOT CARRY OVER TO P^2.

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Fermat's Little Theorm can be shown by induction: 1^p ==1, (x+1)^p ==x^p+1, but this is because all the intermediate terms in the binomial expansion contain p, a prime, which does not divide out. BUT THIS DOES NOT CARRY OVER TO P^2.

yes, agreed, I know that, I didn't say the proof was complete, in fact I thought it would take only a few steps more, but that's not the case (I was trying to complete the proof here using what I've done, but gave me trouble, it is more complicated, I think)

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