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Quick problem?

  1. Jan 24, 2011 #1
    I was just browsing for some small problems the other day and came across this problem and I am unsure if it should be obvious and have a quick answer. In any case, I couldn't figure it out.

    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].
  2. jcsd
  3. Jan 24, 2011 #2
    Nevermind I figured it out using binomial theorem. But it looked like to me that [tex]p[/tex] doesn't necessarily have to be prime. Am I correct?
  4. Jan 24, 2011 #3
    I have a solution. Let a = b + n*p

    [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 pp, and since p >= 2, this is evenly divisible by p2. For the remaining terms, 1 <= k <= p-1, and from that inequality, the binomial coefficient is divisible by p, pk is also divisible by p, and thus, all these terms are divisible by p2.

    Thus, all the terms are divisible by p2, implying that the whole expression is also.
  5. Jan 24, 2011 #4
    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.
  6. Jan 24, 2011 #5
    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.
    Last edited: Jan 25, 2011
  7. Jan 25, 2011 #6
    robert, i like your expansion of a^p - b^p....but I'm afraid I don't understand why exactly it has to be an odd integer. Are you saying that there will not be p terms on the right hand side something if p is even? You started with p odd prime, does that mean the original question does not work for p = 2?
  8. Jan 25, 2011 #7
    It's very easy to test. Let a = b + 2*k. I find

    a2 - b2 = (b+2*k)2 - b2 = 4*k2 + 4*k*b = 4*k*(k+b)

    4 = 22, of course.

    ETA: robert Ihnot's argument will work for p being any integer >= 2.
  9. Jan 25, 2011 #8
    Right that's what I thought, which is why I got confused when he said that we only need odd integers.
  10. Jan 25, 2011 #9
    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:


    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)

    Last edited: Jan 25, 2011
  11. Jan 25, 2011 #10
    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 fermat

    [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
  12. Jan 25, 2011 #11
    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.
    Last edited: Jan 25, 2011
  13. Jan 25, 2011 #12
    you can use the identity [tex]a^p-b^p\equiv (a-b)^P\ mod\ p[/tex]

    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
  14. Jan 27, 2011 #13
    A trivial point, but...
    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
  15. Jan 27, 2011 #14
    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.
    Last edited: Jan 27, 2011
  16. Jan 27, 2011 #15
    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.

    Last edited: Jan 28, 2011
  17. Jan 27, 2011 #16
    p^2 divides p^(a - b), which is congruent to p^a - p^b.

    By Fermat's Litle Theorm, Modulus p. Fermat's Little Theorem says nothing about modulus P^2.
  18. Jan 28, 2011 #17
    This is a weird coincidence since i was trying to solve a similar problem which is:

    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,
  19. Jan 28, 2011 #18
    Last edited: Jan 28, 2011
  20. Jan 28, 2011 #19
    Robert, what you want to prove is: [tex]a^p\equiv\ b^p\ mod\ p^2[/tex]

    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]
  21. Jan 28, 2011 #20
    Right above!
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