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Math Help - Proof involving primes

  1. #1
    Junior Member
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    Proof involving primes

    I need to show that if p is an odd prime not equal to 5, then either p^2-1 or p^2+1 is divisible by 10. It was trivial to show that they were both divisible by 2, so all that's left is to show one of them must be divisible by 5. I tried this a couple of different ways. First, I tried to assume one was not divisible by 5, and prove that the other one was. I also multiplied them together and tried to show that p^4-1 was divisible by 5, since there is a theorem that states if p is prime and p divides ab, then p divides a or p divides b. However, I got stuck using both methods. Any help would be greatly appreciated.
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  2. #2
    Super Member PaulRS's Avatar
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    By Fermat's Little Theorem a^{4}-1 is divisible by 5, if a is not divisible by 5

    Fermat's Little Theorem states that:
    Given a prime p and a natural number a coprime to p
    a^{p-1}\equiv{1}(\bmod.p)
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  3. #3
    Senior Member JaneBennet's Avatar
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    Consider the sequence of integers p-2,p-1,p,p+1,p+2. These are 5 consecutive integers, so exactly one of them is divisible by 5. Hence their product, which is p(p^2-1)(p^2-4), is divisible by 5. But p\ne5. Therefore (p^2-1)(p^2-4) is divisible by 5. Since 5 is prime, this means that one of p^2-1 and p^2-4 must be divisible by 5. In the latter case, p^2+1=(p^2-4)+5 is divisible by 5.

    And as youve pointed out, theyre both divisible by 2; hence whichever of them is divisible by 5 is divisible by 10.
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  4. #4
    Super Member PaulRS's Avatar
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    Examining every possible situation

    p^4-1=(p-1)\cdot(p+1)\cdot{(p^2+1)} (p is not 5)

    Now p=5\cdot{a}+b where a and b are natural numbers, 1\leq{b}<5

    So p^4-1=(5a+b-1)\cdot(5a+b+1)\cdot{(25a^2+5\cdot{ab}+b^2+1)}

    Now you have to show that one of this factors is divisible by five for b=1,2,3,4

    Indeed if b=1, then 5a+b-1=5a which is divisible by 5

    If b=2 then 25a+5ab+2+1= 25a+5ab+5 also divisible by 5

    If b=3 then 25a+5ab+3+1= 25a+5ab+10 divisible by 5

    If b=4 then 5a+b+1= 5a+5 divisible by 5

    We have shown that p^4-1 is divisible by 5 whenever p is not divisible by 5
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