Re: Abstract Algebra Question

On 12 abr, 18:05, "Jack Schmidt" <Jack.Schmidt.SciM...@xxxxxxxxx>
wrote:
On Apr 11, 10:41 am, tlici...@xxxxxxxxx wrote:

This is from the sample GRE subject test:

What is the largest m that can divide (p^4) - 1 for all p = primes
greater than 5 ? Use Fermat's little theorem somehow?

Since x^4=1 mod 16 for all odd x, y^4=1 mod 3 for all y not divisible
by 3, and z^4=1 mod 5 for all z not divisible by 5, 2^4*3*5 = 240
divides p^4-1 for all p relatively prime to 2*3*5. On the other hand,
gcd(7^4-1,11^4-1) = 240, so there is no larger number dividing all
p^4-1 for primes p, p>5.
**************************************************
p^4 - 1 = (p-1)(p+1)(p^2 + 1) ==> all these factors are even numbers,
and exactly one of the first two is divisible by three, and exactly
one of the first two, again, divisible by at least 2^2 = 4 ==> the
number 2^4*3 = 48 divides p^4 - 1 for any prime p > 5.
Since 11^1 - 1 = (2^4)*3*5*61, 48 is the largest such divisor for all
primes p.
Regards
Tonio

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