Why is 48 the largest integer that will evenly divide p^4 - 1 where p is a prime greater than 3?

Answer 1

WHAT BAIROG is talking about is not a definite number,
(p^4-1)/2 changes for every p.

I understand you are asking why is 48 the number that ALWAYS divides p^4-1. Right?
That 48 divides any p^4-1 I will prove.
That no other number divides any p^4-1 given any p prime is harder and at the moment I don't know.

p^4-1=(p^2+1)(p^2-1)=(p^2+1)(p-1)(p+1)

Now:
If p is prime it must be odd. So p-1 is divisible by 2, p+1 is also divisible by 2, that means so far p^4-1 is divisible by 4.
If p is prime it is not divisible by 3 so p=3k+1 or 3k +2. Again in this case p-1 is 3k or p+1=3(k+1) so (p-1)(p+1) must be divisible by 3. So p^4-1 must be divisible by 3*4=12
Also if p is odd then p^2 is odd and p^2+1 is even. So p^4-1 is divisible by 12*2=24.
But there is one more thing. p is prime so it is not divisible by 4.
So p is 4k+1 or 4k+2 or 4k+3. If it is 4k+2 it would be even, so not possible. If p=4k+1 then p-1=4k, divisible by 4. But we already considered it divisible by 2, so instead of 24 p will be divisible by 48. If p=4k+3 then same story with p+1.
So p^4-1 must be divisible by 48.

Answer 2

Answering this question requires you to know quite a bit of number theory, including the structure of residue class mod prime powers and Dirichlet's theorem on primes in arithmetic progressions. Actually, the question becomes more interesting if you ask why 5*48=240 is the largest number that will evenly divide p^4-1 when p is a prime greater than 5.

As was mentioned by other respondants, showing that p^4-1 is divisible by 48 for any such p is not too involved To prove that 48 is the largest such number is difficult without knowledge of some group theory. If you don't know group theory, then you are out of luck (but go read some -- it's good for you!)

It turns out that given a prime power q^n, where q is some prime, the set of congruence classes of integers mod q^n that are relatively prime to q^n forms a group under multiplication of representatives of these classes. When q is odd, these groups are cyclic of order p^(n-1)*(p-1), for any n (when n=1, this is a special case of the theorem that a finite subgroup of the multiplicative group of a field is cyclic, but for n>1, it requires a seperate proof, showing the existence of a "primitive root" mod q^n, i.e., a generator for the group). However, the set of congruences classes mod 2^n of integers relatively prime to 2 is cyclic only when n=1 and n=2 (generators are 1 and 3 respectively). When n>2, the group is isomorphic to Z/2Z + Z/2^(n-2)Z where the + denotes direct sum (generators for each piece are -1 and 5).

Now back to the problem: if you choose a prime p greater than 3, then it is relatively prime to 16. Thus, p represents a class in the group of classes mod 2^4 that are relatively prime to 2, i.e., an element of the group above for 2^4, which is isomorphic to Z/2Z + Z/4Z, and every element of this group has order dividing 4. Thus, raising p to the 4th power (i.e., multiplying its image in this group by 4) gives the identity element of the group, i.e., the class of 1 mod 16, so 16 divides p^4-1. Similarly, p is relatively prime to 3, so represents a class in the group of classes prime to 3 mod 3, which is a group of order 2. Thus, p^4 is in the identity class, i.e., is congruent to 1 mod 3, so 3 divides p^4-1. Since 16 and 3 are relatively prime, the Chinese Remainder Theorem in fact shows that p^4 is congruent to 1 mod 16*3=48, i.e., p^4-1 is divisible by 48.

Now assume that p1^a1 * p2^a2 * ... * pr^ar is the prime decomposition of some number which evenly divides p^4-1 for any prime p greater than 3. In fact, we can even assume that it divides p^4-1 only for primes p>3 distinct from p1, p2, ..., pr and we will still find that 240 is the largest such number. Then the statement implies that p^4 is congruent to 1 mod pm^am for m=1, 2, ..., r, for all primes p>3, in other words, p has order dividing 4 in the multiplicative groups mod pm^am for each i. Now if some pm is greater than 5, then given a primitive root g mod pm^am, by Dirichlet's theorem on primes in arithmetic progressions, there exists a prime p congruent to g mod pm^am. But then this prime p has order pi^(am-1)*(pm-1) > 4 in the multiplicative group of units mod pi^ai since pm>5. Then p^4 is not congruent to 1 mod pm^am, a contradiction. Furthermore, if pm=5, then 5^4-1 is not divisible by pm^am.

Thus, the only possibility for our number is that its factorization is of the form 2^a * 3^b for some a and b. Now if b is greater than 1, the multiplicative group mod 3^b is cyclic of order >=6, so that choosing a prime p congruent toa primitive root mod 3^b, it follows that p^4 is not congruent to 1 mod 3^b. Thus, b can be at most 1. Finally, if a>4, then the multiplicative group mod 2^a has exponent 2^(a-2), so that if we take p=5, which will actually, it turns out, have order 2^(a-2)>4 in this group, so 5^4 is not 1 mod 2^a in this case. Thus, the largest a can be is 4. In other words, the largest possible number satisfying your conditions is 2^4*3=48.

As I said above, you should restate it so that p is an aribtrary prime greater than 5, and then the same type of proof will show that 240 is the largest such number. Here, the extra information comes from the fact that now, the multiplicative group of classes mod 5 has order 4, so any prime > 5 will have order 4 in this group, i.e., p^4-1 will be divisible by 5 in this case as well, and hence, by the Chinese remainder theorem, p^4-1 will be divisible by 240.

Answer 3

p^4 - 1 = (p² + 1)(p + 1)(p - 1)

Every prime greater than 3 is either one more or one less than a multiple of 6.
Let p = 6k ± 1, for positive integer k.

(p² + 1)(p + 1)(p - 1)
= [(6k ± 1)² + 1][(6k ± 1) + 1][(6k ± 1) - 1]
= (36k² ± 12k + 2)·(6k ± 2)·(6k)
= 2(18k² ± 6k + 1)·2(3k ± 1)·(6k)
= 24·k·(18k² ± 6k + 1)·(3k ± 1)

24 therefore divides any (p^4 - 1) for primes p > 3.
But, if k is even, k has a factor of 2, so 48 divides (p^4 - 1) for even values of k.
If k is odd, then (3k ± 1) is even and has a factor of 2, so 48 also divides (p^4 - 1) for odd values of k.

Therefore, 48 divides any (p^4 - 1) for primes p > 3.

Answer 4

It is not. Examle, take 53 (prime)

53^4 - 1 = 7890481 - 1
= 7890480

7890480 / 80 = 98631. 80 is an integer larger than 48.