1) p-1, p , p+1 are 3 consecutive Natural Numbers with p prime and >3.

All primes > 3 are odd so (p-1) and (p+1) are consecutive even numbers.

Therefore both will be even with exactly one of them a multiple of 4

So now we have a factor of 8 in (p-1)(p+1)

(30,31,32) or (40,41,42) would be examples

2) Since p is odd then p^2 is odd making (p^2 + 1) even, picking up another factor of 2

Now we have (p - 1)(p + 1)(p^2 + 1) containing a factor of 8x2 = 16

3) Since p-1, p , p+1 are 3 consecutive Natural Numbers, exactly one of them must be a multiple of 3. It can't be p since it is prime. So exactly one of (p-1) or (p+1) must be a multiple of 3. Note in the above examples the 30 and the 42 are the multiples of 3. So now we have a factor of 16 and 3 giving a factor of 3x16 = 48

If k is even, the left factor will earn a factor of 2, making 48. If k were odd instead, the right factor will earn a factor of 2, making 48. Let's see how that plays out:

k is even (2n): 24(2n)(54(2n)^3 + 36(2n)^2 + 9(2n) + 1)

48n(432n^3 + 144n^2 + 18n + 1)

k is odd (2n+1): 24(2n+1)(54(2n+1)^3 + 36(2n+1)^2 + 9(2n+1) + 1)

This time a different argument applies. Since all the values of k are over on the right, we can forget about the coefficient 24 in front. We just need to see that the rest is even.

We can ignore all the even coefficients further because those terms will always be even. That leaves us with 225k^2 + 125k. If k is even, both terms will be even and add to an even sum. If k is odd instead, both terms will be odd, and again add to an even sum.

Therefore we have that missing factor of 2 we need to complete our 48.

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

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

1)

p-1, p , p+1are 3 consecutive Natural Numbers with p prime and >3.All primes > 3 are odd so (p-1) and (p+1) are consecutive even numbers.

Therefore

both will be evenwith exactlyone of them a multiple of 4So now we have a

factor of 8 in (p-1)(p+1)(30,31,32) or (40,41,42) would be examples

2) Since p is odd then

p^2 is oddmaking(p^2 + 1) even, picking upanother factor of 2Now we have (p - 1)(p + 1)(p^2 + 1) containing a factor of 8x2 = 16

3) Since

p-1, p , p+1are 3 consecutive Natural Numbers,exactly one of them must be a multiple of 3. It can't be p since it is prime. Soexactly one of (p-1) or (p+1) must be a multiple of 3. Note in the above examples the 30 and the 42 are the multiples of 3. So now we have a factor of 16 and 3 giving a factor of 3x16 = 48p^4 - 1 is divisible by 48.

Since p cannot be divisible by 2 or 3 (it's bigger than 3 so it cant be divisible by those) all the rest of the primes fall into either

6k+1or6k+5.Let's check:

(6k+1)^4 - 1 = 1296k^4 + 864k^3 + 216k^2 + 24k + 1 - 1

=24k(54k^3 + 36k^2 + 9k + 1)

If k is

even, the left factor will earn a factor of 2, making 48. If k wereoddinstead, the right factor will earn a factor of 2, making 48. Let's see how that plays out:k is even (2n): 24(2n)(54(2n)^3 + 36(2n)^2 + 9(2n) + 1)

48n(432n^3 + 144n^2 + 18n + 1)k is odd (2n+1): 24(2n+1)(54(2n+1)^3 + 36(2n+1)^2 + 9(2n+1) + 1)

24(2n+1)(432n^3 + 792n^2 + 486n + 100)

48(2n+1)(216n^3 + 396n^2 + 243n + 50)On the other hand:

(6k+5)^4 - 1 = 1296k^4 + 4320k^3 + 5400k^2 + 3000k + 625 - 1

= 24(54k^4 + 180k^3 + 225k^2 + 125k + 26)

This time a different argument applies. Since all the values of k are over on the right, we can forget about the coefficient 24 in front. We just need to see that the rest is even.

We can ignore all the even coefficients further because those terms will always be even. That leaves us with

225k^2 + 125k. If k is even, both terms will be even and add to an even sum. If k is odd instead, both terms will be odd, and again add to an even sum.Therefore we have that missing factor of 2 we need to complete our 48.

QED.

Kinda messy but eh.

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