Figure this one out...

[[m'kay]]

with math, dummy

[ProfessorDreadNaught]

(p^4-1) = (p^2+1)(p-1)(p+1)
16*3*5=240

[kjeopardy]

(p^4-1) = (p^2+1)(p-1)(p+1)
16*3*5=240

Good try, but not quite...just check: 5^4 - 1 = 624.

624 isn't divisible by 240...

[(=DK=)Samonuh]

8.

[kjeopardy]

8.

8 works, but it's not the biggest...

[Doves]

(2x^2+y^2i)(2x^2-y^2i)
i = square root of -1 and is not part of the exponent

[kjeopardy]

(2x^2+y^2i)(2x^2-y^2i)
i = square root of -1 and is not part of the exponent

That works, but requires imaginary numbers...we really only want real numbers in the factorization, although I didn't say so. What you did is the only way to factor something like x^2 + y^2.

In this case, we can get a valid factoring that uses only real numbers, so that's the better answer: although yours is correct.

Regards,
3.141592653...

[Nik]

Is it 48?

[Bryant]

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

Every prime number is odd, so each factor above is even. Given that p > 3 the lowest p is 5.
p=2n+3 where n=1,2,4,... where p is prime
((2n+3)^2 +1)((2n+3)-1)((2n+3)+1) = (4n^2+12n+10)(2n+2)(2n+4) = 8(2n^2+6n+5)(n+1)(n+2)

if n is odd n = 2k-1 where k = 1,... for whatever n
8[(2k-1)^2+6(2k-1)+5][2k-1+1][2k-1+2] = 8[4k^2+8k][2k][2k+1] = 64k^2[k+2][2k+1]
-3 conditions when diving by 3 (remainders of 0,1,2) - the first factor (k^2) covers R=0, the second factor (k+2) covers R=1, nothing covers R=2
-however, if k = 3x+2 (ie R=2), then n = 6x-3 and p = 12x-3 which is divisible by 3 and thus not prime
-so factors are 64*3

if n is even n=2k where k=1,...
8[2(2k)^2+6(2k)+5][(2k)+1][(2k)+2] = 8[8k^2+12k+5][2k+1][2k+2] = 16[8k^2+12k+5][2k+1][k+1]
-3 conditions when diving by 3 (remainders of 0,1,2) - the second factor (2k+1) covers R=1, the third factor covers R=2, nothing covers R=0
-however, if k=3x, then n=6x and p=12x+3 which is divisible by 3 and thus not prime
-so factors are 16*3

So the answer is 48, I brute forced some of the terms to validate this.

[kjeopardy]

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

Every prime number is odd, so each factor above is even. Given that p > 3 the lowest p is 5.
p=2n+3 where n=1,2,4,... where p is prime
((2n+3)^2 +1)((2n+3)-1)((2n+3)+1) = (4n^2+12n+10)(2n+2)(2n+4) = 8(2n^2+6n+5)(n+1)(n+2)

if n is odd n = 2k-1 where k = 1,... for whatever n
8[(2k-1)^2+6(2k-1)+5][2k-1+1][2k-1+2] = 8[4k^2+8k][2k][2k+1] = 64k^2[k+2][2k+1]
-3 conditions when diving by 3 (remainders of 0,1,2) - the first factor (k^2) covers R=0, the second factor (k+2) covers R=1, nothing covers R=2
-however, if k = 3x+2 (ie R=2), then n = 6x-3 and p = 12x-3 which is divisible by 3 and thus not prime
-so factors are 64*3

if n is even n=2k where k=1,...
8[2(2k)^2+6(2k)+5][(2k)+1][(2k)+2] = 8[8k^2+12k+5][2k+1][2k+2] = 16[8k^2+12k+5][2k+1][k+1]
-3 conditions when diving by 3 (remainders of 0,1,2) - the second factor (2k+1) covers R=1, the third factor covers R=2, nothing covers R=0
-however, if k=3x, then n=6x and p=12x+3 which is divisible by 3 and thus not prime
-so factors are 16*3

So the answer is 48, I brute forced some of the terms to validate this.

You are correct! Although you're solution is long and complicated and involves variables. Here is a prettier answer. No variables or stuff...