Question

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Answer 1

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n = 2k+1, since all primes > 5 are odd.

n^4 - 1 = (2k)^4 + 4(2k)^3 + 6(2k)^2 + 4(2k)

166^4 + 32^3 + 24K^2 + 8k = S 240 = 16 * 3 *5

If k i seven, S is divisible by 16 = 2 * 8.

If k is odd, 24 (k^2)+ 8 k = 8 k (3k+1),

and 3k+1 is even, so S is divisible by 16 = 2 * 8. Therefore S is divisible by 16. k mod 3 = 0 => S divisible by 3 k mod 3 = 1 => (2k+1) = 0 mod 3 => n not prime

k mod 3 = 2 => (2k+1) = 2 mod 3, 2^4 = 1 mod 3,

so n^4 - 1 is divisible by 3. n mod 5 = 0 => n not prime

n mod 5 = 1 => (n^4 - 1) = 0 mod 5

n mod 5 = 2 => (n^4 - 1) = 0 mod 5 (16-1 = 15) n mod 5 = 3 => (n^4 - 1) = 0 mod 5 (81 -1 = 80)

n mod 5 = 4 => (n^4 - 1) = 0 mod 5 (256 - 1 = 255)

Therefore n^4 - 1 divisible by 16, by 3, and by 5, so by 240.

Answer 2

Answer:

n = 2k+1, since all primes > 5 are odd.

n^4 - 1 = (2k)^4 + 4(2k)^3 + 6(2k)^2 + 4(2k)

166^4 + 32^3 + 24K^2 + 8k = S 240 = 16 * 3 *5

If k i seven, S is divisible by 16 = 2 * 8.

If k is odd, 24 (k^2)+ 8 k = 8 k (3k+1),

and 3k+1 is even, so S is divisible by 16 = 2 * 8. Therefore S is divisible by 16. k mod 3 = 0 => S divisible by 3 k mod 3 = 1 => (2k+1) = 0 mod 3 => n not prime

k mod 3 = 2 => (2k+1) = 2 mod 3, 2^4 = 1 mod 3,

so n^4 - 1 is divisible by 3. n mod 5 = 0 => n not prime

n mod 5 = 1 => (n^4 - 1) = 0 mod 5

n mod 5 = 2 => (n^4 - 1) = 0 mod 5 (16-1 = 15) n mod 5 = 3 => (n^4 - 1) = 0 mod 5 (81 -1 = 80)

n mod 5 = 4 => (n^4 - 1) = 0 mod 5 (256 - 1 = 255)

Therefore n^4 - 1 divisible by 16, by 3, and by 5, so by 240.