Let n be an integer greater than 6. Prove that if n-1 and n+1 are both prime, then n2(n2+16) is divisible by 720.
If n-1 and n+1 are both prime, then n is even.
Since neither n-1 nor n+1 is divisible by 3, n is divisible by 3.
Therefore 6 divides n.
So 36 divides n^2.
(n^2+16) is divisible by 4; so we're up to 144
We need one more factor of 5.
If N ends in 0, then N^2 ends in 0.
If N ends in 2 or 8, N^2 ends in 4 and adding 16 makes a final digit of 0.
If N ends in 4 or 6, then either N+1 or N-1 ends in 5 and we have our factor of 5, provided N>6
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Posted by Larry
on 2024-09-05 13:51:57 |
How about ...
