Clearly if n is even the expression is even and so not prime.
If n = 1 the expression is prime.
Suppose n = 3k (and k is odd). Then the expression is 2^3k + (3k)^222 = (2^k)^3 + (3k^74)^3 which is the sum of cubes, and is therefore divisible by (2^k + 3k^74)
Suppose n = 3k +/- 1 (and k is even). Then n^2 = (3k +/-1)^2 = 1 mod 3 and so n^222 = (n^2)^111 = 1 mod 3. But n is odd, and so 2^n = -1 mod 3 and so the sum is a multiple of 3.
So to summarize:
* n = 1 is a solution
* n even means the expression is even
* n odd > 1 and not a multiple of 3 means the expression is divisible by 3
* n odd and a multiple of 3 means the expression is divisible by (2^(n/3) + (n/3)^74)
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Posted by Paul
on 2024-07-16 12:09:56 |
Solution
