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 Prime candidates (Posted on 2016-12-20)
For what values of integer n are n^n + 1 and (2n)^(2n) + 1 both prime numbers?

 Submitted by Ady TZIDON No Rating Solution: (Hide) Paul's solution (slightly modified): n = 1 is a solution since 1^1 + 1 = 2 which is prime and 2^2 + 1 = 5 which is prime. Let us consider n>1. Clearly, when n is odd - x^n + 1 is even, and therefore not prime. Now for odd values of n: x^n + 1 is divisible by x+1 when n is odd, so the expression n^n + 1 has a factor (n+1) when n is odd. For example, 5^5 + 1 is divisible by 6. So no odd n is a solution. But if n is even and has an odd factor f, then we can write n = kf, and (kf)^(kf) + 1 = ((kf)^k)^f + 1 which by the same logic is divisible by (kf)^k + 1. For example, 6^6 + 1 = 36^3 + 1 is divisible by 37. So the only plausible values of n are those with no odd factors -- powers of 2. But wait, there's more! Suppose n = 2^p where p has an odd factor f. Then n = 2^(kf) and n^n + 1 = (2^fk)^(2^fk) = ((2^k)^(2^fk))^f + 1 which therefore has 2^(k*2^kf) + 1 as a factor. for example, 8 = 2^3 (k=1, f=3) so 8^8 + 1 is divisible by 2^(2^3) + 1 = 257. So the only possible cases are those where n = 2^(2^k), or where p = 0 which recovers the n=1 solution above. Almost done. For a full solution, both n and 2n must show this property. Now, if n is 2^k, then 2n = 2^(k+1), but it must also be the case that their exponents are both powers of two. Since exponents differ by 1, we need powers of two that are adjacent. 1 and 2 are the only such, so n = 2^1 is the only potential solution that obeys all of the criteria. And indeed, 2^2 + 1 = 5 is prime and 4^4 + 1 = 257 is also prime. So the complete list of allowed n is {1, 2}

 Subject Author Date re(2): solution -@ Paul Paul 2016-12-23 12:03:22 re: solution -@ Paul Ady TZIDON 2016-12-20 22:34:05 re: solution Steve Herman 2016-12-20 18:23:33 computer exploration Charlie 2016-12-20 13:34:08 solution Paul 2016-12-20 13:28:38

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