Nt1310 Unit 1 Lab Report
Lemma: If n is a positive integer, [pic]
proof: [pic]
[pic]
[pic]
= an − bn.
Theorem: If 2n + 1 is an odd prime, then n is a power of 2. proof: If n is a positive integer but not a power of 2, then n = rs where [pic], [pic]and s is odd.
By the preceding lemma, for positive integer m, [pic]
where [pic]means "evenly divides". Substituting a = 2r, b = − 1, and m = s and using that s is odd, [pic]
and thus [pic]
Because 1 < 2r + 1 < 2n + 1, it follows that 2n + 1 is not prime. Therefore, by contraposition n must be a power of 2.
Well my thought is suppose n isn't a power of 2.
Suppose it's 3.
a^3 + 1
No, that's not prime. if n is odd and n>=3, then …show more content…
a^6 + 1
Well now I can just write that as (a^2)^3 + 1 which we know will factorize the same way:
((a^2) + 1) ( (a^2)^2 - (a^2) + 1)
The same idea will work for any number with an odd factor of 3 or more.
Therefore the only numbers n that can possibly lead to primes are those that have no odd factors greater than …show more content…
Read more: If a^n+1 is prime for some number a>=2 and n>=1, show that n must be a power of 2 | Answerbag http://www.answerbag.com/q_view/1122154#ixzz1k6wypEZ9
>(b) if a^n + 1 is prime, show that a is even and that n is a power
> of 2.
If n is not a power of two, then it has an odd prime factor m.
So we can write n = m*r where we know for sure that m is odd. Since m is odd, we can write m = 2*k + 1 for some integer k.
Thus n = 2*k*r + r
Now exhibit a nontrivial factorization of a^n + 1:
a^n + 1 = a^(2*k*r + r) + 1 =
(a^r + 1) *
(a^(2*k*r) - a^(2*k*r - r) + a^(2*k*r - 2*r) - ... + a^(2*r) - a^r +
1)
To see that a has to be even note that if a were odd, then we could write: a = 1 mod 2.
We know that n has to be a power of two, so n is even. So we can write n = 2*q for some integer q.
Then
a^n = a^(2*q) = 1^(2*q) = 1 mod 2.
So if a is odd and n is even, a^n will always be odd.
Then a^n + 1 will always be even and hence cannot be prime.
So if a^n + 1 is to be prime, we cannot have a odd.
I hope this helps. Please write back if you'd like to talk about this some
proof: [pic]
[pic]
[pic]
= an − bn.
Theorem: If 2n + 1 is an odd prime, then n is a power of 2. proof: If n is a positive integer but not a power of 2, then n = rs where [pic], [pic]and s is odd.
By the preceding lemma, for positive integer m, [pic]
where [pic]means "evenly divides". Substituting a = 2r, b = − 1, and m = s and using that s is odd, [pic]
and thus [pic]
Because 1 < 2r + 1 < 2n + 1, it follows that 2n + 1 is not prime. Therefore, by contraposition n must be a power of 2.
Well my thought is suppose n isn't a power of 2.
Suppose it's 3.
a^3 + 1
No, that's not prime. if n is odd and n>=3, then …show more content…
a^6 + 1
Well now I can just write that as (a^2)^3 + 1 which we know will factorize the same way:
((a^2) + 1) ( (a^2)^2 - (a^2) + 1)
The same idea will work for any number with an odd factor of 3 or more.
Therefore the only numbers n that can possibly lead to primes are those that have no odd factors greater than …show more content…
Read more: If a^n+1 is prime for some number a>=2 and n>=1, show that n must be a power of 2 | Answerbag http://www.answerbag.com/q_view/1122154#ixzz1k6wypEZ9
>(b) if a^n + 1 is prime, show that a is even and that n is a power
> of 2.
If n is not a power of two, then it has an odd prime factor m.
So we can write n = m*r where we know for sure that m is odd. Since m is odd, we can write m = 2*k + 1 for some integer k.
Thus n = 2*k*r + r
Now exhibit a nontrivial factorization of a^n + 1:
a^n + 1 = a^(2*k*r + r) + 1 =
(a^r + 1) *
(a^(2*k*r) - a^(2*k*r - r) + a^(2*k*r - 2*r) - ... + a^(2*r) - a^r +
1)
To see that a has to be even note that if a were odd, then we could write: a = 1 mod 2.
We know that n has to be a power of two, so n is even. So we can write n = 2*q for some integer q.
Then
a^n = a^(2*q) = 1^(2*q) = 1 mod 2.
So if a is odd and n is even, a^n will always be odd.
Then a^n + 1 will always be even and hence cannot be prime.
So if a^n + 1 is to be prime, we cannot have a odd.
I hope this helps. Please write back if you'd like to talk about this some