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…
I can factorize like this:
(a + 1) (a^2 - a + 1) and both numbers are greater than one, so a^3+1 can't be prime.
What if n is 6?
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…
1.
In other words, they are exactly the powers of 2.
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
more.
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…
I can factorize like this:
(a + 1) (a^2 - a + 1) and both numbers are greater than one, so a^3+1 can't be prime.
What if n is 6?
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…
1.
In other words, they are exactly the powers of 2.
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
more.