How many distinct prime factors does the following number have:
2 3 2 + 2 1 7 + 1 ?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Great use identifying the Fermat's prime.
Fun fact: 2 1 6 + 1 = 6 5 5 3 7 is the largest known Fermat's prime.
2^32+2^7+1 = (2^16+1)^2 = 65537^2: 65537 is a prime, So 1 prime number
It's a perfect square of ((2^16) +(1)).. which is a fermet prime number
2 3 2 + 2 1 7 + 1 = ( 2 1 6 ) 2 + 2 ∗ ( 2 1 6 ) ∗ 1 + 1 2 = ( 2 1 6 + 1 ) 2 Since the number is a perfect square, the exponents of the decomposition into prime factors are even, and then the number of divisors must be the product of odd numbers, and then an odd number.
You still need to show that 2 1 6 + 1 is prime.
how 2^16+1 will be prime number?
Log in to reply
You don't know if 2 1 6 + 1 is a prime number, but all perfect square numbers have an odd number of divisors and 1 is the only odd posibility.
Log in to reply
2 1 6 is a perfect square, but 2 1 6 + 1 is not a perfect square.
Log in to reply
@Chung Kevin – Sorry, ( 2 1 6 + 1 ) 2 is a perfect square.
Log in to reply
@Soldà Federico – Do you know why it is perfect square ?
It is Fermat prime
Actually, 1 is not a prime number
4295098369 is 2^32+2^17+1 from what I can tell is a prime number and that is 1
No, that number is not a prime.
Problem Loading...
Note Loading...
Set Loading...
2 3 2 + 2 1 7 + 1 = ( 2 1 6 + 1 ) 2
And 2 1 6 + 1 is a Fermat prime so the answer is 1 .