The number 2^48 -1 has 2 factors between 60 and 70, what are they?

Question

This is read as "the number two to the forty-eighth power minus one has two factors"

Answer 1

63 and 65

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Let's walk through the answer
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The most elegant way to do it is pointed out by the previous answers... using the rule x^2 - 1 = (x+1)(x-1)...

2^48 = (2^24)*(2^24)

So 2^24-1 could be written as (2^24)^2 -1 = (2^24+1)(2^24-1).... and so on... continuing, you will end up with the factors (2^6+1)(2^6-1), which is 65 and 63 respectively.

BUT.... even if you don't remember that rule, you can still solve this problem by brute force.

x = 2^48 - 1 = 281474976710655

Find X's prime factors that could multiple to a number between 60 and 70 = 3, 3, 5, 7, 13, 17

You can find the list by dividing prime numbers under 70, which are 2, 3, 5, 7, 11,13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67

But you don't have to test all of them if you apply a few logical rules.
1) immediately eliminated 2 because x is not even, which meant...
2) you can eliminate 29, 31, 37,41, 47, 53, and 59 because the next prime is 3, and any prime greater than 23 would result in product greater than 70.
3) But remember to test 61 and 67 because they, in of themselves, fall into the range.

Next find all the prime factors for each odd number in the range. We can eliminate the even numbers because 2 is not a prime factor under consideration.
61 = nope, it's a prime and not a factor
63 = 3, 3, 7 (bingo, it's a subset of X's prime factors)
65 = 5, 13 (again, a subset of X's prime factors)

You can stop here, but if you're curious to see if there are more than two, you can continue...

67 = another prime, and not a factor
69 = 3, 23

Answer 2

Remember that (x^2 - 1) factors as (x+1) (x-1) I'm going to repeat that several times.

So, (2^48 - 1) = (2^24 + 1) (2^24 + 1)

= (2^12 - 1) ( 2^12 + 1 ) ( 2^ 24 + 1) =

= (2^6+1)(2^6-1) (2^12 + 1) ( 2^24 + 1)

= (2^3 - 1)(2^3 + 1) (2^6 + 1)(2^12 + 1) (2^24 + 1)

I think the answer you are looking for is 63 and 65. However, the prime factors of those are 3,5,7 and 13.

Answer 3

They are 2^6-1 and 2^6 +1 = 63 and 65

Answer 4

63 and 65.

Using difference of two squares repeatedly:
2^48-1 =
(2^24+1)(2^24-1) =
(2^24+1)(2^12+1)(2^12-1) =
(2^24+1)(2^12+1)(2^6+1)(2^6-1)

2^6-1 = 63 and 2^6+1 = 65.