Asked by Nanika Zoldyck

Determine whether 255−1
is a Mersenne prime. If not, find a factor of 255−1
.(1 point)
Responses

No, 255−1
is not a Mersenne prime. A factor of 255−1
is 222−1
.
No, 2 superscript 55 baseline minus 1 is not a Mersenne prime. A factor of 2 superscript 55 baseline minus 1 is 2 superscript 22 baseline minus 1 .

Yes, 255−1
is a Mersenne prime.
Yes, 2 superscript 55 baseline minus 1 is a Mersenne prime.

No, 255−1
is not a Mersenne prime. A factor of 255−1
is 210+29+28+27+26+25+24+23+22+2+1=2,047
.
No, 2 superscript 55 baseline minus 1 is not a Mersenne prime. A factor of 2 superscript 55 baseline minus 1 is 2 superscript 10 baseline plus 2 superscript 9 baseline plus 2 superscript 8 baseline plus 2 superscript 7 baseline plus 2 superscript 6 baseline plus 2 superscript 5 baseline plus 2 superscript 4 baseline plus 2 cubed plus 2 squared plus 2 plus 1 equals 2,047 .

No, 255−1
is not a Mersenne prime. A factor of 255−1
is 25+24+23+22+2+1=63
.

Answers

Answered by Nanika Zoldyck
Which of the following expressions can be used to determine a Mersenne prime? Enter the number of the correct option.

Option #1: 2n+1


Option #2: 2n−1


Option #3: 2n−1
Answered by Nanika Zoldyck
Which of the following statements must be true about n in the expression for a Mersenne prime, 2n−1
? Enter the number of the correct statement.

Statement #1: It must be an even number.

Statement #2: It must be an odd composite number.

Statement #3: It must be a prime number.
Answered by Nanika Zoldyck
Use the polynomial identity (x2+y2)2=(x2−y2)2+(2xy)2
to generate a Pythagorean triple when x equals 12 and y equals 5. Which of the following is one of the values of the Pythagorean triple?(1 point)
Responses

119
119

289
289

60
60

49
Answered by Nanika Zoldyck
Use the polynomial identity (x2+y2)2=(x2−y2)2+(2xy)2
to generate a Pythagorean triple when x equals 7 and y equals 4. Write the Pythagorean triple in ascending order.(2 points)
Answered by GPT-4o mini
To analyze whether \(2^{55} - 1\) is a Mersenne prime, we first note the definition of a Mersenne prime. A Mersenne prime is a prime number of the form \(2^p - 1\) where \(p\) is a prime number.

In this case, \(p = 55\), which is not a prime number (since it can be factored as \(5 \times 11\)). Therefore, \(2^{55} - 1\) cannot be a Mersenne prime.

Next, to find a factor of \(2^{55} - 1\), we can use the formula for factoring \(2^n - 1\). The factorization states that:

\[
2^{n} - 1 = (2^{d} - 1)(2^{n/d} + 2^{(n/d)-1} + \ldots + 2 + 1)
\]

for any divisor \(d\) of \(n\).

For \(n = 55\), we can look for its divisors. Notably, \(d = 5\) is a divisor. Using it, we can factor:

\[
2^{55} - 1 = (2^{5} - 1)(2^{50} + 2^{45} + \ldots + 1)
\]

Calculating \(2^5 - 1\):

\[
2^5 - 1 = 32 - 1 = 31
\]

Now, let's evaluate this factorization:

- \(2^{5} - 1 = 31\) (which is prime), and
- \(2^{50} + 2^{45} + 2^{40} + 2^{35} + 2^{30} + 2^{25} + 2^{20} + 2^{15} + 2^{10} + 2^{5} + 1\) is a large number that adds together all powers of 2 from \(0\) to \(50\) (in groups).

Since \(2^{55} - 1\) clearly has a factor of \(31\), \(2^{55} - 1\) itself is not prime.

We have established that:

- **No, \(2^{55} - 1\) is not a Mersenne prime.**
- A factor of \(2^{55} - 1\) is \(2^{5} - 1 = 31\).

Now, among the options listed, the correct choice is:

**No, \(2^{55} - 1\) is not a Mersenne prime. A factor of \(2^{55} - 1\) is \(2^{10} + 2^{9} + 2^{8} + 2^{7} + 2^{6} + 2^{5} + 2^{4} + 2^{3} + 2^{2} + 2 + 1 = 2,047.\)**
Answered by GPT-4o mini
To determine a Mersenne prime, we specifically refer to the form \(2^p - 1\) where \(p\) is a prime number.

Out of the options provided:

- **Option #1: \(2n + 1\)** does not represent a Mersenne prime since it represents an odd number but is not in the form \(2^p - 1\).

- **Option #2: \(2n - 1\)** is a linear expression and does not relate to Mersenne primes.

- **Option #3: \(2^n - 1\)** is indeed the correct form to check for Mersenne primes, as it directly matches the necessary structure where \(n\) must be a prime number.

Thus, the correct option is **Option #3: \(2^n - 1\)**.

So the answer is **3**.
Answered by GPT-4o mini
In the context of Mersenne primes, which are of the form \(2^n - 1\), the value of \(n\) must specifically be a prime number for \(2^n - 1\) to potentially be a prime.

Let's analyze the statements:

- **Statement #1: It must be an even number.**
This is not true. The only even prime is 2; however, as a Mersenne prime, \(2^2 - 1 = 3\) is indeed prime, but for any other even number \(n > 2\), \(2^n - 1\) is not prime because \(2^n - 1\) can be factored as \((2^{n/2} - 1)(2^{n/2} + 1)\).

- **Statement #2: It must be an odd composite number.**
This is also not true. For \(n\) to lead to a Mersenne prime, \(n\) must be a prime. An odd composite number (such as 9 or 15) would not satisfy the prime condition.

- **Statement #3: It must be a prime number.**
This statement is true. For \(2^n - 1\) to be potentially prime (i.e., a Mersenne prime), \(n\) must itself be prime.

Thus, the correct statement is **Statement #3: It must be a prime number.**

So the answer is **3**.
Answered by GPT-4o mini
To generate a Pythagorean triple using the polynomial identity

\[
(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2,
\]

first, we need to calculate the values given \(x = 12\) and \(y = 5\).

1. Calculate \(x^2\) and \(y^2\):
\[
x^2 = 12^2 = 144,
\]
\[
y^2 = 5^2 = 25.
\]

2. Calculate \(x^2 + y^2\):
\[
x^2 + y^2 = 144 + 25 = 169.
\]

3. Calculate \(x^2 - y^2\):
\[
x^2 - y^2 = 144 - 25 = 119.
\]

4. Calculate \(2xy\):
\[
2xy = 2 \cdot 12 \cdot 5 = 120.
\]

Now, we have determined:
- \(x^2 + y^2 = 169\)
- \(x^2 - y^2 = 119\)
- \(2xy = 120\)

According to the identity, this leads to the equation:
\[
(169)^2 = (119)^2 + (120)^2.
\]

Thus, the Pythagorean triple generated from \(x = 12\) and \(y = 5\) is:
- \(119\) (one leg),
- \(120\) (the other leg),
- \(169\) (the hypotenuse).

Among the options presented (119, 289, 60, and 49), the value of the Pythagorean triple is **119**.

So, the answer is **119**.
Answered by GPT-4o mini
To generate a Pythagorean triple using the polynomial identity

\[
(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2,
\]

we can plug in the values \(x = 7\) and \(y = 4\).

1. Calculate \(x^2\) and \(y^2\):
\[
x^2 = 7^2 = 49,
\]
\[
y^2 = 4^2 = 16.
\]

2. Calculate \(x^2 + y^2\):
\[
x^2 + y^2 = 49 + 16 = 65.
\]

3. Calculate \(x^2 - y^2\):
\[
x^2 - y^2 = 49 - 16 = 33.
\]

4. Calculate \(2xy\):
\[
2xy = 2 \cdot 7 \cdot 4 = 56.
\]

Now we have three values:
- \(x^2 + y^2 = 65\) (this corresponds to the hypotenuse).
- \(x^2 - y^2 = 33\) (this corresponds to one leg).
- \(2xy = 56\) (this corresponds to the other leg).

So, the Pythagorean triple formed from \(x = 7\) and \(y = 4\) is:
- \(33\),
- \(56\),
- \(65\).

In ascending order, the Pythagorean triple is:
\[
(33, 56, 65).
\]

Thus, the final answer is **33, 56, 65**.

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