Verified answer

★ Concept :-

Here the concept of Prime Numbers has been used. We see that we have to find the value of n such that 4n⁴ + 1 is equal to a prime number. This can be proved by two ways. Here the main application method is shown in main answer. Another method is shown in 'More to know' section. For main answer, we will firstly simplify the given equation and then find it's solution using Factor Theorem.

______________________________________

★ Solution :-

Given,

» 4n⁴ + 1

Let's now factorise this.

→ 4n⁴ + 1 = (2n²)² + (1)² [since 1² = 1]

By identity, we know that {a² + b² = (a + b)² - 2ab}, by using this

→ 4n⁴ + 1 = (2n² + 1)² - 2(2n²)(1)

→ 4n⁴ + 1 = (2n² + 1)² - 4n²

Here the term 4n² can be further simplified.

→ 4n⁴ + 1 = (2n² + 1)² - (2n)²

By identity, we know that {a² - b² = (a + b)(a - b)}

→ 4n⁴ + 1 = (2n² + 1 - 2n)(2n² + 1 + 2n)

Now comes the main logic into picture.

• Prime Numbers : These are the numbers whose factors are 1 and the number themselves.

Here since we factorised to find the factors, so factors must be equal to 1. So,

➜ (2n² + 1 - 2n)(2n² + 1 + 2n) = 1

Here since both terms are being multiplied. So, either (2n² + 1 - 2n) = 1 or (2n² + 1 + 2n) = 1

➜ 2n² + 1 - 2n = 1 or 2n² + 1 + 2n = 1

Clearly here (2n² + 1 + 2n) ≠ 1 because n ∈ N (Natural numbers). So any value of n, will make the LHS more than 1. So,

➜ 2n² + 1 - 2n = 1

➜ 2n² + 1 - 2n - 1 = 0

➜ 2n² - 2n = 0

➜ 2n(n - 1) = 0

Here either 2n = 0 or (n - 1) = 0. So,

➜ 2n = 0 or n - 1 = 0

➜ n = 0 or n = 1

Clearly n ≠ 0 because n ∈ N where N is natural number.

This means the value of n = 1. So now let's apply this value and check our answer.

>> 4n⁴ + 1 = 4(1)⁴ + 1 = 4(1) + 1 = 4 + 1 = 5

[n = 1]

We know that 5 is a prime number since the factors of 5 are 1 and 5 itself. So, our anserr is correct.

When n = 1, then 4n⁴ + 1 is a prime number.

________________________________

★ More to know :-

• Alternative method ::

We know that the value of n belongs to the numbers among Natural Numbers.

So now let's divide the natural numbers as odd and even. [Only the eding digits of the numbers]

• Odd : 2,4,6,8,...

• Even : 1,3,5,7,9,...

• For Even ::

We see that among even numbers except for 4, all when raised to the power 4 give ending digits as 6. Now since 6 is the ending digit then multiplying it with 4 will give ending digit as 4 and adding 1 will give ending digit as 5. This means the number is divisible by 5 also. Hence it is not a prime number.

• For Odd ::

For now we shall leave 1 as it is. For other numbers, if we raise them to power 4 will give the last digit as 1 and now multiplying it with 4 will give last digit as 4 and adding one again will give the last digit as 5 which again will make the numbers as composite. The left digit is number is 1 hence it is the value of n.

So, n = 1.

Answer:

★ Concept :-

Here the concept of Prime Numbers has been used. We see that we have to find the value of n such that 4n⁴ + 1 is equal to a prime number. This can be proved by two ways. Here the main application method is shown in main answer. Another method is shown in 'More to know' section. For main answer, we will firstly simplify the given equation and then find it's solution using Factor Theorem.

______________________________________

★ Solution :-

Given,

» 4n⁴ + 1

Let's now factorise this.

→ 4n⁴ + 1 = (2n²)² + (1)² [since 1² = 1]

By identity, we know that {a² + b² = (a + b)² - 2ab}, by using this

→ 4n⁴ + 1 = (2n² + 1)² - 2(2n²)(1)

→ 4n⁴ + 1 = (2n² + 1)² - 4n²

Here the term 4n² can be further simplified.

→ 4n⁴ + 1 = (2n² + 1)² - (2n)²

By identity, we know that {a² - b² = (a + b)(a - b)}

→ 4n⁴ + 1 = (2n² + 1 - 2n)(2n² + 1 + 2n)

Now comes the main logic into picture.

• Prime Numbers : These are the numbers whose factors are 1 and the number themselves.

Here since we factorised to find the factors, so factors must be equal to 1. So,

➜ (2n² + 1 - 2n)(2n² + 1 + 2n) = 1

Here since both terms are being multiplied. So, either (2n² + 1 - 2n) = 1 or (2n² + 1 + 2n) = 1

➜ 2n² + 1 - 2n = 1 or 2n² + 1 + 2n = 1

Clearly here (2n² + 1 + 2n) ≠ 1 because n ∈ N (Natural numbers). So any value of n, will make the LHS more than 1. So,

➜ 2n² + 1 - 2n = 1

➜ 2n² + 1 - 2n - 1 = 0

➜ 2n² - 2n = 0

➜ 2n(n - 1) = 0

Here either 2n = 0 or (n - 1) = 0. So,

➜ 2n = 0 or n - 1 = 0

➜ n = 0 or n = 1

Clearly n ≠ 0 because n ∈ N where N is natural number.

This means the value of n = 1. So now let's apply this value and check our answer.

>> 4n⁴ + 1 = 4(1)⁴ + 1 = 4(1) + 1 = 4 + 1 = 5

[n = 1]

We know that 5 is a prime number since the factors of 5 are 1 and 5 itself. So, our anserr is correct.

\;\underline{\boxed{\tt{Value\;\:of\;\:n\;=\;\bf{\purple{1}}}}}

Valueofn=1

When n = 1, then 4n⁴ + 1 is a prime number.

________________________________

★ More to know :-

• Alternative method ::

We know that the value of n belongs to the numbers among Natural Numbers.

So now let's divide the natural numbers as odd and even. [Only the eding digits of the numbers]

• Odd : 2,4,6,8,...

• Even : 1,3,5,7,9,...

• For Even ::

We see that among even numbers except for 4, all when raised to the power 4 give ending digits as 6. Now since 6 is the ending digit then multiplying it with 4 will give ending digit as 4 and adding 1 will give ending digit as 5. This means the number is divisible by 5 also. Hence it is not a prime number.

• For Odd ::

For now we shall leave 1 as it is. For other numbers, if we raise them to power 4 will give the last digit as 1 and now multiplying it with 4 will give last digit as 4 and adding one again will give the last digit as 5 which again will make the numbers as composite. The left digit is number is 1 hence it is the value of n.

So, n = 1.