Hello I am stuck on a homework problem for the question of trying to prove that for any integer n,n^2 +5 is not divisible by 4. I looked at a step by step guide and i'm confused by the answer.The step by step solution resulted in 2 cases with even resulting inn^2 + 5 = 4m+1 for n is even as proof that n^2 + 5 is not divisible by 4. Shouldn’t the 4m indicate that it is divisible? Thank you I am somewhat confused by this math concept. ### Explanation of Solution#### Given information:Any integer \( n \).#### Formula used:- \( x \) is even if and only if there exists an integer \( k \) such that \( x = 2k \).- \( x \) is odd if and only if there exists an integer \( k \) such that \( x = 2k + 1 \).#### Proof:Suppose integer \( n \).We divide into cases accordingly:#### Case 1: \( n \) is even.If \( n \) is even, then \( n \) is divisible by 2.So, \( n = 2k \), where \( k \) is any integer.When \( n \) is divisible by 2, then \( n^2 \) is also divisible by 2.\[n^2 + 5 = (2k)^2 + 5\]\[n^2 + 5 = 4k^2 + 5 = 4k^2 + 4 + 1 = 4(k^2 + 1) + 1\]let \( k^2 + 1 = m \)\[n^2 + 5 = 4m + 1\]When \( n \) is even, we can see that \( n^2 + 5 \) is not divisible by 4.#### Case 2: \( n \) is oddBy the definition of odd, for integer \( k \) such that,\[n = 2k + 1\]\[n^2 + 5 = (2k + 1)^2 + 5\]\[n^2 + 5 = 4k^2 + 4k + 1 + 5 = 4k^2 + 4k + 6 = 4k^2 + 4k + 4 + 2 = 4(k^2 + k + 1) + 2\]let \( m = k^2 + k + 1 \)\[n^2 + 5 = 4m + 2\]When \( n \) is odd, we can see that \( n^2 + 5 \) is not divisible by 4.Hence for any integer \( n \), \( n^2

Even number: A number that is divisible by two. All even numbers finish with one of the digits: 0,…

Answered: Hello I am stuck on a homework problem…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Hello I am stuck on a homework problem for the question of trying to prove that for any integer n,n^2 +5 is not divisible by 4. I looked at a step by step guide and i'm confused by the answer.

The step by step solution resulted in 2 cases with even resulting in

n^2 + 5 = 4m+1 for n is even as proof that n^2 + 5 is not divisible by 4. Shouldn’t the 4m indicate that it is divisible? Thank you I am somewhat confused by this math concept.

### Explanation of Solution

#### Given information:
Any integer \( n \).

#### Formula used:
- \( x \) is even if and only if there exists an integer \( k \) such that \( x = 2k \).
- \( x \) is odd if and only if there exists an integer \( k \) such that \( x = 2k + 1 \).

#### Proof:
Suppose integer \( n \).

We divide into cases accordingly:

#### Case 1: \( n \) is even.
If \( n \) is even, then \( n \) is divisible by 2.
So, \( n = 2k \), where \( k \) is any integer.

When \( n \) is divisible by 2, then \( n^2 \) is also divisible by 2.
\[
n^2 + 5 = (2k)^2 + 5
\]
\[
n^2 + 5 = 4k^2 + 5 = 4k^2 + 4 + 1 = 4(k^2 + 1) + 1
\]
let \( k^2 + 1 = m \)
\[
n^2 + 5 = 4m + 1
\]
When \( n \) is even, we can see that \( n^2 + 5 \) is not divisible by 4.

#### Case 2: \( n \) is odd
By the definition of odd, for integer \( k \) such that,
\[
n = 2k + 1
\]
\[
n^2 + 5 = (2k + 1)^2 + 5
\]
\[
n^2 + 5 = 4k^2 + 4k + 1 + 5 = 4k^2 + 4k + 6 = 4k^2 + 4k + 4 + 2 = 4(k^2 + k + 1) + 2
\]
let \( m = k^2 + k + 1 \)
\[
n^2 + 5 = 4m + 2
\]
When \( n \) is odd, we can see that \( n^2 + 5 \) is not divisible by 4.
Hence for any integer \( n \), \( n^2

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For the previous problem how did you calculate for 1,5,17,and 37 for proving that n^2 + 1 is not divisible by 4. Thanks sorry i'm still confused on how to check for divisibility of a proofs.

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