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 SolutionGiven information:Any integer n.Formula used:x is even if and only if there exists an integer k such that xx is odd if and only if there exists an integer k such that xProof:Suppose integer n.We divide into cases accordingly:Case 1:nis 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 than n² is also divisible by 2.n² + 5 = (2k)² + 5n² + 5 4k² + 5==4k² + 4 + 1 = 4 (k² + 1) + 1let k² + 1 = mn² + 5 = 4m + 1When n is even, we can see that n² + 5 is not divisible by 4.Case 2: n is oddBy the definition of odd, for integer k such that,n = 2k + 1=(2k + 1)² + 5n² + 5n² + 5 =4k² + 1 + 4k + 5n² + 5=4k² + 4k + 6 = 4k² + 4k + 4 + 2 = 4 (k² + k + 1) + 2let m =k² + k + 1n² + 5= 4m + 2When n is odd, we can see that n² + 5 is not divisible by 4.Hence for any integer n, n² + 5 is not divisible by 4.= 2k.= 2k + 1.

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

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College Algebra (MindTap Course List)

12th Edition

ISBN:9781305652231

Author:R. David Gustafson, Jeff Hughes

Publisher:R. David Gustafson, Jeff Hughes

Chapter8: Sequences, Series, And Probability

Section8.5: Mathematical Induction

Problem 42E

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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
x is odd if and only if there exists an integer k such that x
Proof:
Suppose integer n.
We divide into cases accordingly:
Case 1:nis 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 than n² is also divisible by 2.
n² + 5 = (2k)² + 5
n² + 5 4k² + 5
=
=
4k² + 4 + 1 = 4 (k² + 1) + 1
let k² + 1 = m
n² + 5 = 4m + 1
When n is even, we can see that n² + 5 is not divisible by 4.
Case 2: n is odd
By the definition of odd, for integer k such that,
n = 2k + 1
=
(2k + 1)² + 5
n² + 5
n² + 5 =
4k² + 1 + 4k + 5
n² + 5
=
4k² + 4k + 6 = 4k² + 4k + 4 + 2 = 4 (k² + k + 1) + 2
let m =
k² + k + 1
n² + 5
= 4m + 2
When n is odd, we can see that n² + 5 is not divisible by 4.
Hence for any integer n, n² + 5 is not divisible by 4.
= 2k.
= 2k + 1.

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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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