Claim. Among any 4 integers, there are two, a and b, such that a² – 6² is divisible by 5. Proof. We create three boxes labeled B₁, B₁, and B4, so that integer n is placed in B; if n² mod 5 = i. Now given any 4 integers, when they are placed among these 3 boxes, some box (say B;) will contain at least two integers (call them a and b) by the Pigeonhole Principle. Since a and b are both in box B₂, we know that a² mod 5 i and 6² mod 5 = i. This means a² = 5K + i and b² = 5Q + ¿ for integers K and Q. From this it follows that a² − b² = 5(K – Q); that is, a²- 6² is divisible by 5. = = -
Claim. Among any 4 integers, there are two, a and b, such that a² – 6² is divisible by 5. Proof. We create three boxes labeled B₁, B₁, and B4, so that integer n is placed in B; if n² mod 5 = i. Now given any 4 integers, when they are placed among these 3 boxes, some box (say B;) will contain at least two integers (call them a and b) by the Pigeonhole Principle. Since a and b are both in box B₂, we know that a² mod 5 i and 6² mod 5 = i. This means a² = 5K + i and b² = 5Q + ¿ for integers K and Q. From this it follows that a² − b² = 5(K – Q); that is, a²- 6² is divisible by 5. = = -
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Please show your work,
How the number of created/defined boxes is related to Pigeonhole Principal?
How the number of created/defined boxes is related to given in the problem integers?
![**Instructions:**
Read the proof below. If the numbers given in the second sentence are \(\{64, 69, 97, 118\}\), then which boxes end up with at least two numbers? (Select all)
**Box Selection:**
- Box 0: [ ]
- Box 1: [✔]
- Box 4: [ ]
**Selection Result:**
- Box 1: Box 1 and Box 4
---
**Mathematical Claim:**
Among any 4 integers, there are two, \(a\) and \(b\), such that \(a^2 - b^2\) is divisible by 5.
**Proof:**
We create three boxes labeled \(B_0\), \(B_1\), and \(B_4\), so that integer \(n\) is placed in \(B_i\) if \(n^2 \mod 5 = i\).
Given any 4 integers, when they are placed among these 3 boxes, some box (say \(B_i\)) will contain at least two integers (call them \(a\) and \(b\)) by the Pigeonhole Principle. Since \(a\) and \(b\) are both in box \(B_i\), we know that \(a^2 \mod 5 = i\) and \(b^2 \mod 5 = i\). This means \(a^2 = 5K + i\) and \(b^2 = 5Q + i\) for integers \(K\) and \(Q\). From this, it follows that \(a^2 - b^2 = 5(K - Q)\); that is, \(a^2 - b^2\) is divisible by 5.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb97f1139-ddfd-4008-99cf-70dde3f1fabc%2Fa68d320e-383a-4754-af64-a532d80e37f4%2Fe2hh3kh_processed.png&w=3840&q=75)
Transcribed Image Text:**Instructions:**
Read the proof below. If the numbers given in the second sentence are \(\{64, 69, 97, 118\}\), then which boxes end up with at least two numbers? (Select all)
**Box Selection:**
- Box 0: [ ]
- Box 1: [✔]
- Box 4: [ ]
**Selection Result:**
- Box 1: Box 1 and Box 4
---
**Mathematical Claim:**
Among any 4 integers, there are two, \(a\) and \(b\), such that \(a^2 - b^2\) is divisible by 5.
**Proof:**
We create three boxes labeled \(B_0\), \(B_1\), and \(B_4\), so that integer \(n\) is placed in \(B_i\) if \(n^2 \mod 5 = i\).
Given any 4 integers, when they are placed among these 3 boxes, some box (say \(B_i\)) will contain at least two integers (call them \(a\) and \(b\)) by the Pigeonhole Principle. Since \(a\) and \(b\) are both in box \(B_i\), we know that \(a^2 \mod 5 = i\) and \(b^2 \mod 5 = i\). This means \(a^2 = 5K + i\) and \(b^2 = 5Q + i\) for integers \(K\) and \(Q\). From this, it follows that \(a^2 - b^2 = 5(K - Q)\); that is, \(a^2 - b^2\) is divisible by 5.
Expert Solution
Step 1
This is a simple consequence of using the division algorithm. The solution has been attached in step 2.
Step by step
Solved in 2 steps with 1 images
Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question
For 97, why would it be Box 4? Isn't it should belong to box 1 too?
Solution
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