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

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