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What mistakes are found in this proof.

What mistakes are found in this proof.

Advanced Engineering Mathematics
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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What mistakes are found in this proof.

5.74. Below is given a proof of a result. What result is proved? Proof Let a = 2 (mod 4) and b =1 (mod 4) and assume, to the contrary, that 4 (a² + 2b) . Since a = 2 (mod 4) and b =1 (mod 4) , it follows that a = 4r + 2 and b = 4s +1, where r, s E Z. Therefore, a? + 26 = (4r + 2)? + 2 (4s+ 1) = (16r² + 16r + 4) + (8s + 2) 16r2 + 16r + 8s + 6. Since 4 (a? + 2b) , we have a? + 2b = 4t, wheret e Z. So, 16r2 + 16r + 8s + 6 = 4t and 6 = 4t – 16r2 – 16r – 8s = 4 (t – 4r2 – 4r – 2s). We've updated our read aloud feature! Since t – 4r2 – 4r – 2s is an integer, 4|6, which is a contradiction. Give it a try here
Transcribed Image Text:5.74. Below is given a proof of a result. What result is proved? Proof Let a = 2 (mod 4) and b =1 (mod 4) and assume, to the contrary, that 4 (a² + 2b) . Since a = 2 (mod 4) and b =1 (mod 4) , it follows that a = 4r + 2 and b = 4s +1, where r, s E Z. Therefore, a? + 26 = (4r + 2)? + 2 (4s+ 1) = (16r² + 16r + 4) + (8s + 2) 16r2 + 16r + 8s + 6. Since 4 (a? + 2b) , we have a? + 2b = 4t, wheret e Z. So, 16r2 + 16r + 8s + 6 = 4t and 6 = 4t – 16r2 – 16r – 8s = 4 (t – 4r2 – 4r – 2s). We've updated our read aloud feature! Since t – 4r2 – 4r – 2s is an integer, 4|6, which is a contradiction. Give it a try here
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