Question 4 Let a and b be natural numbers. Give all solutions to the following equation: 2ª + 1 = 6² Hint: Rewrite the above such that it talks about multiplication () instead of addition (+).

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Question 4
Let a and b be natural numbers.
Give all solutions to the following equation:
2a + 1 = 6²
Hint: Rewrite the above such that it talks about multiplication () instead of addition (+).
Transcribed Image Text:Question 4 Let a and b be natural numbers. Give all solutions to the following equation: 2a + 1 = 6² Hint: Rewrite the above such that it talks about multiplication () instead of addition (+).
Expert Solution
Step 1

Solution:

   2+ 1 = b

  => 2= b- 1 

            = (b - 1)(b + 1) 

=> both (b - 1) and (b + 1) are of the form 2for some integers k. 

   i.e, b - 1 = 2m, say

         b + 1 = 2, say. 

=> m + n = a

    Also, 2= 2- 2

            => 2m = 2(2n-1 - 1) 

            => 2m-1 = 2n-1 - 1

 

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Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question
Comviqlar
Q SEARCH
Step 1
Solution:
20+1=b2
=
ASK
2a = b2 - 1
= (b -1) (b
=> both (b - 1)
form 2k for som
i.e, b - 1 = 2m, sa
b+1=2, s
=>m+n=a
Also, 2m 2n-2
N065 %
=> 2m = 2(2n-1-1)
=> 2m-1 = 2n-1 - 1
=> 1 + 1 = 2 = 2n-1
=> n-1=1
=> n = 1+1=2
So, a = 1 + 2 = 3
Then, b² = 2³ + 1 = 9
-
=> b=3
CHAT
How did we get
this step
Which is 2^m =
2^n - 2
Can you explain?
11:16
Step 2
This implies 2m-1 is odd, which is only
possible if 2m-1 = 1
=> m - 1 = 0
=> m = 1
Then, 1 = 2n-1 - 1
Transcribed Image Text:Comviqlar Q SEARCH Step 1 Solution: 20+1=b2 = ASK 2a = b2 - 1 = (b -1) (b => both (b - 1) form 2k for som i.e, b - 1 = 2m, sa b+1=2, s =>m+n=a Also, 2m 2n-2 N065 % => 2m = 2(2n-1-1) => 2m-1 = 2n-1 - 1 => 1 + 1 = 2 = 2n-1 => n-1=1 => n = 1+1=2 So, a = 1 + 2 = 3 Then, b² = 2³ + 1 = 9 - => b=3 CHAT How did we get this step Which is 2^m = 2^n - 2 Can you explain? 11:16 Step 2 This implies 2m-1 is odd, which is only possible if 2m-1 = 1 => m - 1 = 0 => m = 1 Then, 1 = 2n-1 - 1
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