Question

prove that any odd prime number is either of form 4a+1 or 4a-1 where 'a' is a positive integer​

Answer 1

Given :- prove that any odd prime number is either of form 4a+1 or 4a-1 where 'a' is a positive integer .

Solution :-

Let us assume that, a is any odd prime number .

dividing p by 4 we get,

→ p = 4a + r { where 0 ≤ r < 4 }

so,

→ value of r will be = 0, 1, 2 or 3 .

putting r = 0,

→ p = 4a

→ p = 2 * (2a) .

since p is a factor of 2, it is not a odd number . so, r = 0 not possible .

putting r = 1,

→ p = 4a + 1 .

since (4q + )1 is an odd number . p will be in the form of (4a + 1) .

putting r = 2 ,

→ p = 4a + 2

→ p = 2(2a + 1)

since p is a factor of 2, it is not a odd number . so, r = 2 not possible .

putting r = 3,

→ p = 4a + 3

→ p = 4a + 3 - 4 + 4

→ p = (4a + 4) + (3 - 4)

→ p = 4(a + 1) - 1

→ p = 4b - 1

where b = (a + 1)

since (4b - 1) is an odd number .

therefore, we can conclude that, any odd prime number is either of form (4a + 1) or (4a - 1) .

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Answer 2

Given : odd prime number

To Find   : prove that any odd prime number is either of form 4a+1 or 4a-1 where 'a' is a positive integer​

Solution:

Without loosing generality any natural number can be written in form

4q , 4q+ 1 , 4q  + 2  , 4q  + 3

4q  = 2 ( 2q)  is even  number  

4q  + 1     = 2(2q )  + 1   odd  number

=>   4a  + 1   odd  number

4q + 2  = 2( 2q + 2)  even  number

4q  + 3    =   2(2q + 1)  + 1     odd  number

4q  + 3    =  4q  + 4  - 1  

= 4(q  + 1)  - 1

= 4a  - 1          odd number  

Hence   any odd prime number is either of form 4a+1 or 4a-1  

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