Question

show that any positive odd integer is of the form 4p+1 or 4p+3

Answer 1

Let be any positive integer

We know by Euclid's algorithm, if a and b are two positive integers, there exist unique integers q and r satisfying, where

Take



Since 0 ≤ r < 4, the possible remainders are 0, 1, 2 and 3.

That is, can be , whereq is the quotient.

Since  is odd,  cannot be 4q or 4q + 2 as they are both divisible by 2.

Therefore, any odd integer is of the form 4q + 1 or 4q + 3.

Answer 2

hi mate here is ur answer

to show that any positive odd integer is of form 4p+1 or 4p+3

euclid division algorithm : a= bq+ r

here b is 4

applying euclid division algorithm on a and 4

so possible remainders less than 4 are 0,1,2,3

when remainder is 0

a = 4q + 0

a = 2( 2q) + 0 (let 2q be p)

a= 2(p) this is the form of an even number '

when remainder is 1

a = 4q + 1

a = 2(2q) +1(let 2q be p)

a = 2(p) + 1 this in a form of odd no.

when remainder is 2

a = 4q + 2

a = 2(2q+1) (let 2q+1 be p)

a = 2(p) it is an even integer

when remainder is 3

a = 4q+3

a = 2(2q + 1) + 1 let 2q+ 1 be p

a = 2(p) + 1 it is an odd no.

hence proved that any odd no. is of form 4p+1 or 4p+3

hope it helps!!!

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