Using the idea of EuclidSolutionAny integer n when divides b

Using the idea of Euclid

Solution

Any integer n when divides by 4 gives remainder either 0,1,2,3. There is no other possibility.

Or n is of the form either remainder 0 or 1 or 2 or 3

This can be written as

n = 4k or 4k+1 or 4k+2 or 4k+3

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b) If n is odd prime, then n can give remainders only 1 and 3.

Hence n can be only as 4k+1 or 4k+3

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c) if m = 4k+1 and n = 4p+1

mn = 16kp+4(k+p)+1 = 4(integer)+1

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d) This can be extended to taking m and mn and again mn and m^2 n like that any number of times.