# Let n be a natural number. Show that n^2 has always the (smallest positive) remainder 0 or 1 by division of 4. In formula n^2 = 0 (mod 4) or n^2 =1 (mod 4)

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If `n` is even, so that `n=2k` for some integer `k,` then

`n^2=(2k)^2=4k^2,` so `n^2-=0 (mod 4)` if `n` is even.

If `n` is odd, then `n=2j+1` for some integer `j.` In that case,

`n^2=(2j+1)^2=4j^2+4j+1=4(j^2+j)+1,`

so `n^2-=1(mod 4)` if `n` is odd. These are the only two possibilities, so the proof is complete.

To prove n^2 is either divisible by 4 or leaves remainder as 1. Natural number are in two set odd and even

let n is even number i.e n=2m ,m is a nutural number.

n^2=(2m)^2=4m^2 ,which is multiple of 4 so is divisible by 4 and remainder will 0.

Let n is odd number i.e. n=2m+1, m is natural no.

n^2=(2m+1)^2=4m^2+4m+1=4(m^2+m)+1

first term in above expression is multiple of 4 so divisble by 4 ,and seconder term leaves remainder 1.

This statement is not true for n=1 ,1 is natural no.