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Euclid's division lemma states that for any two positive integers, aa and bb, there exist unique integers qq and rr such that:
a=bq+ra=bq+r
where 0≤r<b0≤r<b.
Let's prove the given statement using Euclid's division lemma.
Consider any positive integer nn. We want to show that n2n2 can be expressed in the form 3m3m or 3m+13m+1 for some integer mm.
First, let's divide nn by 33 using Euclid's division lemma:
n=3q+rn=3q+r
where 0≤r<30≤r<3.
Now, let's square both sides:
n2=(3q+r)2n2=(3q+r)2
Expanding the right side:
n2=9q2+6qr+r2n2=9q2+6qr+r2
Now, consider the possible values of rr:
If r=0r=0, then n2=9q2n2=9q2. Since 9q29q2 is divisible by 33 (because each term 9q29q2 is divisible by 33), we can express n2n2 in the form 3m3m, where m=3q2m=3q2.
If r=1r=1, then n2=9q2+6q+1=3(3q2+2q)+1n2=9q2+6q+1=3(3q2+2q)+1. Here, n2n2 is of the form 3m+13m+1, where m=3q2+2qm=3q2+2q.
If r=2r=2, then n2=9q2+12q+4=3(3q2+4q+1)+1n2=9q2+12q+4=3(3q2+4q+1)+1. Here, n2n2 is of the form 3m+13m+1, where m=3q2+4q+1m=3q2+4q+1.
In all cases, n2n2 is either of the form 3m3m or 3m+13m+1 for some integer mm, which completes the proof.
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