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Prove that every perfect square is \(0\) or \(1\) modulo 4.
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Is this another even/odd proof?
by ft of a all exponents of prime factorization are even, so if it contains a 2 it contains a 4 now the question is, suppose it is odd, then what?
yes
Hmm...\[p^2 = (2n + 1)^2 = 4n^2 + 4n + 1\]
when even remainder is 0 and when odd remainder is 1
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every no.is always of the form 4n ,4n+1,4n+2 or 4n+3 sq each, you shhould get what you seek..
What now?
@shubhamsrg as the method if \(n^2\) is odd
**"has the method"
Ohh!
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It's already in front of me.
mod 4 4(n)+1 is 1
Yes :)
Thanks all
@satellite73: My MSE username is Dumb Cow in case you wonder... :P
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