Modular Congruences of Integer Squares

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prove that for any integer n, n^{2} \cong 0 or 1 (mod 3), and n^{2} \cong 0,1,4(mod 5)
 
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And what have you tried...?
 
The only thing i found was that if you can prove n\congm mod 3 than n^{2} \cong m^{2} mod 3

but i couldn't prove n \cong 0 mod 3 so i gave up
 
You only have to consider n^2 (mod 3) when n=0,1,2. A same type of comment applies mod 5. (Why?)
 

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