phyguy321 Messages 45 Reaction score 0 Thread starter Oct 2, 2008 #1 prove that for any integer n, n[tex]^{2}[/tex] [tex]\cong[/tex] 0 or 1 (mod 3), and n[tex]^{2}[/tex] [tex]\cong[/tex] 0,1,4(mod 5)
prove that for any integer n, n[tex]^{2}[/tex] [tex]\cong[/tex] 0 or 1 (mod 3), and n[tex]^{2}[/tex] [tex]\cong[/tex] 0,1,4(mod 5)
morphism Science Advisor Homework Helper Messages 2,014 Reaction score 4 Oct 2, 2008 #2 And what have you tried...?
phyguy321 Messages 45 Reaction score 0 Oct 3, 2008 #3 The only thing i found was that if you can prove n[tex]\cong[/tex]m mod 3 than n[tex]^{2}[/tex] [tex]\cong[/tex] m[tex]^{2}[/tex] mod 3 but i couldn't prove n [tex]\cong[/tex] 0 mod 3 so i gave up
The only thing i found was that if you can prove n[tex]\cong[/tex]m mod 3 than n[tex]^{2}[/tex] [tex]\cong[/tex] m[tex]^{2}[/tex] mod 3 but i couldn't prove n [tex]\cong[/tex] 0 mod 3 so i gave up
morphism Science Advisor Homework Helper Messages 2,014 Reaction score 4 Oct 4, 2008 #4 You only have to consider n^2 (mod 3) when n=0,1,2. A same type of comment applies mod 5. (Why?)