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Shadowys
 one year ago
Best ResponseYou've already chosen the best response.1there are only five possible ways for n. first case: \(n \equiv 0(mod 5)\), then \(n^2 \equiv 0(mod 5)\) second case: \(n \equiv 1(mod 5)\)), then \(n^2 \equiv 1(mod 5)\) third case:\(n \equiv 2(mod 5)\)), then \(n^2 \equiv 4(mod 5)\) fourth case:\(n \equiv 3(mod 5)\)), then \(n^2 \equiv 6(mod 5)\), which is equal to \(n^2 \equiv 1(mod 5)\), as \(6 \equiv 1(mod 5)\) fifth case:\(n \equiv 4(mod 5)\)), then \(n^2 \equiv 16(mod 5)\) which is equal to \(n^2 \equiv 1(mod 5)\), as \(16 \equiv 1(mod 5)\) so we conclude that if n is an integer, n^2 mod 5 is either 0,1, or 4

Shadowys
 one year ago
Best ResponseYou've already chosen the best response.1sorry, for fourth case, it's \(n^2 \equiv 9 (mod 5)\) and it's equal to \(n^2 \equiv 4 (mod 5)\)

gmer
 one year ago
Best ResponseYou've already chosen the best response.0Excellent, but how do we know that there are only five possible values for n?

Shadowys
 one year ago
Best ResponseYou've already chosen the best response.1because the remainder for any integer k divided by an integer n has only n1 remainders. e.g. for 2, you can only get 0 or 1 as remainder. any other remainder means that the division is incomplete.

Shadowys
 one year ago
Best ResponseYou've already chosen the best response.1sorry it'snot n1, it's n. but the range is 0,1,2,3,...,n1
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