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This is more common in math than in e.g. computer science but it is worth keeping in mind because it simplifies a lot. This is not to say that the choice of 0, 1, 2, ..., b-1 as representatives of a value of a mod b (when a is an integer and "a mod b" is a function output) is not a natural one, which may be what is going on here.
1 mod 3 equals 1, since 1/3 = 0 with a remainder of 1. To find 1 mod 3 using the modulus method, we first find the highest multiple of the divisor, 3 that is equal to or less than the dividend, 1. Then, we subtract the highest multiple from the dividend to get the answer to 1 mod 3. Multiples of 3 are 0, 3, 6, 9, etc. and the highest multiple ...
There seems to be a fundamental difference between $2\pmod3$ and $1\pmod3$ in this way (similarly, between $3\pmod4$ and $1\pmod4$). $\endgroup$ – Greg Martin Commented Apr 28, 2017 at 20:07
Let n be even, then n2n + 1 ≡ n(1) + 1 (mod3) ≡ n + 1 (mod 3) We want this to be 0 mod3, so n ≡ 2 (mod3). We have the following conditions on n, n ≡ 0 (mod 2) and n ≡ 2 (mod 3). This can be consolidated as n ≡ 2 (mod 6). Likewise you can deal with the other case. This was very helpful, thanks.
We can check that $0^2\cong0\pmod 3$, $1^2\cong1\pmod 3$, and $2^2\cong4\cong1 \pmod3$. There are only three cases to check. There are only three cases to check. Share
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1. Assume that x, y are integers, and f(x, y) = x4 − 2x3y + x2y2 − y4 is a prime number greater than 3. Prove that f(x, y) ≡ 1 (mod3). For example, f(34, 21) = 883 is a prime and 883 ≡ 1 (mod 3). [ Hint: x4 − 2x3y + x2y2 − y4 = (x2 − xy + y2)(x2 − xy − y2). Ive tried many of the modular arithmetic tricks I know but I can't get ...
Suppose $p = 1 \bmod 3$, prove the following statements: prove that $x^2 + x + 1 = 0 \mod p$ has a solution ...
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