From here, Phil concludes that the solution set to
2x + 6 ≡ 4 (mod 8) is {x; x ≡ −1 (mod 8)}.

(a) Is Phil's answer correct? If not, which step in Phil's reasoning is
incorrect? Explain what is wrong with it and find the (correct) solutions for the equation 2x+6 ≡ 4 (mod 8) .

(b) Find the condition for the modulus n, for which the congruence
2x+6 ≡ 4 (mod 8) actually does have the solution set {x; x ≡ −1 (mod 8)}. (i.e.
it is the set of all solutions). Prove your answer.

...mod 6) 3ñ2 ≡ 1 (mod 7) 9ñ3 ≡ 1 (mod 11) By ... we see that ñ1 = 5, ñ2 = 5, ñ3 = 5 are solutions. Consequently , a solution of the system is xo = c1n1ñ1 ...

... This solution is comprised of a detailed explanation for solving ... step-by-step explanation for finding all solutions of each ... a) x2 + x +1 ≡ 0 (mod 11) (b ...

... m|p-1 (1<m<p-1). How many integral solutions are there of the congruence x^m - g ≡ (mod p ... Please see the attached file for the complete solution. ...

... Find all solutions to the system of congruences. x = 2(mod 3) x = 1(mod 4) x = 3(mod 5). ... 1) Problem Solution: (solution for an even number problem) see below... ...

... We conclude that all solutions are congruent to 2201 ( mod 315), and that the least ... have a simultaneous solution that is unique modulo M. Compairing (A) and ...

... of the Euclid's Proposition VII.30, (2) has a solution. ... 105 = lcm(3, 5, 7). Thus we have solutions 23, 128 ... then 2t = 3 (mod 7). From division tables modulo 7, 3 ...

... Therefore, b is quadratic residue (mod p). ...please ... This solution provides a step-by-step ... given problems regarding positive roots and quadratic congruences. ...

... congruent to -1 mod p. The more detail problem is in the attached file. The solution is in the attached file. Ring Theory 3 Solving Congruences By :- Thokchom ...