Suppose that a and b are integers, a ≡ 11(mod 19), and b ≡ 3(mod 19). Find the integer c with 0 <= c <= 18 such that c ≡ 2a^(2)+3b^(2)(mod 19)

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Definitions
Division algorithm Let a be an integer and d a positive integer. Then there are unique integers q and r with 0 <= r < d such that a=dq+r
q is called the quotient and r is called the remainder
q=a div d
r=a mod d
Theorem 5 Let m be a positive integer. If a ≡ b(mod m) and c ≡ d(mod m), then a+c ≡ b+d(mod m) and ac ≡ bd(mod m).
Solution
a=11(mod 19)
b ≡ 3(mod 19)
0 <= c <= 18
Use theorem 5:
c ≡ 2a^(2)+3b^(2)(mod 19)
=2*11^(2)+3*3^(2)(mod 19)
=2*121+3*9(mod 19)
=242+27(mod 19)
=269(mod 19)
=3(mod 19)
We then obtain c=3 with 0 <= c <= 18.