Question

Find the solutions of the congruence 15x2 + 19x ≡ 5 (mod 11). [Hint: Show the congruence is equivalent to the congruence 15x2 + 19x + 6 ≡ 0 (mod 11). Factor the left-hand side of the congruence; show that a solution of the quadratic congruence is a solution of one of the two different linear congruences.]

Answer 1

Answer:

It is mainly necessary to obtain a new way of expressing congruence through the 'simplification of the polynomial'.

15x^2 + 19x = 5 (mod 11)15x

2

+19x=5(mod11)

15x^2 + 19x + 6 = (6+ 5) (mod 11)15x

2

+19x+6=(6+5)(mod11)

15x^2 + 19x + 6 = 15x^2+ 10x+9x+6 = 5x(3x+2) + 3(3x+2)15x

2

+19x+6=15x

2

+10x+9x+6=5x(3x+2)+3(3x+2)

(5x+3)(3x+2) = 0 mod (11)(5x+3)(3x+2)=0mod(11)

When we have the congruence, we solve it. That is

\begin{gathered}(5x+3) = 0 mod (11)\\5x = -3 mod 11\\5x = 8 mod 11\end{gathered}

(5x+3)=0mod(11)

5x=−3mod11

5x=8mod11

We know as well that the operation 5*9 is 45, but is equal to 1 mod 11, that's mean, we need to multiply both sides for 9,

\begin{gathered}x = 9*8 mod 11\\x = 72 mod 11\\x = 6 mod 11.\end{gathered}

x=9∗8mod11

x=72mod11

x=6mod11.

For the other hand we solve the another congruence, \begin{gathered}(3x+2) = 0 mod (11)\\3x = -2 mod 11\\3x = 9 mod 11\end{gathered}

(3x+2)=0mod(11)

3x=−2mod11

3x=9mod11

So the solution is {3, 6}