Thread: Congruence Equations

Could someone help me find the solutions to the following:

(x^2)+x=0 in Z5.
(x^2)+x=0 in Z6.

Thanks. MK

Hello, MK!

These are of the same type . . . I'll explain the first one.

x² + x .= .0 .(mod 5)

We can apply standard algebraic methods.

Factor: .x(x + 1) .= .0 .(mod 5)

Then: .x .= .0 .(mod 5) . → . x .= .5m, for some integer m.

And: .x+1 .= .0 .(mod 5) . → . x .= .5n - 1, for some integer n.

Originally Posted by Soroban

Hello, MK!

These are of the same type . . . I'll explain the first one.


We can apply standard algebraic methods.

Factor: .x(x + 1) .= .0 .(mod 5)

Then: .x .= .0 .(mod 5) . → . x .= .5m, for some integer m.

And: .x+1 .= .0 .(mod 5) . → . x .= .5n - 1, for some integer n.

One thing Soroban did not mention is that, the ring of integers multiplication modulo n>1 has no zero-divisors meaning that if, (where n is a prime or power of a prime).
xy=0
Then,
x=0 or y=0

Originally Posted by MKLyon

(x^2)+x=0 in Z6.

Same idea but more complex because the ring,
Z6 has zero divisors.

Thus,
x(x+1)=0
Thus,
x=0, thus, x=0
x+1=0, thus, x=5
x=2 and x+1=3, thus, x=2
x=3 and x+1=2 no solution.