Solutions to congruence using Chinese remainder theorem

May 2012
25
1
A place
f(x) = x^5 + 3x^2 + 4
Find all solutions to the congruence f(x) congruent to 0 mod 12 by using the Chinese remainder theorem.

I'm not sure exactly how to do this properly.

I started by creating two congruences since 12 = 4 x 3.
I manually tried entries 0 to 3 for mod 4 and entries 0 to 2 for mod 3.
The only common solution is x=2, so I tried x = 2 as well as x= 5, 8, 11 (because of mod 3) and x= 6, 10 (because of mod 4) in the original polynomial.
My final answer is x = 2, 5, or 8.
1) Is this right?
2) Whether or not it's right, how do I work this problem and present the solution properly?
Thanks in advance.
 
Jun 2013
1,113
590
Lebanon
\(\displaystyle x \equiv 2 \pmod 3\)

and

\(\displaystyle x \equiv 0,1,2 \pmod 4\)

Combining these we get 3 systems each of which can be solved using CRT

your answer is correct
 
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May 2012
25
1
A place
Thank you for the reply.

"Combining these we get 3 systems each of which can be solved using CRT"

So, the three systems are:
1) x= 2mod3 and x= 2mod4 which gives the answer x = 2
2) x= 2mod3 and x= 1mod4 which gives the answer x = 5
3) x= 2mod3 and x= 0mod4 which gives the answer x = 8

Right?
 
Last edited:
Jun 2013
1,113
590
Lebanon
Correct

\(\displaystyle x \equiv 2,5,8 \pmod {12}\)