# One Congruence Problem

• Sep 29th 2010, 09:14 PM
Janu42
One Congruence Problem
I'm struggling with this one:

How many incongruent solutions are there to the congruence \$\displaystyle x^5 + x - 6 = 0(mod 144)\$?

I know I have to use Hensel's Lemma here. But the examples I've seen the mod is easily factored (e.g. \$\displaystyle 27 = 3^3\$, or \$\displaystyle 25 = 5^2\$) so those examples aren't exactly the same as what is presented here I'm guessing since 144 is \$\displaystyle 2^4 * 3^2\$
• Sep 29th 2010, 09:27 PM
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Quote:

Originally Posted by Janu42
I'm struggling with this one:

How many incongruent solutions are there to the congruence \$\displaystyle x^5 + x - 6 = 0(mod 144)\$?

I know I have to use Hensel's Lemma here. But the examples I've seen the mod is easily factored (e.g. \$\displaystyle 27 = 3^3\$, or \$\displaystyle 25 = 5^2\$) so those examples aren't exactly the same as what is presented here I'm guessing since 144 is \$\displaystyle 2^4 * 3^2\$

If you solve mod 2^4 and mod 3^2 then you can combine with the Chinese Remainder Theorem.
• Sep 29th 2010, 09:41 PM
Janu42
OK but it's asking me how many incongruent solutions there are, so how does that work? There's not necessarily just one solution right?
• Sep 29th 2010, 09:51 PM
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Quote:

Originally Posted by Janu42
OK but it's asking me how many incongruent solutions there are, so how does that work? There's not necessarily just one solution right?

Oh right, you only have to count the solutions, not find them.

Well if you count n distinct (that is, incongruent) solutions mod 2^4 and m distinct solutions mod 3^2, then the number of distinct solutions mod 144 is nm by CRT.

To be honest I've only worked a little bit with Hensel's lemma so if you have a hard time finding n and m I may not be of much help, in that case someone else will probably step in.