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Number Theory solution using number theory

suvadip

Member
Feb 21, 2013
69
Show that 2x^3+x^2+3x-1 = 0 (mod 5)
has exactly three solutions


How to proceed with it?
 
Last edited:

ThePerfectHacker

Well-known member
Jan 26, 2012
236
Substitute $x=0,1,2,3,4$.
 

soroban

Well-known member
Feb 2, 2012
409
Hello, suvadip!

I suppose someone wants to see all the steps.



[tex]\text{Show that }\,2x^3+x^2+3x-1\:\equiv\:0\text{ (mod 5)}[/tex]
[tex]\text{has exactly three solutions.}[/tex]

We have: .[tex]2x^3+x^2 + 3x-1\:\equiv\:0\text{ (mod 5)}[/tex]

Then: .[tex]2x^3 + x^2 - 2x - 1 \:\equiv\:0\text{ (mod 5)}[/tex]

Factor: .[tex]x^2(2x+1) - (2x+1) \:\equiv\;0\text{ (mod 5)}[/tex]

Factor: .[tex](2x+1)(x^2-1) \:\equiv\:0\text{ (mod 5)}[/tex]

Factor: .[tex](2x+1)(x-1)(x+1) \:\equiv\:0\text{ (mod 5)}[/tex]


[tex]2x+1\:\equiv\:0 \text{ (mod 5)}\quad\Rightarrow\quad 2x \:\equiv\:-1 \text{ (mod 5)}[/tex]

. . [tex]2x \:\equiv\:4 \text{ (mod 5)} \quad\Rightarrow\quad \boxed{x \:\equiv\:2\text{ (mod 5)}} [/tex]


[tex]x-1\:\equiv\:0\text{ (mod 5)} \quad\Rightarrow\quad \boxed{x \:\equiv\:1\text{ (mod 5)}}[/tex]


[tex]x+1\:\equiv\:0\text{ (mod 5)} \quad\Rightarrow\quad x \:\equiv\:-1\text{ (mod 5)}[/tex]

. . [tex]\boxed{x\:\equiv\:4\text{ (mod 5)}}[/tex]