MHB Is there also an other way to solve the equation?

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The discussion focuses on solving the equation \(x^3 + 4x + 8 = 0 \pmod{15}\), where initial checks reveal no solutions within the set \(\{0, 1, 2, \ldots, 14\}\). An optimization approach suggests that if a root exists, it can be factored, leading to the conclusion that potential roots must be coprime to 15. The conversation also explores using the Chinese Remainder Theorem to find solutions by checking modulo 3 and modulo 5, confirming that while a solution exists modulo 3, none are found modulo 5. Ultimately, the discussion highlights the importance of modular arithmetic and theorems in solving polynomial equations.
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Hey! :cool:

I am looking at the exercise: Solve $x^3+4x+8=0 \pmod {15}$.

I checked all the numbers from the set $\{ 0,1,2 \dots, 14 \}$ and noticed that there are no solutions.
But could I also solve the equation,with an other way,maybe using a theorem? (Thinking)
 
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evinda said:
Hey! :cool:

I am looking at the exercise: Solve $x^3+4x+8=0 \pmod {15}$.

I checked all the numbers from the set $\{ 0,1,2 \dots, 14 \}$ and noticed that there are no solutions.
But could I also solve the equation,with an other way,maybe using a theorem? (Thinking)

Hi! ;)

Small optimization.

If there is a root $a$, it must be possible to factorize into $(x-a)(x^2+bx+c)$.
The last term is $-ac$ and it must be equivalent to $8 \pmod{15}$.
Since $(8,15)=1$, this is only possible if $(a,15)=1$, so you can skip $0,3,5,6,9,10,12 \pmod{15}$.
 
as 15 = 3 * 5 we can check for modulo 3 and modulo 5

for checking for modulo 3 we have
$x^3 + x + 2 = 0$
x = 0 and 1 does not satisfy but x = 2 does satisfy

for checking for mod 5
$x^3- x + 3= 0$
or x(x+1)(x-1) = 3

x = 0 or 1 or 4 give 1st term zero so we need to check for 2 and 3 and do not get a solution
 
I like Serena said:
Hi! ;)

Small optimization.

If there is a root $a$, it must be possible to factorize into $(x-a)(x^2+bx+c)$.
The last term is $-ac$ and it must be equivalent to $8 \pmod{15}$.
Since $(8,15)=1$, this is only possible if $(a,15)=1$, so you can skip $0,3,5,6,9,10,12 \pmod{15}$.

So,if we have for example $a \equiv b \pmod m$ and $m \nmid b$ then do we know that $m \nmid a $ ? (Thinking)

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kaliprasad said:
as 15 = 3 * 5 we can check for modulo 3 and modulo 5

for checking for modulo 3 we have
$x^3 + x + 2 = 0$
x = 0 and 1 does not satisfy but x = 2 does satisfy

for checking for mod 5
$x^3- x + 3= 0$
or x(x+1)(x-1) = 3

x = 0 or 1 or 4 give 1st term zero so we need to check for 2 and 3 and do not get a solution

Is there a theorem that says that if we are looking for solutions of an equation in $\mathbb{Z}_{m \cdot n}$ and $(m,n)=1$,then we can check for the solutions in $\mathbb{Z}_m$ and $\mathbb{Z}_n$ and the common solutions of $\mathbb{Z}_m$ and $\mathbb{Z}_n$ are the solutions of the equation in $\mathbb{Z}_{m \cdot n}$ ? :confused:
 
evinda said:
So,if we have for example $a \equiv b \pmod m$ and $m \nmid b$ then do we know that $m \nmid a $ ? (Thinking)

Erm... yes, that is correct. (Wondering)

Remember that $a \equiv b \pmod m$ means that $a=b+mk$.

So suppose $a$ is not divisible by $m$.
Since $mk$ is divisible by $m$ that implies that $b$ is indeed not divisible by $m$.
 
evinda said:
Is there a theorem that says that if we are looking for solutions of an equation in $\mathbb{Z}_{m \cdot n}$ and $(m,n)=1$,then we can check for the solutions in $\mathbb{Z}_m$ and $\mathbb{Z}_n$ and the common solutions of $\mathbb{Z}_m$ and $\mathbb{Z}_n$ are the solutions of the equation in $\mathbb{Z}_{m \cdot n}$ ? :confused:

this is known as per Chinese remainder theorem
 
I like Serena said:
Erm... yes, that is correct. (Wondering)

Remember that $a \equiv b \pmod m$ means that $a=b+mk$.

So suppose $a$ is not divisible by $m$.
Since $mk$ is divisible by $m$ that implies that $b$ is indeed not divisible by $m$.

I understand..thank you very much! (Cool)

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kaliprasad said:
this is known as per Chinese remainder theorem

A ok..thank you! :)
 
evinda said:
A ok..thank you! :)

You actually stated the Chinese Remainder Theorem quite nicely! (Angel)
 
I like Serena said:
You actually stated the Chinese Remainder Theorem quite nicely! (Angel)

(Blush) (Clapping) (Mmm)
 
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