Is there also an other way to solve the equation?

[evinda]

Hey!

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)

 

[I like Serena]

Hey!

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 $0,3,5,6,9,10,12 \pmod{15}$.

 

[kaliprasad]

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

 

[evinda]

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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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 $\mathbb{Z}_{m \cdot n}$ ?

 

[I like Serena]

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 $m$.

 

[kaliprasad]

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 $\mathbb{Z}_{m \cdot n}$ ?

this is known as per Chinese remainder theorem

 

[evinda]

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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this is known as per Chinese remainder theorem

A ok..thank you! :)

 

[I like Serena]

A ok..thank you! :)

You actually stated the Chinese Remainder Theorem quite nicely! (Angel)

 

[evinda]

You actually stated the Chinese Remainder Theorem quite nicely! (Angel)

(Blush) (Clapping) (Mmm)