Find X in the Equation












0












$begingroup$


I'm not a mathematician and I have forgotten about some basics in mathematics.
I have this equation:



$$x^y pmod z = w$$



Given $y, z,$ and $w,$ how will I find $x$? How will I get the equation for $x$?






modular-arithmetic linear-transformations education prime-factorization







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    (old joke): I found $x$, it is inside the parentheses: $$(boxed{x}^y) mod z = w$$
    $endgroup$
    – Michael
    Sep 17 '18 at 4:22








  • 1




    $begingroup$
    @Michael Lol.Anyways how will I get the equation for x?
    $endgroup$
    – Eliyah
    Sep 17 '18 at 4:28






  • 1




    $begingroup$
    Please refer to how to ask a good question to help you improve formatting and your tags
    $endgroup$
    – Chase Ryan Taylor
    Sep 17 '18 at 5:00
















0












$begingroup$


I'm not a mathematician and I have forgotten about some basics in mathematics.
I have this equation:



$$x^y pmod z = w$$



Given $y, z,$ and $w,$ how will I find $x$? How will I get the equation for $x$?






modular-arithmetic linear-transformations education prime-factorization







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    (old joke): I found $x$, it is inside the parentheses: $$(boxed{x}^y) mod z = w$$
    $endgroup$
    – Michael
    Sep 17 '18 at 4:22








  • 1




    $begingroup$
    @Michael Lol.Anyways how will I get the equation for x?
    $endgroup$
    – Eliyah
    Sep 17 '18 at 4:28






  • 1




    $begingroup$
    Please refer to how to ask a good question to help you improve formatting and your tags
    $endgroup$
    – Chase Ryan Taylor
    Sep 17 '18 at 5:00














0












0








0





$begingroup$


I'm not a mathematician and I have forgotten about some basics in mathematics.
I have this equation:



$$x^y pmod z = w$$



Given $y, z,$ and $w,$ how will I find $x$? How will I get the equation for $x$?






modular-arithmetic linear-transformations education prime-factorization







share|cite|improve this question











$endgroup$




I'm not a mathematician and I have forgotten about some basics in mathematics.
I have this equation:



$$x^y pmod z = w$$



Given $y, z,$ and $w,$ how will I find $x$? How will I get the equation for $x$?






modular-arithmetic linear-transformations education prime-factorization




modular-arithmetic linear-transformations education prime-factorization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 13:40









amWhy

192k28225439




192k28225439










asked Sep 17 '18 at 4:13









EliyahEliyah

1236




1236








  • 1




    $begingroup$
    (old joke): I found $x$, it is inside the parentheses: $$(boxed{x}^y) mod z = w$$
    $endgroup$
    – Michael
    Sep 17 '18 at 4:22








  • 1




    $begingroup$
    @Michael Lol.Anyways how will I get the equation for x?
    $endgroup$
    – Eliyah
    Sep 17 '18 at 4:28






  • 1




    $begingroup$
    Please refer to how to ask a good question to help you improve formatting and your tags
    $endgroup$
    – Chase Ryan Taylor
    Sep 17 '18 at 5:00














  • 1




    $begingroup$
    (old joke): I found $x$, it is inside the parentheses: $$(boxed{x}^y) mod z = w$$
    $endgroup$
    – Michael
    Sep 17 '18 at 4:22








  • 1




    $begingroup$
    @Michael Lol.Anyways how will I get the equation for x?
    $endgroup$
    – Eliyah
    Sep 17 '18 at 4:28






  • 1




    $begingroup$
    Please refer to how to ask a good question to help you improve formatting and your tags
    $endgroup$
    – Chase Ryan Taylor
    Sep 17 '18 at 5:00








1




1




$begingroup$
(old joke): I found $x$, it is inside the parentheses: $$(boxed{x}^y) mod z = w$$
$endgroup$
– Michael
Sep 17 '18 at 4:22






$begingroup$
(old joke): I found $x$, it is inside the parentheses: $$(boxed{x}^y) mod z = w$$
$endgroup$
– Michael
Sep 17 '18 at 4:22






1




1




$begingroup$
@Michael Lol.Anyways how will I get the equation for x?
$endgroup$
– Eliyah
Sep 17 '18 at 4:28




$begingroup$
@Michael Lol.Anyways how will I get the equation for x?
$endgroup$
– Eliyah
Sep 17 '18 at 4:28




1




1




$begingroup$
Please refer to how to ask a good question to help you improve formatting and your tags
$endgroup$
– Chase Ryan Taylor
Sep 17 '18 at 5:00




$begingroup$
Please refer to how to ask a good question to help you improve formatting and your tags
$endgroup$
– Chase Ryan Taylor
Sep 17 '18 at 5:00










1 Answer
1






active

oldest

votes


















2












$begingroup$

I assume here that $x,y,z,w$ are integers (and $y$ is a positive integer). Then your equation means:
$$ x^y = rz+w$$
for some integer $r$. Now note that if $x$ has this property, then so does $x+kz$ for any integer $k$. So it suffices to search for $x in {0, 1, 2, ..., z-1}$.



Example:
$$x^{11} mod 5 = 2 $$



Try $x in {0, 1, 2, 3, 4}$:
begin{align}
0^{11} = 0 : (mod 5)\
1^{11} = 1 : (mod 5) \
2^{11} = 3 : (mod 5) \
boxed{3^{11} = 2} : (mod 5) \
4^{11} = 4 : (mod 5)
end{align}
So then solutions are $x=3+5k$ for all integers $k$.




The $mod z$ operation is equal to the remainder when we divide by $z$. So for example




  • $Rem[26/5] = 1$ if and only if $26-1$ is divisible by $5$.

  • $Rem[83/9] = 2$ if and only if $83-2$ is divisible by $9$.

  • $Rem[26/7] = 5$ if and only if $26 - 5$ is divisible by $7$.

  • $Rem[n/z] = w$ if and only if $n-w$ is divisible by $z$.


In particular $Rem[n/z]=w$ if and only if $n-w= rz$ for some integer $r$.



If we take $n=x^y$ then we see that $Rem[x^y/z]=w$ if and only if $x^y - w = rz$ for some integer $r$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much but how did u get the rz term. Or what is the opposite operation of a modulo? Can you explain it step by step. Thank you again.
    $endgroup$
    – Eliyah
    Sep 17 '18 at 4:49








  • 1




    $begingroup$
    See my added content on this.
    $endgroup$
    – Michael
    Sep 17 '18 at 4:56











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1 Answer
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1 Answer
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2












$begingroup$

I assume here that $x,y,z,w$ are integers (and $y$ is a positive integer). Then your equation means:
$$ x^y = rz+w$$
for some integer $r$. Now note that if $x$ has this property, then so does $x+kz$ for any integer $k$. So it suffices to search for $x in {0, 1, 2, ..., z-1}$.



Example:
$$x^{11} mod 5 = 2 $$



Try $x in {0, 1, 2, 3, 4}$:
begin{align}
0^{11} = 0 : (mod 5)\
1^{11} = 1 : (mod 5) \
2^{11} = 3 : (mod 5) \
boxed{3^{11} = 2} : (mod 5) \
4^{11} = 4 : (mod 5)
end{align}
So then solutions are $x=3+5k$ for all integers $k$.




The $mod z$ operation is equal to the remainder when we divide by $z$. So for example




  • $Rem[26/5] = 1$ if and only if $26-1$ is divisible by $5$.

  • $Rem[83/9] = 2$ if and only if $83-2$ is divisible by $9$.

  • $Rem[26/7] = 5$ if and only if $26 - 5$ is divisible by $7$.

  • $Rem[n/z] = w$ if and only if $n-w$ is divisible by $z$.


In particular $Rem[n/z]=w$ if and only if $n-w= rz$ for some integer $r$.



If we take $n=x^y$ then we see that $Rem[x^y/z]=w$ if and only if $x^y - w = rz$ for some integer $r$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much but how did u get the rz term. Or what is the opposite operation of a modulo? Can you explain it step by step. Thank you again.
    $endgroup$
    – Eliyah
    Sep 17 '18 at 4:49








  • 1




    $begingroup$
    See my added content on this.
    $endgroup$
    – Michael
    Sep 17 '18 at 4:56
















2












$begingroup$

I assume here that $x,y,z,w$ are integers (and $y$ is a positive integer). Then your equation means:
$$ x^y = rz+w$$
for some integer $r$. Now note that if $x$ has this property, then so does $x+kz$ for any integer $k$. So it suffices to search for $x in {0, 1, 2, ..., z-1}$.



Example:
$$x^{11} mod 5 = 2 $$



Try $x in {0, 1, 2, 3, 4}$:
begin{align}
0^{11} = 0 : (mod 5)\
1^{11} = 1 : (mod 5) \
2^{11} = 3 : (mod 5) \
boxed{3^{11} = 2} : (mod 5) \
4^{11} = 4 : (mod 5)
end{align}
So then solutions are $x=3+5k$ for all integers $k$.




The $mod z$ operation is equal to the remainder when we divide by $z$. So for example




  • $Rem[26/5] = 1$ if and only if $26-1$ is divisible by $5$.

  • $Rem[83/9] = 2$ if and only if $83-2$ is divisible by $9$.

  • $Rem[26/7] = 5$ if and only if $26 - 5$ is divisible by $7$.

  • $Rem[n/z] = w$ if and only if $n-w$ is divisible by $z$.


In particular $Rem[n/z]=w$ if and only if $n-w= rz$ for some integer $r$.



If we take $n=x^y$ then we see that $Rem[x^y/z]=w$ if and only if $x^y - w = rz$ for some integer $r$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you so much but how did u get the rz term. Or what is the opposite operation of a modulo? Can you explain it step by step. Thank you again.
    $endgroup$
    – Eliyah
    Sep 17 '18 at 4:49








  • 1




    $begingroup$
    See my added content on this.
    $endgroup$
    – Michael
    Sep 17 '18 at 4:56














2












2








2





$begingroup$

I assume here that $x,y,z,w$ are integers (and $y$ is a positive integer). Then your equation means:
$$ x^y = rz+w$$
for some integer $r$. Now note that if $x$ has this property, then so does $x+kz$ for any integer $k$. So it suffices to search for $x in {0, 1, 2, ..., z-1}$.



Example:
$$x^{11} mod 5 = 2 $$



Try $x in {0, 1, 2, 3, 4}$:
begin{align}
0^{11} = 0 : (mod 5)\
1^{11} = 1 : (mod 5) \
2^{11} = 3 : (mod 5) \
boxed{3^{11} = 2} : (mod 5) \
4^{11} = 4 : (mod 5)
end{align}
So then solutions are $x=3+5k$ for all integers $k$.




The $mod z$ operation is equal to the remainder when we divide by $z$. So for example




  • $Rem[26/5] = 1$ if and only if $26-1$ is divisible by $5$.

  • $Rem[83/9] = 2$ if and only if $83-2$ is divisible by $9$.

  • $Rem[26/7] = 5$ if and only if $26 - 5$ is divisible by $7$.

  • $Rem[n/z] = w$ if and only if $n-w$ is divisible by $z$.


In particular $Rem[n/z]=w$ if and only if $n-w= rz$ for some integer $r$.



If we take $n=x^y$ then we see that $Rem[x^y/z]=w$ if and only if $x^y - w = rz$ for some integer $r$.






share|cite|improve this answer











$endgroup$



I assume here that $x,y,z,w$ are integers (and $y$ is a positive integer). Then your equation means:
$$ x^y = rz+w$$
for some integer $r$. Now note that if $x$ has this property, then so does $x+kz$ for any integer $k$. So it suffices to search for $x in {0, 1, 2, ..., z-1}$.



Example:
$$x^{11} mod 5 = 2 $$



Try $x in {0, 1, 2, 3, 4}$:
begin{align}
0^{11} = 0 : (mod 5)\
1^{11} = 1 : (mod 5) \
2^{11} = 3 : (mod 5) \
boxed{3^{11} = 2} : (mod 5) \
4^{11} = 4 : (mod 5)
end{align}
So then solutions are $x=3+5k$ for all integers $k$.




The $mod z$ operation is equal to the remainder when we divide by $z$. So for example




  • $Rem[26/5] = 1$ if and only if $26-1$ is divisible by $5$.

  • $Rem[83/9] = 2$ if and only if $83-2$ is divisible by $9$.

  • $Rem[26/7] = 5$ if and only if $26 - 5$ is divisible by $7$.

  • $Rem[n/z] = w$ if and only if $n-w$ is divisible by $z$.


In particular $Rem[n/z]=w$ if and only if $n-w= rz$ for some integer $r$.



If we take $n=x^y$ then we see that $Rem[x^y/z]=w$ if and only if $x^y - w = rz$ for some integer $r$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 17 '18 at 5:01

























answered Sep 17 '18 at 4:42









MichaelMichael

12.8k11429




12.8k11429












  • $begingroup$
    Thank you so much but how did u get the rz term. Or what is the opposite operation of a modulo? Can you explain it step by step. Thank you again.
    $endgroup$
    – Eliyah
    Sep 17 '18 at 4:49








  • 1




    $begingroup$
    See my added content on this.
    $endgroup$
    – Michael
    Sep 17 '18 at 4:56


















  • $begingroup$
    Thank you so much but how did u get the rz term. Or what is the opposite operation of a modulo? Can you explain it step by step. Thank you again.
    $endgroup$
    – Eliyah
    Sep 17 '18 at 4:49








  • 1




    $begingroup$
    See my added content on this.
    $endgroup$
    – Michael
    Sep 17 '18 at 4:56
















$begingroup$
Thank you so much but how did u get the rz term. Or what is the opposite operation of a modulo? Can you explain it step by step. Thank you again.
$endgroup$
– Eliyah
Sep 17 '18 at 4:49






$begingroup$
Thank you so much but how did u get the rz term. Or what is the opposite operation of a modulo? Can you explain it step by step. Thank you again.
$endgroup$
– Eliyah
Sep 17 '18 at 4:49






1




1




$begingroup$
See my added content on this.
$endgroup$
– Michael
Sep 17 '18 at 4:56




$begingroup$
See my added content on this.
$endgroup$
– Michael
Sep 17 '18 at 4:56


















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