Does there exist a prime p for each natural n such that all polynomials in Zp of nth degree or less with coefficients in [-n,n]∩Z split?
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Does there exist a prime $p$ for each natural number $n$ such that all $n$th degree or less polynomials in $mathbbZ_p$ with coefficients in $[-n,n] cap mathbbZ$ split?
Motivation:
If you have a finite field and want to add roots of various polynomials, an easy way to do this is to construct an extension field that gives you those new roots; however, if you instead demand that the field be of the form $mathbbZ_p$ for some prime $p,$ then this no longer works. Instead, you must find a new prime that has roots for the additional polynomials.
If you know of a proof of the statement above or a link to a resource discussing this problem or related problems, please let me know. Thanks.
based on the discussion so far it appears that this is not possible for all n, so I therefore would like to know what the largest such n is, Thanks
Regarding the choice of p1, it is easy to show this will never work, consider the polynomial of degree 2 x^2-k, this will only split if k is a square, however in Zp for any prime p there are (p-1)/2 non squares and all k must be considered because [-n,n] is equivalent to [0,p-1]. Therefore we must consider p greater than n if we hope to find such a p. Like I said above, based on the conversation so far it seems like this is not possible for all n so please try to find an n for which this can't be done.
number-theory finite-fields
asked Aug 17 at 3:41
mathew
380114
 |Â
show 7 more comments
up vote
2
down vote
favorite
Does there exist a prime $p$ for each natural number $n$ such that all $n$th degree or less polynomials in $mathbbZ_p$ with coefficients in $[-n,n] cap mathbbZ$ split?
Motivation:
If you have a finite field and want to add roots of various polynomials, an easy way to do this is to construct an extension field that gives you those new roots; however, if you instead demand that the field be of the form $mathbbZ_p$ for some prime $p,$ then this no longer works. Instead, you must find a new prime that has roots for the additional polynomials.
If you know of a proof of the statement above or a link to a resource discussing this problem or related problems, please let me know. Thanks.
based on the discussion so far it appears that this is not possible for all n, so I therefore would like to know what the largest such n is, Thanks
Regarding the choice of p1, it is easy to show this will never work, consider the polynomial of degree 2 x^2-k, this will only split if k is a square, however in Zp for any prime p there are (p-1)/2 non squares and all k must be considered because [-n,n] is equivalent to [0,p-1]. Therefore we must consider p greater than n if we hope to find such a p. Like I said above, based on the conversation so far it seems like this is not possible for all n so please try to find an n for which this can't be done.
number-theory finite-fields
asked Aug 17 at 3:41
mathew
380114
I am not sure what is meant by "add[ing] roots to various polynomials." Every finite field is order $p^k$ for some prime $p$ and some positive integer $k.$ Generally, if you adjoin a root $alpha$ of an irreducible polynomial of degree $d$ to the finite field $F_q,$ you obtain a field of order $q^d.$ (Here, $q$ is prime-power.)
– Dylan_Carlo_Beck
Aug 17 at 3:52
Furthermore, there are irreducible polynomials of every degree in $mathbbZ_p$ for every prime $p,$ so given a prime $p$ and a positive integer $n,$ there exists an irreducible polynomial of degree $n$ in $mathbbZ_p,$ and therefore, this polynomial cannot split in $mathbbZ_p.$
– Dylan_Carlo_Beck
Aug 17 at 3:54
sorry, I meant adding roots of various polynomials, I just fixed that
– mathew
Aug 17 at 4:12
I realize that all finite fields have irreducible polynomials, what I'm asking is that all polynomials of a particular kind be split-able, that is, polynomials of with degree n or less with coefficients in the interval [-n,n]
– mathew
Aug 17 at 4:14
1
Do you mean for your degree $n$ and the $n$ in your set $-n, -n+1, dots, n-1, n $ to coincide?
– Dylan_Carlo_Beck
Aug 17 at 4:16
 |Â
show 7 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Does there exist a prime $p$ for each natural number $n$ such that all $n$th degree or less polynomials in $mathbbZ_p$ with coefficients in $[-n,n] cap mathbbZ$ split?
Motivation:
If you have a finite field and want to add roots of various polynomials, an easy way to do this is to construct an extension field that gives you those new roots; however, if you instead demand that the field be of the form $mathbbZ_p$ for some prime $p,$ then this no longer works. Instead, you must find a new prime that has roots for the additional polynomials.
If you know of a proof of the statement above or a link to a resource discussing this problem or related problems, please let me know. Thanks.
based on the discussion so far it appears that this is not possible for all n, so I therefore would like to know what the largest such n is, Thanks
Regarding the choice of p1, it is easy to show this will never work, consider the polynomial of degree 2 x^2-k, this will only split if k is a square, however in Zp for any prime p there are (p-1)/2 non squares and all k must be considered because [-n,n] is equivalent to [0,p-1]. Therefore we must consider p greater than n if we hope to find such a p. Like I said above, based on the conversation so far it seems like this is not possible for all n so please try to find an n for which this can't be done.
number-theory finite-fields
asked Aug 17 at 3:41
mathew
380114
Does there exist a prime $p$ for each natural number $n$ such that all $n$th degree or less polynomials in $mathbbZ_p$ with coefficients in $[-n,n] cap mathbbZ$ split?
Motivation:
If you have a finite field and want to add roots of various polynomials, an easy way to do this is to construct an extension field that gives you those new roots; however, if you instead demand that the field be of the form $mathbbZ_p$ for some prime $p,$ then this no longer works. Instead, you must find a new prime that has roots for the additional polynomials.
If you know of a proof of the statement above or a link to a resource discussing this problem or related problems, please let me know. Thanks.
based on the discussion so far it appears that this is not possible for all n, so I therefore would like to know what the largest such n is, Thanks
Regarding the choice of p1, it is easy to show this will never work, consider the polynomial of degree 2 x^2-k, this will only split if k is a square, however in Zp for any prime p there are (p-1)/2 non squares and all k must be considered because [-n,n] is equivalent to [0,p-1]. Therefore we must consider p greater than n if we hope to find such a p. Like I said above, based on the conversation so far it seems like this is not possible for all n so please try to find an n for which this can't be done.
number-theory finite-fields
asked Aug 17 at 3:41
mathew
380114
edited Aug 17 at 4:58
asked Aug 17 at 3:41
mathew
380114
asked Aug 17 at 3:41
mathew
380114
asked Aug 17 at 3:41
mathew
380114
380114
I am not sure what is meant by "add[ing] roots to various polynomials." Every finite field is order $p^k$ for some prime $p$ and some positive integer $k.$ Generally, if you adjoin a root $alpha$ of an irreducible polynomial of degree $d$ to the finite field $F_q,$ you obtain a field of order $q^d.$ (Here, $q$ is prime-power.)
– Dylan_Carlo_Beck
Aug 17 at 3:52
Furthermore, there are irreducible polynomials of every degree in $mathbbZ_p$ for every prime $p,$ so given a prime $p$ and a positive integer $n,$ there exists an irreducible polynomial of degree $n$ in $mathbbZ_p,$ and therefore, this polynomial cannot split in $mathbbZ_p.$
– Dylan_Carlo_Beck
Aug 17 at 3:54
sorry, I meant adding roots of various polynomials, I just fixed that
– mathew
Aug 17 at 4:12
I realize that all finite fields have irreducible polynomials, what I'm asking is that all polynomials of a particular kind be split-able, that is, polynomials of with degree n or less with coefficients in the interval [-n,n]
– mathew
Aug 17 at 4:14
1
Do you mean for your degree $n$ and the $n$ in your set $-n, -n+1, dots, n-1, n $ to coincide?
– Dylan_Carlo_Beck
Aug 17 at 4:16
 |Â
show 7 more comments
I am not sure what is meant by "add[ing] roots to various polynomials." Every finite field is order $p^k$ for some prime $p$ and some positive integer $k.$ Generally, if you adjoin a root $alpha$ of an irreducible polynomial of degree $d$ to the finite field $F_q,$ you obtain a field of order $q^d.$ (Here, $q$ is prime-power.)
– Dylan_Carlo_Beck
Aug 17 at 3:52
Furthermore, there are irreducible polynomials of every degree in $mathbbZ_p$ for every prime $p,$ so given a prime $p$ and a positive integer $n,$ there exists an irreducible polynomial of degree $n$ in $mathbbZ_p,$ and therefore, this polynomial cannot split in $mathbbZ_p.$
– Dylan_Carlo_Beck
Aug 17 at 3:54
sorry, I meant adding roots of various polynomials, I just fixed that
– mathew
Aug 17 at 4:12
I realize that all finite fields have irreducible polynomials, what I'm asking is that all polynomials of a particular kind be split-able, that is, polynomials of with degree n or less with coefficients in the interval [-n,n]
– mathew
Aug 17 at 4:14
1
Do you mean for your degree $n$ and the $n$ in your set $-n, -n+1, dots, n-1, n $ to coincide?
– Dylan_Carlo_Beck
Aug 17 at 4:16
I am not sure what is meant by "add[ing] roots to various polynomials." Every finite field is order $p^k$ for some prime $p$ and some positive integer $k.$ Generally, if you adjoin a root $alpha$ of an irreducible polynomial of degree $d$ to the finite field $F_q,$ you obtain a field of order $q^d.$ (Here, $q$ is prime-power.)
– Dylan_Carlo_Beck
Aug 17 at 3:52
I am not sure what is meant by "add[ing] roots to various polynomials." Every finite field is order $p^k$ for some prime $p$ and some positive integer $k.$ Generally, if you adjoin a root $alpha$ of an irreducible polynomial of degree $d$ to the finite field $F_q,$ you obtain a field of order $q^d.$ (Here, $q$ is prime-power.)
– Dylan_Carlo_Beck
Aug 17 at 3:52
Furthermore, there are irreducible polynomials of every degree in $mathbbZ_p$ for every prime $p,$ so given a prime $p$ and a positive integer $n,$ there exists an irreducible polynomial of degree $n$ in $mathbbZ_p,$ and therefore, this polynomial cannot split in $mathbbZ_p.$
– Dylan_Carlo_Beck
Aug 17 at 3:54
Furthermore, there are irreducible polynomials of every degree in $mathbbZ_p$ for every prime $p,$ so given a prime $p$ and a positive integer $n,$ there exists an irreducible polynomial of degree $n$ in $mathbbZ_p,$ and therefore, this polynomial cannot split in $mathbbZ_p.$
– Dylan_Carlo_Beck
Aug 17 at 3:54
sorry, I meant adding roots of various polynomials, I just fixed that
– mathew
Aug 17 at 4:12
sorry, I meant adding roots of various polynomials, I just fixed that
– mathew
Aug 17 at 4:12
I realize that all finite fields have irreducible polynomials, what I'm asking is that all polynomials of a particular kind be split-able, that is, polynomials of with degree n or less with coefficients in the interval [-n,n]
– mathew
Aug 17 at 4:14
I realize that all finite fields have irreducible polynomials, what I'm asking is that all polynomials of a particular kind be split-able, that is, polynomials of with degree n or less with coefficients in the interval [-n,n]
– mathew
Aug 17 at 4:14
1
1
Do you mean for your degree $n$ and the $n$ in your set $-n, -n+1, dots, n-1, n $ to coincide?
– Dylan_Carlo_Beck
Aug 17 at 4:16
Do you mean for your degree $n$ and the $n$ in your set $-n, -n+1, dots, n-1, n $ to coincide?
– Dylan_Carlo_Beck
Aug 17 at 4:16
 |Â
show 7 more comments
1 Answer
1
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oldest
votes
up vote
4
down vote
accepted
This follows from Chebotarev's density theorem.
Let $f_n(x)$ be the product of all those (finitely many for a fixed $n$) polynomials, viewed as a polynomial with integer coefficients. Let $L_n$ be the splitting field of $f_n(x)$ over the rationals. If $G_n=Gal(L_n/BbbQ)$, then Chebotarev's theorem implies that the density of primes $p$ that split totally in $L_n$ is $1/|G_n|$. The total splitting of a prime $p$ implies that $f_n(x)$ splits into linear factors modulo $p$.
So there will exist infinitely many primes with the prescribed property, but finding one may take a while.
answered Aug 17 at 5:03
Jyrki Lahtonen
105k12161358
There are known error terms to Chebotarev's density formula, but I don't remember any. Those would tell how high you need to go to find a suitable prime $p$.
– Jyrki Lahtonen
Aug 17 at 5:05
I am in a discussion with @ Dylan_Carlo_Beck right now, I'm not sure how you would join us if you want to but if you know how and would like to please do
– mathew
Aug 17 at 5:17
I don't completely understand your answer, I concerned that your looking at spliting fields over the rationals, I am concerned only with fields of the form Zp for some prime p which is of course not characteristic 0 so the rationals shouldn't come into this as far as I can tell, but I'm probably missing your point
– mathew
Aug 17 at 5:18
@mathew You need a few pieces from algebraic number theory to see the connection. A prime ideal $p$ splits completely in the field $BbbQ(alpha)$, if the minimal polynomial of $alpha$ splits completely modulo $p$. We cannot arrange $f_n(x)$ to be the minimal polynomial of anything, but that doesn't matter, because the splitting of a prime ideal only depends on the field.
– Jyrki Lahtonen
Aug 17 at 5:23
ok thanks for your answer, I will need to examine it a bit more to fully understand
– mathew
Aug 17 at 5:31
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
This follows from Chebotarev's density theorem.
Let $f_n(x)$ be the product of all those (finitely many for a fixed $n$) polynomials, viewed as a polynomial with integer coefficients. Let $L_n$ be the splitting field of $f_n(x)$ over the rationals. If $G_n=Gal(L_n/BbbQ)$, then Chebotarev's theorem implies that the density of primes $p$ that split totally in $L_n$ is $1/|G_n|$. The total splitting of a prime $p$ implies that $f_n(x)$ splits into linear factors modulo $p$.
So there will exist infinitely many primes with the prescribed property, but finding one may take a while.
answered Aug 17 at 5:03
Jyrki Lahtonen
105k12161358
There are known error terms to Chebotarev's density formula, but I don't remember any. Those would tell how high you need to go to find a suitable prime $p$.
– Jyrki Lahtonen
Aug 17 at 5:05
I am in a discussion with @ Dylan_Carlo_Beck right now, I'm not sure how you would join us if you want to but if you know how and would like to please do
– mathew
Aug 17 at 5:17
I don't completely understand your answer, I concerned that your looking at spliting fields over the rationals, I am concerned only with fields of the form Zp for some prime p which is of course not characteristic 0 so the rationals shouldn't come into this as far as I can tell, but I'm probably missing your point
– mathew
Aug 17 at 5:18
@mathew You need a few pieces from algebraic number theory to see the connection. A prime ideal $p$ splits completely in the field $BbbQ(alpha)$, if the minimal polynomial of $alpha$ splits completely modulo $p$. We cannot arrange $f_n(x)$ to be the minimal polynomial of anything, but that doesn't matter, because the splitting of a prime ideal only depends on the field.
– Jyrki Lahtonen
Aug 17 at 5:23
ok thanks for your answer, I will need to examine it a bit more to fully understand
– mathew
Aug 17 at 5:31
 |Â
show 1 more comment
up vote
4
down vote
accepted
This follows from Chebotarev's density theorem.
Let $f_n(x)$ be the product of all those (finitely many for a fixed $n$) polynomials, viewed as a polynomial with integer coefficients. Let $L_n$ be the splitting field of $f_n(x)$ over the rationals. If $G_n=Gal(L_n/BbbQ)$, then Chebotarev's theorem implies that the density of primes $p$ that split totally in $L_n$ is $1/|G_n|$. The total splitting of a prime $p$ implies that $f_n(x)$ splits into linear factors modulo $p$.
So there will exist infinitely many primes with the prescribed property, but finding one may take a while.
answered Aug 17 at 5:03
Jyrki Lahtonen
105k12161358
There are known error terms to Chebotarev's density formula, but I don't remember any. Those would tell how high you need to go to find a suitable prime $p$.
– Jyrki Lahtonen
Aug 17 at 5:05
I am in a discussion with @ Dylan_Carlo_Beck right now, I'm not sure how you would join us if you want to but if you know how and would like to please do
– mathew
Aug 17 at 5:17
I don't completely understand your answer, I concerned that your looking at spliting fields over the rationals, I am concerned only with fields of the form Zp for some prime p which is of course not characteristic 0 so the rationals shouldn't come into this as far as I can tell, but I'm probably missing your point
– mathew
Aug 17 at 5:18
@mathew You need a few pieces from algebraic number theory to see the connection. A prime ideal $p$ splits completely in the field $BbbQ(alpha)$, if the minimal polynomial of $alpha$ splits completely modulo $p$. We cannot arrange $f_n(x)$ to be the minimal polynomial of anything, but that doesn't matter, because the splitting of a prime ideal only depends on the field.
– Jyrki Lahtonen
Aug 17 at 5:23
ok thanks for your answer, I will need to examine it a bit more to fully understand
– mathew
Aug 17 at 5:31
 |Â
show 1 more comment
up vote
4
down vote
accepted
up vote
4
down vote
accepted
This follows from Chebotarev's density theorem.
Let $f_n(x)$ be the product of all those (finitely many for a fixed $n$) polynomials, viewed as a polynomial with integer coefficients. Let $L_n$ be the splitting field of $f_n(x)$ over the rationals. If $G_n=Gal(L_n/BbbQ)$, then Chebotarev's theorem implies that the density of primes $p$ that split totally in $L_n$ is $1/|G_n|$. The total splitting of a prime $p$ implies that $f_n(x)$ splits into linear factors modulo $p$.
So there will exist infinitely many primes with the prescribed property, but finding one may take a while.
answered Aug 17 at 5:03
Jyrki Lahtonen
105k12161358
This follows from Chebotarev's density theorem.
Let $f_n(x)$ be the product of all those (finitely many for a fixed $n$) polynomials, viewed as a polynomial with integer coefficients. Let $L_n$ be the splitting field of $f_n(x)$ over the rationals. If $G_n=Gal(L_n/BbbQ)$, then Chebotarev's theorem implies that the density of primes $p$ that split totally in $L_n$ is $1/|G_n|$. The total splitting of a prime $p$ implies that $f_n(x)$ splits into linear factors modulo $p$.
So there will exist infinitely many primes with the prescribed property, but finding one may take a while.
answered Aug 17 at 5:03
Jyrki Lahtonen
105k12161358
edited Aug 23 at 7:53
answered Aug 17 at 5:03
Jyrki Lahtonen
105k12161358
answered Aug 17 at 5:03
Jyrki Lahtonen
105k12161358
answered Aug 17 at 5:03
Jyrki Lahtonen
105k12161358
105k12161358
There are known error terms to Chebotarev's density formula, but I don't remember any. Those would tell how high you need to go to find a suitable prime $p$.
– Jyrki Lahtonen
Aug 17 at 5:05
I am in a discussion with @ Dylan_Carlo_Beck right now, I'm not sure how you would join us if you want to but if you know how and would like to please do
– mathew
Aug 17 at 5:17
I don't completely understand your answer, I concerned that your looking at spliting fields over the rationals, I am concerned only with fields of the form Zp for some prime p which is of course not characteristic 0 so the rationals shouldn't come into this as far as I can tell, but I'm probably missing your point
– mathew
Aug 17 at 5:18
@mathew You need a few pieces from algebraic number theory to see the connection. A prime ideal $p$ splits completely in the field $BbbQ(alpha)$, if the minimal polynomial of $alpha$ splits completely modulo $p$. We cannot arrange $f_n(x)$ to be the minimal polynomial of anything, but that doesn't matter, because the splitting of a prime ideal only depends on the field.
– Jyrki Lahtonen
Aug 17 at 5:23
ok thanks for your answer, I will need to examine it a bit more to fully understand
– mathew
Aug 17 at 5:31
 |Â
show 1 more comment
There are known error terms to Chebotarev's density formula, but I don't remember any. Those would tell how high you need to go to find a suitable prime $p$.
– Jyrki Lahtonen
Aug 17 at 5:05
I am in a discussion with @ Dylan_Carlo_Beck right now, I'm not sure how you would join us if you want to but if you know how and would like to please do
– mathew
Aug 17 at 5:17
I don't completely understand your answer, I concerned that your looking at spliting fields over the rationals, I am concerned only with fields of the form Zp for some prime p which is of course not characteristic 0 so the rationals shouldn't come into this as far as I can tell, but I'm probably missing your point
– mathew
Aug 17 at 5:18
@mathew You need a few pieces from algebraic number theory to see the connection. A prime ideal $p$ splits completely in the field $BbbQ(alpha)$, if the minimal polynomial of $alpha$ splits completely modulo $p$. We cannot arrange $f_n(x)$ to be the minimal polynomial of anything, but that doesn't matter, because the splitting of a prime ideal only depends on the field.
– Jyrki Lahtonen
Aug 17 at 5:23
ok thanks for your answer, I will need to examine it a bit more to fully understand
– mathew
Aug 17 at 5:31
There are known error terms to Chebotarev's density formula, but I don't remember any. Those would tell how high you need to go to find a suitable prime $p$.
– Jyrki Lahtonen
Aug 17 at 5:05
There are known error terms to Chebotarev's density formula, but I don't remember any. Those would tell how high you need to go to find a suitable prime $p$.
– Jyrki Lahtonen
Aug 17 at 5:05
I am in a discussion with @ Dylan_Carlo_Beck right now, I'm not sure how you would join us if you want to but if you know how and would like to please do
– mathew
Aug 17 at 5:17
I am in a discussion with @ Dylan_Carlo_Beck right now, I'm not sure how you would join us if you want to but if you know how and would like to please do
– mathew
Aug 17 at 5:17
I don't completely understand your answer, I concerned that your looking at spliting fields over the rationals, I am concerned only with fields of the form Zp for some prime p which is of course not characteristic 0 so the rationals shouldn't come into this as far as I can tell, but I'm probably missing your point
– mathew
Aug 17 at 5:18
I don't completely understand your answer, I concerned that your looking at spliting fields over the rationals, I am concerned only with fields of the form Zp for some prime p which is of course not characteristic 0 so the rationals shouldn't come into this as far as I can tell, but I'm probably missing your point
– mathew
Aug 17 at 5:18
@mathew You need a few pieces from algebraic number theory to see the connection. A prime ideal $p$ splits completely in the field $BbbQ(alpha)$, if the minimal polynomial of $alpha$ splits completely modulo $p$. We cannot arrange $f_n(x)$ to be the minimal polynomial of anything, but that doesn't matter, because the splitting of a prime ideal only depends on the field.
– Jyrki Lahtonen
Aug 17 at 5:23
@mathew You need a few pieces from algebraic number theory to see the connection. A prime ideal $p$ splits completely in the field $BbbQ(alpha)$, if the minimal polynomial of $alpha$ splits completely modulo $p$. We cannot arrange $f_n(x)$ to be the minimal polynomial of anything, but that doesn't matter, because the splitting of a prime ideal only depends on the field.
– Jyrki Lahtonen
Aug 17 at 5:23
ok thanks for your answer, I will need to examine it a bit more to fully understand
– mathew
Aug 17 at 5:31
ok thanks for your answer, I will need to examine it a bit more to fully understand
– mathew
Aug 17 at 5:31
 |Â
show 1 more comment
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I am not sure what is meant by "add[ing] roots to various polynomials." Every finite field is order $p^k$ for some prime $p$ and some positive integer $k.$ Generally, if you adjoin a root $alpha$ of an irreducible polynomial of degree $d$ to the finite field $F_q,$ you obtain a field of order $q^d.$ (Here, $q$ is prime-power.)
– Dylan_Carlo_Beck
Aug 17 at 3:52
Furthermore, there are irreducible polynomials of every degree in $mathbbZ_p$ for every prime $p,$ so given a prime $p$ and a positive integer $n,$ there exists an irreducible polynomial of degree $n$ in $mathbbZ_p,$ and therefore, this polynomial cannot split in $mathbbZ_p.$
– Dylan_Carlo_Beck
Aug 17 at 3:54
sorry, I meant adding roots of various polynomials, I just fixed that
– mathew
Aug 17 at 4:12
I realize that all finite fields have irreducible polynomials, what I'm asking is that all polynomials of a particular kind be split-able, that is, polynomials of with degree n or less with coefficients in the interval [-n,n]
– mathew
Aug 17 at 4:14
1
Do you mean for your degree $n$ and the $n$ in your set $-n, -n+1, dots, n-1, n $ to coincide?
– Dylan_Carlo_Beck
Aug 17 at 4:16