Differences of Prime Numbers












10












$begingroup$


Let $a<b<c$ be primes such that $c-a$, $c-b$, and $b-a$ are also prime. It is rather simple to show that $(2,5,7)$ is the only triple that satisfies these conditions:



Proof Sketch:




  1. The case $a>2$ reduces to a system of equations with no solution after realizing that each difference must be equal to $2$.


  2. The case $a=2$ reduces to the existence of a prime $p$ such that $p+2$ and $p+4$ are also prime. The only such tuple is $p=3rightarrow (3,5,7)$. A modular argument takes care of uniqueness.


  3. The tuple $(a,b,c)=(2,5,7)$ follows from $p=3$.



I was wondering if there is a more elegant approach utilizing many number theoretic tools (i.e. elliptic curves, algebraic number theory, ect.)? I realize that this is entirely unnecessary as we can appeal to the most atomic of theory to solve this problem; however, I don't spend much time with number theory and was looking for some application of "modern techniques".






number-theory prime-numbers







share|cite|improve this question









$endgroup$












  • $begingroup$
    Probably not. Sorry.
    $endgroup$
    – Qiaochu Yuan
    Sep 27 '15 at 2:30






  • 1




    $begingroup$
    Care to elaborate? As I stated, I have little "feel" for such techniques.
    $endgroup$
    – Joseph Zambrano
    Sep 27 '15 at 2:32






  • 1




    $begingroup$
    @QiaochuYuan Really. Op writes an entire post, well made too, and you respond with an opinion worth ten cents? I mean, what is the point in doing that? You don't even say it can't be done you just vacillate about the possibility of being wrong...
    $endgroup$
    – Zach466920
    Sep 27 '15 at 4:34






  • 5




    $begingroup$
    @Zach466920: look, it's a comment specifically because it's not an answer. It's very hard to rule out the possibility of a technique being applied to a problem, but it's unlikely in this case. Problems of this general type belong to an area sometimes called "additive number theory" and the techniques there just look very different from the techniques used in algebraic number theory, etc. In any case, there's a perfectly nice elementary solution that makes the problem very clear; I don't see a need to go hunting for more sophisticated techniques on this particular problem relative to others.
    $endgroup$
    – Qiaochu Yuan
    Sep 27 '15 at 5:01












  • $begingroup$
    @QiaochuYaun Thank you for the clarification. I will certainly look into some additive number theoretic techniques. To clarify, this isn't an attempt to hunt for overly complicated techniques; rather, I am interested in juxtaposing the elementary proof with something "higher level" to see what such techniques have to offer (perhaps some deeper context). Like I said, I have spent little time in this domain so I thought this might be a good example to do just that.
    $endgroup$
    – Joseph Zambrano
    Sep 27 '15 at 13:24
















10












$begingroup$


Let $a<b<c$ be primes such that $c-a$, $c-b$, and $b-a$ are also prime. It is rather simple to show that $(2,5,7)$ is the only triple that satisfies these conditions:



Proof Sketch:




  1. The case $a>2$ reduces to a system of equations with no solution after realizing that each difference must be equal to $2$.


  2. The case $a=2$ reduces to the existence of a prime $p$ such that $p+2$ and $p+4$ are also prime. The only such tuple is $p=3rightarrow (3,5,7)$. A modular argument takes care of uniqueness.


  3. The tuple $(a,b,c)=(2,5,7)$ follows from $p=3$.



I was wondering if there is a more elegant approach utilizing many number theoretic tools (i.e. elliptic curves, algebraic number theory, ect.)? I realize that this is entirely unnecessary as we can appeal to the most atomic of theory to solve this problem; however, I don't spend much time with number theory and was looking for some application of "modern techniques".






number-theory prime-numbers







share|cite|improve this question









$endgroup$












  • $begingroup$
    Probably not. Sorry.
    $endgroup$
    – Qiaochu Yuan
    Sep 27 '15 at 2:30






  • 1




    $begingroup$
    Care to elaborate? As I stated, I have little "feel" for such techniques.
    $endgroup$
    – Joseph Zambrano
    Sep 27 '15 at 2:32






  • 1




    $begingroup$
    @QiaochuYuan Really. Op writes an entire post, well made too, and you respond with an opinion worth ten cents? I mean, what is the point in doing that? You don't even say it can't be done you just vacillate about the possibility of being wrong...
    $endgroup$
    – Zach466920
    Sep 27 '15 at 4:34






  • 5




    $begingroup$
    @Zach466920: look, it's a comment specifically because it's not an answer. It's very hard to rule out the possibility of a technique being applied to a problem, but it's unlikely in this case. Problems of this general type belong to an area sometimes called "additive number theory" and the techniques there just look very different from the techniques used in algebraic number theory, etc. In any case, there's a perfectly nice elementary solution that makes the problem very clear; I don't see a need to go hunting for more sophisticated techniques on this particular problem relative to others.
    $endgroup$
    – Qiaochu Yuan
    Sep 27 '15 at 5:01












  • $begingroup$
    @QiaochuYaun Thank you for the clarification. I will certainly look into some additive number theoretic techniques. To clarify, this isn't an attempt to hunt for overly complicated techniques; rather, I am interested in juxtaposing the elementary proof with something "higher level" to see what such techniques have to offer (perhaps some deeper context). Like I said, I have spent little time in this domain so I thought this might be a good example to do just that.
    $endgroup$
    – Joseph Zambrano
    Sep 27 '15 at 13:24














10












10








10


2



$begingroup$


Let $a<b<c$ be primes such that $c-a$, $c-b$, and $b-a$ are also prime. It is rather simple to show that $(2,5,7)$ is the only triple that satisfies these conditions:



Proof Sketch:




  1. The case $a>2$ reduces to a system of equations with no solution after realizing that each difference must be equal to $2$.


  2. The case $a=2$ reduces to the existence of a prime $p$ such that $p+2$ and $p+4$ are also prime. The only such tuple is $p=3rightarrow (3,5,7)$. A modular argument takes care of uniqueness.


  3. The tuple $(a,b,c)=(2,5,7)$ follows from $p=3$.



I was wondering if there is a more elegant approach utilizing many number theoretic tools (i.e. elliptic curves, algebraic number theory, ect.)? I realize that this is entirely unnecessary as we can appeal to the most atomic of theory to solve this problem; however, I don't spend much time with number theory and was looking for some application of "modern techniques".






number-theory prime-numbers







share|cite|improve this question









$endgroup$




Let $a<b<c$ be primes such that $c-a$, $c-b$, and $b-a$ are also prime. It is rather simple to show that $(2,5,7)$ is the only triple that satisfies these conditions:



Proof Sketch:




  1. The case $a>2$ reduces to a system of equations with no solution after realizing that each difference must be equal to $2$.


  2. The case $a=2$ reduces to the existence of a prime $p$ such that $p+2$ and $p+4$ are also prime. The only such tuple is $p=3rightarrow (3,5,7)$. A modular argument takes care of uniqueness.


  3. The tuple $(a,b,c)=(2,5,7)$ follows from $p=3$.



I was wondering if there is a more elegant approach utilizing many number theoretic tools (i.e. elliptic curves, algebraic number theory, ect.)? I realize that this is entirely unnecessary as we can appeal to the most atomic of theory to solve this problem; however, I don't spend much time with number theory and was looking for some application of "modern techniques".






number-theory prime-numbers




number-theory prime-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 27 '15 at 2:00









Joseph ZambranoJoseph Zambrano

623415




623415












  • $begingroup$
    Probably not. Sorry.
    $endgroup$
    – Qiaochu Yuan
    Sep 27 '15 at 2:30






  • 1




    $begingroup$
    Care to elaborate? As I stated, I have little "feel" for such techniques.
    $endgroup$
    – Joseph Zambrano
    Sep 27 '15 at 2:32






  • 1




    $begingroup$
    @QiaochuYuan Really. Op writes an entire post, well made too, and you respond with an opinion worth ten cents? I mean, what is the point in doing that? You don't even say it can't be done you just vacillate about the possibility of being wrong...
    $endgroup$
    – Zach466920
    Sep 27 '15 at 4:34






  • 5




    $begingroup$
    @Zach466920: look, it's a comment specifically because it's not an answer. It's very hard to rule out the possibility of a technique being applied to a problem, but it's unlikely in this case. Problems of this general type belong to an area sometimes called "additive number theory" and the techniques there just look very different from the techniques used in algebraic number theory, etc. In any case, there's a perfectly nice elementary solution that makes the problem very clear; I don't see a need to go hunting for more sophisticated techniques on this particular problem relative to others.
    $endgroup$
    – Qiaochu Yuan
    Sep 27 '15 at 5:01












  • $begingroup$
    @QiaochuYaun Thank you for the clarification. I will certainly look into some additive number theoretic techniques. To clarify, this isn't an attempt to hunt for overly complicated techniques; rather, I am interested in juxtaposing the elementary proof with something "higher level" to see what such techniques have to offer (perhaps some deeper context). Like I said, I have spent little time in this domain so I thought this might be a good example to do just that.
    $endgroup$
    – Joseph Zambrano
    Sep 27 '15 at 13:24


















  • $begingroup$
    Probably not. Sorry.
    $endgroup$
    – Qiaochu Yuan
    Sep 27 '15 at 2:30






  • 1




    $begingroup$
    Care to elaborate? As I stated, I have little "feel" for such techniques.
    $endgroup$
    – Joseph Zambrano
    Sep 27 '15 at 2:32






  • 1




    $begingroup$
    @QiaochuYuan Really. Op writes an entire post, well made too, and you respond with an opinion worth ten cents? I mean, what is the point in doing that? You don't even say it can't be done you just vacillate about the possibility of being wrong...
    $endgroup$
    – Zach466920
    Sep 27 '15 at 4:34






  • 5




    $begingroup$
    @Zach466920: look, it's a comment specifically because it's not an answer. It's very hard to rule out the possibility of a technique being applied to a problem, but it's unlikely in this case. Problems of this general type belong to an area sometimes called "additive number theory" and the techniques there just look very different from the techniques used in algebraic number theory, etc. In any case, there's a perfectly nice elementary solution that makes the problem very clear; I don't see a need to go hunting for more sophisticated techniques on this particular problem relative to others.
    $endgroup$
    – Qiaochu Yuan
    Sep 27 '15 at 5:01












  • $begingroup$
    @QiaochuYaun Thank you for the clarification. I will certainly look into some additive number theoretic techniques. To clarify, this isn't an attempt to hunt for overly complicated techniques; rather, I am interested in juxtaposing the elementary proof with something "higher level" to see what such techniques have to offer (perhaps some deeper context). Like I said, I have spent little time in this domain so I thought this might be a good example to do just that.
    $endgroup$
    – Joseph Zambrano
    Sep 27 '15 at 13:24
















$begingroup$
Probably not. Sorry.
$endgroup$
– Qiaochu Yuan
Sep 27 '15 at 2:30




$begingroup$
Probably not. Sorry.
$endgroup$
– Qiaochu Yuan
Sep 27 '15 at 2:30




1




1




$begingroup$
Care to elaborate? As I stated, I have little "feel" for such techniques.
$endgroup$
– Joseph Zambrano
Sep 27 '15 at 2:32




$begingroup$
Care to elaborate? As I stated, I have little "feel" for such techniques.
$endgroup$
– Joseph Zambrano
Sep 27 '15 at 2:32




1




1




$begingroup$
@QiaochuYuan Really. Op writes an entire post, well made too, and you respond with an opinion worth ten cents? I mean, what is the point in doing that? You don't even say it can't be done you just vacillate about the possibility of being wrong...
$endgroup$
– Zach466920
Sep 27 '15 at 4:34




$begingroup$
@QiaochuYuan Really. Op writes an entire post, well made too, and you respond with an opinion worth ten cents? I mean, what is the point in doing that? You don't even say it can't be done you just vacillate about the possibility of being wrong...
$endgroup$
– Zach466920
Sep 27 '15 at 4:34




5




5




$begingroup$
@Zach466920: look, it's a comment specifically because it's not an answer. It's very hard to rule out the possibility of a technique being applied to a problem, but it's unlikely in this case. Problems of this general type belong to an area sometimes called "additive number theory" and the techniques there just look very different from the techniques used in algebraic number theory, etc. In any case, there's a perfectly nice elementary solution that makes the problem very clear; I don't see a need to go hunting for more sophisticated techniques on this particular problem relative to others.
$endgroup$
– Qiaochu Yuan
Sep 27 '15 at 5:01






$begingroup$
@Zach466920: look, it's a comment specifically because it's not an answer. It's very hard to rule out the possibility of a technique being applied to a problem, but it's unlikely in this case. Problems of this general type belong to an area sometimes called "additive number theory" and the techniques there just look very different from the techniques used in algebraic number theory, etc. In any case, there's a perfectly nice elementary solution that makes the problem very clear; I don't see a need to go hunting for more sophisticated techniques on this particular problem relative to others.
$endgroup$
– Qiaochu Yuan
Sep 27 '15 at 5:01














$begingroup$
@QiaochuYaun Thank you for the clarification. I will certainly look into some additive number theoretic techniques. To clarify, this isn't an attempt to hunt for overly complicated techniques; rather, I am interested in juxtaposing the elementary proof with something "higher level" to see what such techniques have to offer (perhaps some deeper context). Like I said, I have spent little time in this domain so I thought this might be a good example to do just that.
$endgroup$
– Joseph Zambrano
Sep 27 '15 at 13:24




$begingroup$
@QiaochuYaun Thank you for the clarification. I will certainly look into some additive number theoretic techniques. To clarify, this isn't an attempt to hunt for overly complicated techniques; rather, I am interested in juxtaposing the elementary proof with something "higher level" to see what such techniques have to offer (perhaps some deeper context). Like I said, I have spent little time in this domain so I thought this might be a good example to do just that.
$endgroup$
– Joseph Zambrano
Sep 27 '15 at 13:24










1 Answer
1






active

oldest

votes


















0












$begingroup$

So I actually not giving the mathematical proof but I am describing the reason of not having more triplets.



As a$<$b$<$c are primes such that c-a, c-b and b-a are also primes.



So we can't take all primes as odd one because there does't exist such odd primes a,b&c which will satisfy our condition because the values of c-a, c-b and b-a are even numbers and one of these values must greater than 2, which will not be a prime.



Now it is clear that out of these 3 primes (a,b&c) one should be 2 and it must be 'a' than only we can get 'b' and 'c' greater than 'a'.



Suppose if we take 'b' & 'c' as a pair of twin primes and 'a' as 2 than (c-a) is must be prime, (c-b) must be prime but it is not necessary that (b-a) is also prime because as (a=2) so b-2 is not prime as 'b' & 'c' are already twin primes so 'b-2' can't be a twin prime with 'b' except (b=5) therefore 'b-a' or 'b-2' is not a prime number (exception, b=5)



Therefore there is only 1 triplet of your conditions.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    "it is not necessary that (b-a) is also prime." Right, but you need a stronger statement: that (b-a) is only prime if it is equal to 3.
    $endgroup$
    – eyeballfrog
    Dec 18 '18 at 17:33










  • $begingroup$
    @eyeballfrog I edited my answer, is it now nearby correct answer or not?
    $endgroup$
    – Dynamo
    Dec 18 '18 at 17:42










  • $begingroup$
    That works, but it's dancing around the point a little. The point is that one of $b -2,,b,b+2$ must be divisible by 3. Since it's not $b$ or $b+2 = c$, it must be $b-2 = b-a$, and the only prime number divisible by 3 is 3.
    $endgroup$
    – eyeballfrog
    Dec 18 '18 at 17:46












  • $begingroup$
    @eyeballfrog Yes, you are right...
    $endgroup$
    – Dynamo
    Dec 18 '18 at 17:47












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

So I actually not giving the mathematical proof but I am describing the reason of not having more triplets.



As a$<$b$<$c are primes such that c-a, c-b and b-a are also primes.



So we can't take all primes as odd one because there does't exist such odd primes a,b&c which will satisfy our condition because the values of c-a, c-b and b-a are even numbers and one of these values must greater than 2, which will not be a prime.



Now it is clear that out of these 3 primes (a,b&c) one should be 2 and it must be 'a' than only we can get 'b' and 'c' greater than 'a'.



Suppose if we take 'b' & 'c' as a pair of twin primes and 'a' as 2 than (c-a) is must be prime, (c-b) must be prime but it is not necessary that (b-a) is also prime because as (a=2) so b-2 is not prime as 'b' & 'c' are already twin primes so 'b-2' can't be a twin prime with 'b' except (b=5) therefore 'b-a' or 'b-2' is not a prime number (exception, b=5)



Therefore there is only 1 triplet of your conditions.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    "it is not necessary that (b-a) is also prime." Right, but you need a stronger statement: that (b-a) is only prime if it is equal to 3.
    $endgroup$
    – eyeballfrog
    Dec 18 '18 at 17:33










  • $begingroup$
    @eyeballfrog I edited my answer, is it now nearby correct answer or not?
    $endgroup$
    – Dynamo
    Dec 18 '18 at 17:42










  • $begingroup$
    That works, but it's dancing around the point a little. The point is that one of $b -2,,b,b+2$ must be divisible by 3. Since it's not $b$ or $b+2 = c$, it must be $b-2 = b-a$, and the only prime number divisible by 3 is 3.
    $endgroup$
    – eyeballfrog
    Dec 18 '18 at 17:46












  • $begingroup$
    @eyeballfrog Yes, you are right...
    $endgroup$
    – Dynamo
    Dec 18 '18 at 17:47
















0












$begingroup$

So I actually not giving the mathematical proof but I am describing the reason of not having more triplets.



As a$<$b$<$c are primes such that c-a, c-b and b-a are also primes.



So we can't take all primes as odd one because there does't exist such odd primes a,b&c which will satisfy our condition because the values of c-a, c-b and b-a are even numbers and one of these values must greater than 2, which will not be a prime.



Now it is clear that out of these 3 primes (a,b&c) one should be 2 and it must be 'a' than only we can get 'b' and 'c' greater than 'a'.



Suppose if we take 'b' & 'c' as a pair of twin primes and 'a' as 2 than (c-a) is must be prime, (c-b) must be prime but it is not necessary that (b-a) is also prime because as (a=2) so b-2 is not prime as 'b' & 'c' are already twin primes so 'b-2' can't be a twin prime with 'b' except (b=5) therefore 'b-a' or 'b-2' is not a prime number (exception, b=5)



Therefore there is only 1 triplet of your conditions.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    "it is not necessary that (b-a) is also prime." Right, but you need a stronger statement: that (b-a) is only prime if it is equal to 3.
    $endgroup$
    – eyeballfrog
    Dec 18 '18 at 17:33










  • $begingroup$
    @eyeballfrog I edited my answer, is it now nearby correct answer or not?
    $endgroup$
    – Dynamo
    Dec 18 '18 at 17:42










  • $begingroup$
    That works, but it's dancing around the point a little. The point is that one of $b -2,,b,b+2$ must be divisible by 3. Since it's not $b$ or $b+2 = c$, it must be $b-2 = b-a$, and the only prime number divisible by 3 is 3.
    $endgroup$
    – eyeballfrog
    Dec 18 '18 at 17:46












  • $begingroup$
    @eyeballfrog Yes, you are right...
    $endgroup$
    – Dynamo
    Dec 18 '18 at 17:47














0












0








0





$begingroup$

So I actually not giving the mathematical proof but I am describing the reason of not having more triplets.



As a$<$b$<$c are primes such that c-a, c-b and b-a are also primes.



So we can't take all primes as odd one because there does't exist such odd primes a,b&c which will satisfy our condition because the values of c-a, c-b and b-a are even numbers and one of these values must greater than 2, which will not be a prime.



Now it is clear that out of these 3 primes (a,b&c) one should be 2 and it must be 'a' than only we can get 'b' and 'c' greater than 'a'.



Suppose if we take 'b' & 'c' as a pair of twin primes and 'a' as 2 than (c-a) is must be prime, (c-b) must be prime but it is not necessary that (b-a) is also prime because as (a=2) so b-2 is not prime as 'b' & 'c' are already twin primes so 'b-2' can't be a twin prime with 'b' except (b=5) therefore 'b-a' or 'b-2' is not a prime number (exception, b=5)



Therefore there is only 1 triplet of your conditions.






share|cite|improve this answer











$endgroup$



So I actually not giving the mathematical proof but I am describing the reason of not having more triplets.



As a$<$b$<$c are primes such that c-a, c-b and b-a are also primes.



So we can't take all primes as odd one because there does't exist such odd primes a,b&c which will satisfy our condition because the values of c-a, c-b and b-a are even numbers and one of these values must greater than 2, which will not be a prime.



Now it is clear that out of these 3 primes (a,b&c) one should be 2 and it must be 'a' than only we can get 'b' and 'c' greater than 'a'.



Suppose if we take 'b' & 'c' as a pair of twin primes and 'a' as 2 than (c-a) is must be prime, (c-b) must be prime but it is not necessary that (b-a) is also prime because as (a=2) so b-2 is not prime as 'b' & 'c' are already twin primes so 'b-2' can't be a twin prime with 'b' except (b=5) therefore 'b-a' or 'b-2' is not a prime number (exception, b=5)



Therefore there is only 1 triplet of your conditions.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 18 '18 at 17:41

























answered Dec 18 '18 at 16:41









DynamoDynamo

104517




104517








  • 1




    $begingroup$
    "it is not necessary that (b-a) is also prime." Right, but you need a stronger statement: that (b-a) is only prime if it is equal to 3.
    $endgroup$
    – eyeballfrog
    Dec 18 '18 at 17:33










  • $begingroup$
    @eyeballfrog I edited my answer, is it now nearby correct answer or not?
    $endgroup$
    – Dynamo
    Dec 18 '18 at 17:42










  • $begingroup$
    That works, but it's dancing around the point a little. The point is that one of $b -2,,b,b+2$ must be divisible by 3. Since it's not $b$ or $b+2 = c$, it must be $b-2 = b-a$, and the only prime number divisible by 3 is 3.
    $endgroup$
    – eyeballfrog
    Dec 18 '18 at 17:46












  • $begingroup$
    @eyeballfrog Yes, you are right...
    $endgroup$
    – Dynamo
    Dec 18 '18 at 17:47














  • 1




    $begingroup$
    "it is not necessary that (b-a) is also prime." Right, but you need a stronger statement: that (b-a) is only prime if it is equal to 3.
    $endgroup$
    – eyeballfrog
    Dec 18 '18 at 17:33










  • $begingroup$
    @eyeballfrog I edited my answer, is it now nearby correct answer or not?
    $endgroup$
    – Dynamo
    Dec 18 '18 at 17:42










  • $begingroup$
    That works, but it's dancing around the point a little. The point is that one of $b -2,,b,b+2$ must be divisible by 3. Since it's not $b$ or $b+2 = c$, it must be $b-2 = b-a$, and the only prime number divisible by 3 is 3.
    $endgroup$
    – eyeballfrog
    Dec 18 '18 at 17:46












  • $begingroup$
    @eyeballfrog Yes, you are right...
    $endgroup$
    – Dynamo
    Dec 18 '18 at 17:47








1




1




$begingroup$
"it is not necessary that (b-a) is also prime." Right, but you need a stronger statement: that (b-a) is only prime if it is equal to 3.
$endgroup$
– eyeballfrog
Dec 18 '18 at 17:33




$begingroup$
"it is not necessary that (b-a) is also prime." Right, but you need a stronger statement: that (b-a) is only prime if it is equal to 3.
$endgroup$
– eyeballfrog
Dec 18 '18 at 17:33












$begingroup$
@eyeballfrog I edited my answer, is it now nearby correct answer or not?
$endgroup$
– Dynamo
Dec 18 '18 at 17:42




$begingroup$
@eyeballfrog I edited my answer, is it now nearby correct answer or not?
$endgroup$
– Dynamo
Dec 18 '18 at 17:42












$begingroup$
That works, but it's dancing around the point a little. The point is that one of $b -2,,b,b+2$ must be divisible by 3. Since it's not $b$ or $b+2 = c$, it must be $b-2 = b-a$, and the only prime number divisible by 3 is 3.
$endgroup$
– eyeballfrog
Dec 18 '18 at 17:46






$begingroup$
That works, but it's dancing around the point a little. The point is that one of $b -2,,b,b+2$ must be divisible by 3. Since it's not $b$ or $b+2 = c$, it must be $b-2 = b-a$, and the only prime number divisible by 3 is 3.
$endgroup$
– eyeballfrog
Dec 18 '18 at 17:46














$begingroup$
@eyeballfrog Yes, you are right...
$endgroup$
– Dynamo
Dec 18 '18 at 17:47




$begingroup$
@eyeballfrog Yes, you are right...
$endgroup$
– Dynamo
Dec 18 '18 at 17:47


















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