Mixing
Find the values of p and q
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If $p^3+p=q^2+q$ where $p$ and $q$ are prime numbers,
Find all the solutions (p, q)
I tried to solve this exercise using that:
$p^2 = -1(textmod , q)$ and $q = -1(textmod , p)$;
So: $q+1=ap$ and $p^2+1=bq$, where $b$ and $q$ integers.
Then I tried to solve a quadratic equation, but I could not finish the problem
number-theory elementary-number-theory prime-numbers
asked 30 mins ago
Matheus Domingos
856
add a comment |Â
up vote
3
down vote
favorite
If $p^3+p=q^2+q$ where $p$ and $q$ are prime numbers,
Find all the solutions (p, q)
I tried to solve this exercise using that:
$p^2 = -1(textmod , q)$ and $q = -1(textmod , p)$;
So: $q+1=ap$ and $p^2+1=bq$, where $b$ and $q$ integers.
Then I tried to solve a quadratic equation, but I could not finish the problem
number-theory elementary-number-theory prime-numbers
asked 30 mins ago
Matheus Domingos
856
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
If $p^3+p=q^2+q$ where $p$ and $q$ are prime numbers,
Find all the solutions (p, q)
I tried to solve this exercise using that:
$p^2 = -1(textmod , q)$ and $q = -1(textmod , p)$;
So: $q+1=ap$ and $p^2+1=bq$, where $b$ and $q$ integers.
Then I tried to solve a quadratic equation, but I could not finish the problem
number-theory elementary-number-theory prime-numbers
asked 30 mins ago
Matheus Domingos
856
If $p^3+p=q^2+q$ where $p$ and $q$ are prime numbers,
Find all the solutions (p, q)
I tried to solve this exercise using that:
$p^2 = -1(textmod , q)$ and $q = -1(textmod , p)$;
So: $q+1=ap$ and $p^2+1=bq$, where $b$ and $q$ integers.
Then I tried to solve a quadratic equation, but I could not finish the problem
number-theory elementary-number-theory prime-numbers
number-theory elementary-number-theory prime-numbers
asked 30 mins ago
Matheus Domingos
856
asked 30 mins ago
Matheus Domingos
856
edited 19 mins ago
asked 30 mins ago
Matheus Domingos
856
asked 30 mins ago
Matheus Domingos
856
asked 30 mins ago
Matheus Domingos
856
856
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
accepted
$$p^3+p=q^2+q implies p|q^2+q implies p|q mathrmor p|(q+1).$$
Obviously, $pneq q$, so $pnmid q$. Thus, we can set $q=kp-1$. This then reduces to
$$p^2-k^2p+(k+1)=0,$$
which has an integer solution iff $k^4-4k-4$ is a square. Can you see why this is not the case for large $k$, and determine the solutions from there?
answered 19 mins ago
Carl Schildkraut
8,98711238
Oh thanks, now I know how to finish it, thanks
– Matheus Domingos
17 mins ago
Is there a better way to do that? I learned in high school to solve that using k^4-4k-4=a^2 and then I have to study that. Is there a better way to do that?
– Matheus Domingos
24 secs ago
add a comment |Â
up vote
2
down vote
Hint: Since $p$ divides $p^3+p$, also $p$ divides $q(q+1)$, hence $p$ either divides $q$ or $p$ divides $q+1$, because $p$ is prime. Can you finish it?
answered 20 mins ago
Dietrich Burde
75.7k64185
add a comment |Â
up vote
2
down vote
Clearly $pneq q$, and because $p$ and $q$ are prime and
$$p(p^2+1)=p^3+p=q^2+q=q(q+1),$$
we must have $pmid q+1$ and $qmid p^2+1$. Write
$$q+1=apqquadtext and qquad p^2+1=bq,$$
to find that $p^2-abp+b+1=0$. In particular $b+1equiv0pmodp$, say $b=cp-1$, but then
$$p^2+1=(cp-1)q=(cp-1)(ap-1)=acp^2-(a+c)p+1.$$
Note that $a$, $b$ and $c$ are positive integers, and that $a>1$ as otherwise $p=q+1$, so $(p,q)=(3,2)$ which is not a solution. The equation above implies that
$$(ac-1)p=a+c,$$
and as $pgeq2$ clearly we cannot have $a,cgeq2$. Hence $c=1$ and so
$$p^2+1=bq=(cp-1)q=(p-1)q.$$
In particular $p-1mid p^2+1$. As $p-1mid p^2-1$ it follows that $p-1=2$, so $p=3$ and hence $q=5$.
answered 20 mins ago
Servaes
18.8k33684
How do I finish this solution? Finding all the solution?
– Matheus Domingos
8 mins ago
@MatheusDomingos I completed the solution for you.
– Servaes
7 mins ago
1
Thank you, now I understood the limitation. I need to have a+c=>ac-1... What implies that (a-1)(c-1)=>2. Thanks
– Matheus Domingos
1 min ago
I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
– Servaes
35 secs ago
add a comment |Â
up vote
2
down vote
Remember that if $a,b$ are positive integers such that $amid b$ then $aleq b$. I'll be using this frekvently here.
From $$p(p^2+1)= q(q+1)implies pmid q;;;rm or;;;pmid q+1$$
1. case $pmid q$, then $q+1mid p^2+1$. Write $q+1=s$ then we get $$psmid (p^2+1)(s-1) = p^2s-p^2+s-1implies psmid p^2-s+1$$
Since $p^2+1geq s$ we have 2 subcases:
1.1 case $p^2+1>s$, then $psleq p^2-s+1$ so $s(p+1)leq p^2+1$, and thus $$sleq p^2+1over p+1 <pimplies sleq p-1$$
So $q+1leq p-1 leq q-1$ and thus no solution.
1.2 case $p^2+1=s$, then $q^2+1 = q+1$ and again no solution.
2. case $pmid q+1$, then $qmid p^2+1$. Then we get $$pqmid (p^2+1)(q+1) = p^2q+p^2+q+1implies pqmid p^2+q+1$$
so we have $pqleq p^2+q+1$ so $q leq p^2+1over p-1 leq p+2$ if $pgeq 3$.
So if $pgeq 3$ and since $pmid q$ that $qin p,p+1,p+2$ which is easy to finish by hand.
answered 18 mins ago
greedoid
30.7k94184
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$$p^3+p=q^2+q implies p|q^2+q implies p|q mathrmor p|(q+1).$$
Obviously, $pneq q$, so $pnmid q$. Thus, we can set $q=kp-1$. This then reduces to
$$p^2-k^2p+(k+1)=0,$$
which has an integer solution iff $k^4-4k-4$ is a square. Can you see why this is not the case for large $k$, and determine the solutions from there?
answered 19 mins ago
Carl Schildkraut
8,98711238
Oh thanks, now I know how to finish it, thanks
– Matheus Domingos
17 mins ago
Is there a better way to do that? I learned in high school to solve that using k^4-4k-4=a^2 and then I have to study that. Is there a better way to do that?
– Matheus Domingos
24 secs ago
add a comment |Â
up vote
1
down vote
accepted
$$p^3+p=q^2+q implies p|q^2+q implies p|q mathrmor p|(q+1).$$
Obviously, $pneq q$, so $pnmid q$. Thus, we can set $q=kp-1$. This then reduces to
$$p^2-k^2p+(k+1)=0,$$
which has an integer solution iff $k^4-4k-4$ is a square. Can you see why this is not the case for large $k$, and determine the solutions from there?
answered 19 mins ago
Carl Schildkraut
8,98711238
Oh thanks, now I know how to finish it, thanks
– Matheus Domingos
17 mins ago
Is there a better way to do that? I learned in high school to solve that using k^4-4k-4=a^2 and then I have to study that. Is there a better way to do that?
– Matheus Domingos
24 secs ago
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$$p^3+p=q^2+q implies p|q^2+q implies p|q mathrmor p|(q+1).$$
Obviously, $pneq q$, so $pnmid q$. Thus, we can set $q=kp-1$. This then reduces to
$$p^2-k^2p+(k+1)=0,$$
which has an integer solution iff $k^4-4k-4$ is a square. Can you see why this is not the case for large $k$, and determine the solutions from there?
answered 19 mins ago
Carl Schildkraut
8,98711238
$$p^3+p=q^2+q implies p|q^2+q implies p|q mathrmor p|(q+1).$$
Obviously, $pneq q$, so $pnmid q$. Thus, we can set $q=kp-1$. This then reduces to
$$p^2-k^2p+(k+1)=0,$$
which has an integer solution iff $k^4-4k-4$ is a square. Can you see why this is not the case for large $k$, and determine the solutions from there?
answered 19 mins ago
Carl Schildkraut
8,98711238
edited 12 mins ago
answered 19 mins ago
Carl Schildkraut
8,98711238
answered 19 mins ago
Carl Schildkraut
8,98711238
answered 19 mins ago
Carl Schildkraut
8,98711238
8,98711238
Oh thanks, now I know how to finish it, thanks
– Matheus Domingos
17 mins ago
Is there a better way to do that? I learned in high school to solve that using k^4-4k-4=a^2 and then I have to study that. Is there a better way to do that?
– Matheus Domingos
24 secs ago
add a comment |Â
Oh thanks, now I know how to finish it, thanks
– Matheus Domingos
17 mins ago
Is there a better way to do that? I learned in high school to solve that using k^4-4k-4=a^2 and then I have to study that. Is there a better way to do that?
– Matheus Domingos
24 secs ago
Oh thanks, now I know how to finish it, thanks
– Matheus Domingos
17 mins ago
Oh thanks, now I know how to finish it, thanks
– Matheus Domingos
17 mins ago
Is there a better way to do that? I learned in high school to solve that using k^4-4k-4=a^2 and then I have to study that. Is there a better way to do that?
– Matheus Domingos
24 secs ago
Is there a better way to do that? I learned in high school to solve that using k^4-4k-4=a^2 and then I have to study that. Is there a better way to do that?
– Matheus Domingos
24 secs ago
add a comment |Â
up vote
2
down vote
Hint: Since $p$ divides $p^3+p$, also $p$ divides $q(q+1)$, hence $p$ either divides $q$ or $p$ divides $q+1$, because $p$ is prime. Can you finish it?
answered 20 mins ago
Dietrich Burde
75.7k64185
add a comment |Â
up vote
2
down vote
Hint: Since $p$ divides $p^3+p$, also $p$ divides $q(q+1)$, hence $p$ either divides $q$ or $p$ divides $q+1$, because $p$ is prime. Can you finish it?
answered 20 mins ago
Dietrich Burde
75.7k64185
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: Since $p$ divides $p^3+p$, also $p$ divides $q(q+1)$, hence $p$ either divides $q$ or $p$ divides $q+1$, because $p$ is prime. Can you finish it?
answered 20 mins ago
Dietrich Burde
75.7k64185
Hint: Since $p$ divides $p^3+p$, also $p$ divides $q(q+1)$, hence $p$ either divides $q$ or $p$ divides $q+1$, because $p$ is prime. Can you finish it?
answered 20 mins ago
Dietrich Burde
75.7k64185
answered 20 mins ago
Dietrich Burde
75.7k64185
answered 20 mins ago
Dietrich Burde
75.7k64185
answered 20 mins ago
Dietrich Burde
75.7k64185
75.7k64185
add a comment |Â
add a comment |Â
up vote
2
down vote
Clearly $pneq q$, and because $p$ and $q$ are prime and
$$p(p^2+1)=p^3+p=q^2+q=q(q+1),$$
we must have $pmid q+1$ and $qmid p^2+1$. Write
$$q+1=apqquadtext and qquad p^2+1=bq,$$
to find that $p^2-abp+b+1=0$. In particular $b+1equiv0pmodp$, say $b=cp-1$, but then
$$p^2+1=(cp-1)q=(cp-1)(ap-1)=acp^2-(a+c)p+1.$$
Note that $a$, $b$ and $c$ are positive integers, and that $a>1$ as otherwise $p=q+1$, so $(p,q)=(3,2)$ which is not a solution. The equation above implies that
$$(ac-1)p=a+c,$$
and as $pgeq2$ clearly we cannot have $a,cgeq2$. Hence $c=1$ and so
$$p^2+1=bq=(cp-1)q=(p-1)q.$$
In particular $p-1mid p^2+1$. As $p-1mid p^2-1$ it follows that $p-1=2$, so $p=3$ and hence $q=5$.
answered 20 mins ago
Servaes
18.8k33684
How do I finish this solution? Finding all the solution?
– Matheus Domingos
8 mins ago
@MatheusDomingos I completed the solution for you.
– Servaes
7 mins ago
1
Thank you, now I understood the limitation. I need to have a+c=>ac-1... What implies that (a-1)(c-1)=>2. Thanks
– Matheus Domingos
1 min ago
I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
– Servaes
35 secs ago
add a comment |Â
up vote
2
down vote
Clearly $pneq q$, and because $p$ and $q$ are prime and
$$p(p^2+1)=p^3+p=q^2+q=q(q+1),$$
we must have $pmid q+1$ and $qmid p^2+1$. Write
$$q+1=apqquadtext and qquad p^2+1=bq,$$
to find that $p^2-abp+b+1=0$. In particular $b+1equiv0pmodp$, say $b=cp-1$, but then
$$p^2+1=(cp-1)q=(cp-1)(ap-1)=acp^2-(a+c)p+1.$$
Note that $a$, $b$ and $c$ are positive integers, and that $a>1$ as otherwise $p=q+1$, so $(p,q)=(3,2)$ which is not a solution. The equation above implies that
$$(ac-1)p=a+c,$$
and as $pgeq2$ clearly we cannot have $a,cgeq2$. Hence $c=1$ and so
$$p^2+1=bq=(cp-1)q=(p-1)q.$$
In particular $p-1mid p^2+1$. As $p-1mid p^2-1$ it follows that $p-1=2$, so $p=3$ and hence $q=5$.
answered 20 mins ago
Servaes
18.8k33684
How do I finish this solution? Finding all the solution?
– Matheus Domingos
8 mins ago
@MatheusDomingos I completed the solution for you.
– Servaes
7 mins ago
1
Thank you, now I understood the limitation. I need to have a+c=>ac-1... What implies that (a-1)(c-1)=>2. Thanks
– Matheus Domingos
1 min ago
I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
– Servaes
35 secs ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Clearly $pneq q$, and because $p$ and $q$ are prime and
$$p(p^2+1)=p^3+p=q^2+q=q(q+1),$$
we must have $pmid q+1$ and $qmid p^2+1$. Write
$$q+1=apqquadtext and qquad p^2+1=bq,$$
to find that $p^2-abp+b+1=0$. In particular $b+1equiv0pmodp$, say $b=cp-1$, but then
$$p^2+1=(cp-1)q=(cp-1)(ap-1)=acp^2-(a+c)p+1.$$
Note that $a$, $b$ and $c$ are positive integers, and that $a>1$ as otherwise $p=q+1$, so $(p,q)=(3,2)$ which is not a solution. The equation above implies that
$$(ac-1)p=a+c,$$
and as $pgeq2$ clearly we cannot have $a,cgeq2$. Hence $c=1$ and so
$$p^2+1=bq=(cp-1)q=(p-1)q.$$
In particular $p-1mid p^2+1$. As $p-1mid p^2-1$ it follows that $p-1=2$, so $p=3$ and hence $q=5$.
answered 20 mins ago
Servaes
18.8k33684
Clearly $pneq q$, and because $p$ and $q$ are prime and
$$p(p^2+1)=p^3+p=q^2+q=q(q+1),$$
we must have $pmid q+1$ and $qmid p^2+1$. Write
$$q+1=apqquadtext and qquad p^2+1=bq,$$
to find that $p^2-abp+b+1=0$. In particular $b+1equiv0pmodp$, say $b=cp-1$, but then
$$p^2+1=(cp-1)q=(cp-1)(ap-1)=acp^2-(a+c)p+1.$$
Note that $a$, $b$ and $c$ are positive integers, and that $a>1$ as otherwise $p=q+1$, so $(p,q)=(3,2)$ which is not a solution. The equation above implies that
$$(ac-1)p=a+c,$$
and as $pgeq2$ clearly we cannot have $a,cgeq2$. Hence $c=1$ and so
$$p^2+1=bq=(cp-1)q=(p-1)q.$$
In particular $p-1mid p^2+1$. As $p-1mid p^2-1$ it follows that $p-1=2$, so $p=3$ and hence $q=5$.
answered 20 mins ago
Servaes
18.8k33684
edited 7 mins ago
answered 20 mins ago
Servaes
18.8k33684
answered 20 mins ago
Servaes
18.8k33684
answered 20 mins ago
Servaes
18.8k33684
18.8k33684
How do I finish this solution? Finding all the solution?
– Matheus Domingos
8 mins ago
@MatheusDomingos I completed the solution for you.
– Servaes
7 mins ago
1
Thank you, now I understood the limitation. I need to have a+c=>ac-1... What implies that (a-1)(c-1)=>2. Thanks
– Matheus Domingos
1 min ago
I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
– Servaes
35 secs ago
add a comment |Â
How do I finish this solution? Finding all the solution?
– Matheus Domingos
8 mins ago
@MatheusDomingos I completed the solution for you.
– Servaes
7 mins ago
1
Thank you, now I understood the limitation. I need to have a+c=>ac-1... What implies that (a-1)(c-1)=>2. Thanks
– Matheus Domingos
1 min ago
I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
– Servaes
35 secs ago
How do I finish this solution? Finding all the solution?
– Matheus Domingos
8 mins ago
How do I finish this solution? Finding all the solution?
– Matheus Domingos
8 mins ago
@MatheusDomingos I completed the solution for you.
– Servaes
7 mins ago
@MatheusDomingos I completed the solution for you.
– Servaes
7 mins ago
1
1
Thank you, now I understood the limitation. I need to have a+c=>ac-1... What implies that (a-1)(c-1)=>2. Thanks
– Matheus Domingos
1 min ago
Thank you, now I understood the limitation. I need to have a+c=>ac-1... What implies that (a-1)(c-1)=>2. Thanks
– Matheus Domingos
1 min ago
I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
– Servaes
35 secs ago
I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
– Servaes
35 secs ago
add a comment |Â
up vote
2
down vote
Remember that if $a,b$ are positive integers such that $amid b$ then $aleq b$. I'll be using this frekvently here.
From $$p(p^2+1)= q(q+1)implies pmid q;;;rm or;;;pmid q+1$$
1. case $pmid q$, then $q+1mid p^2+1$. Write $q+1=s$ then we get $$psmid (p^2+1)(s-1) = p^2s-p^2+s-1implies psmid p^2-s+1$$
Since $p^2+1geq s$ we have 2 subcases:
1.1 case $p^2+1>s$, then $psleq p^2-s+1$ so $s(p+1)leq p^2+1$, and thus $$sleq p^2+1over p+1 <pimplies sleq p-1$$
So $q+1leq p-1 leq q-1$ and thus no solution.
1.2 case $p^2+1=s$, then $q^2+1 = q+1$ and again no solution.
2. case $pmid q+1$, then $qmid p^2+1$. Then we get $$pqmid (p^2+1)(q+1) = p^2q+p^2+q+1implies pqmid p^2+q+1$$
so we have $pqleq p^2+q+1$ so $q leq p^2+1over p-1 leq p+2$ if $pgeq 3$.
So if $pgeq 3$ and since $pmid q$ that $qin p,p+1,p+2$ which is easy to finish by hand.
answered 18 mins ago
greedoid
30.7k94184
add a comment |Â
up vote
2
down vote
Remember that if $a,b$ are positive integers such that $amid b$ then $aleq b$. I'll be using this frekvently here.
From $$p(p^2+1)= q(q+1)implies pmid q;;;rm or;;;pmid q+1$$
1. case $pmid q$, then $q+1mid p^2+1$. Write $q+1=s$ then we get $$psmid (p^2+1)(s-1) = p^2s-p^2+s-1implies psmid p^2-s+1$$
Since $p^2+1geq s$ we have 2 subcases:
1.1 case $p^2+1>s$, then $psleq p^2-s+1$ so $s(p+1)leq p^2+1$, and thus $$sleq p^2+1over p+1 <pimplies sleq p-1$$
So $q+1leq p-1 leq q-1$ and thus no solution.
1.2 case $p^2+1=s$, then $q^2+1 = q+1$ and again no solution.
2. case $pmid q+1$, then $qmid p^2+1$. Then we get $$pqmid (p^2+1)(q+1) = p^2q+p^2+q+1implies pqmid p^2+q+1$$
so we have $pqleq p^2+q+1$ so $q leq p^2+1over p-1 leq p+2$ if $pgeq 3$.
So if $pgeq 3$ and since $pmid q$ that $qin p,p+1,p+2$ which is easy to finish by hand.
answered 18 mins ago
greedoid
30.7k94184
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Remember that if $a,b$ are positive integers such that $amid b$ then $aleq b$. I'll be using this frekvently here.
From $$p(p^2+1)= q(q+1)implies pmid q;;;rm or;;;pmid q+1$$
1. case $pmid q$, then $q+1mid p^2+1$. Write $q+1=s$ then we get $$psmid (p^2+1)(s-1) = p^2s-p^2+s-1implies psmid p^2-s+1$$
Since $p^2+1geq s$ we have 2 subcases:
1.1 case $p^2+1>s$, then $psleq p^2-s+1$ so $s(p+1)leq p^2+1$, and thus $$sleq p^2+1over p+1 <pimplies sleq p-1$$
So $q+1leq p-1 leq q-1$ and thus no solution.
1.2 case $p^2+1=s$, then $q^2+1 = q+1$ and again no solution.
2. case $pmid q+1$, then $qmid p^2+1$. Then we get $$pqmid (p^2+1)(q+1) = p^2q+p^2+q+1implies pqmid p^2+q+1$$
so we have $pqleq p^2+q+1$ so $q leq p^2+1over p-1 leq p+2$ if $pgeq 3$.
So if $pgeq 3$ and since $pmid q$ that $qin p,p+1,p+2$ which is easy to finish by hand.
answered 18 mins ago
greedoid
30.7k94184
Remember that if $a,b$ are positive integers such that $amid b$ then $aleq b$. I'll be using this frekvently here.
From $$p(p^2+1)= q(q+1)implies pmid q;;;rm or;;;pmid q+1$$
1. case $pmid q$, then $q+1mid p^2+1$. Write $q+1=s$ then we get $$psmid (p^2+1)(s-1) = p^2s-p^2+s-1implies psmid p^2-s+1$$
Since $p^2+1geq s$ we have 2 subcases:
1.1 case $p^2+1>s$, then $psleq p^2-s+1$ so $s(p+1)leq p^2+1$, and thus $$sleq p^2+1over p+1 <pimplies sleq p-1$$
So $q+1leq p-1 leq q-1$ and thus no solution.
1.2 case $p^2+1=s$, then $q^2+1 = q+1$ and again no solution.
2. case $pmid q+1$, then $qmid p^2+1$. Then we get $$pqmid (p^2+1)(q+1) = p^2q+p^2+q+1implies pqmid p^2+q+1$$
so we have $pqleq p^2+q+1$ so $q leq p^2+1over p-1 leq p+2$ if $pgeq 3$.
So if $pgeq 3$ and since $pmid q$ that $qin p,p+1,p+2$ which is easy to finish by hand.
answered 18 mins ago
greedoid
30.7k94184
edited 2 mins ago
answered 18 mins ago
greedoid
30.7k94184
answered 18 mins ago
greedoid
30.7k94184
answered 18 mins ago
greedoid
30.7k94184
30.7k94184
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