Find the values of $p$ and $q$
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:
$p^2 = -1(text{mod} , q)$ and $q = -1(text{mod} , p)$;
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
algebra-precalculus elementary-number-theory polynomials prime-numbers diophantine-equations
edited Nov 28 '18 at 1:50
Servaes
22.4k33793
asked Oct 6 '18 at 18:43
Matheus Domingos
956
add a comment |
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:
$p^2 = -1(text{mod} , q)$ and $q = -1(text{mod} , p)$;
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
algebra-precalculus elementary-number-theory polynomials prime-numbers diophantine-equations
edited Nov 28 '18 at 1:50
Servaes
22.4k33793
asked Oct 6 '18 at 18:43
Matheus Domingos
956
add a comment |
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:
$p^2 = -1(text{mod} , q)$ and $q = -1(text{mod} , p)$;
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
algebra-precalculus elementary-number-theory polynomials prime-numbers diophantine-equations
edited Nov 28 '18 at 1:50
Servaes
22.4k33793
asked Oct 6 '18 at 18:43
Matheus Domingos
956
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:
$p^2 = -1(text{mod} , q)$ and $q = -1(text{mod} , p)$;
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
algebra-precalculus elementary-number-theory polynomials prime-numbers diophantine-equations
algebra-precalculus elementary-number-theory polynomials prime-numbers diophantine-equations
edited Nov 28 '18 at 1:50
Servaes
22.4k33793
asked Oct 6 '18 at 18:43
Matheus Domingos
956
edited Nov 28 '18 at 1:50
Servaes
22.4k33793
asked Oct 6 '18 at 18:43
Matheus Domingos
956
edited Nov 28 '18 at 1:50
Servaes
22.4k33793
edited Nov 28 '18 at 1:50
Servaes
22.4k33793
edited Nov 28 '18 at 1:50
Servaes
22.4k33793
22.4k33793
asked Oct 6 '18 at 18:43
Matheus Domingos
956
asked Oct 6 '18 at 18:43
Matheus Domingos
956
asked Oct 6 '18 at 18:43
Matheus Domingos
956
956
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
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 $b+1equiv0pmod{p}$, 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 $p=q+1$ which implies that $(p,q)=(3,2)$ which is not a solution. The equation above simplifies to
$$(ac-1)p=a+c,$$
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 Oct 6 '18 at 18:53
Servaes
22.4k33793
How do I finish this solution? Finding all the solution?
– Matheus Domingos
Oct 6 '18 at 19:05
@MatheusDomingos I completed the solution for you.
– Servaes
Oct 6 '18 at 19:05
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
Oct 6 '18 at 19:11
1
I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
– Servaes
Oct 6 '18 at 19:12
Understood, more elegant this way,thanks again
– Matheus Domingos
Oct 6 '18 at 19:14
add a comment |
$$p^3+p=q^2+q implies p|q^2+q implies p|q mathrm{or} p|(q+1).$$
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 Oct 6 '18 at 18:54
Carl Schildkraut
11.2k11441
Oh thanks, now I know how to finish it, thanks
– Matheus Domingos
Oct 6 '18 at 18:55
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
Oct 6 '18 at 19:12
@MatheusDomingos Note that $k^4-4k-4$ is very close to $(k^2)^2$. Specifically, for most positive $k$, it is between $(k^2-1)^2$ and $k^2$.
– Carl Schildkraut
Oct 6 '18 at 19:16
1
Note that $k^4=(k^2)^2$ is a square, and that the previous square is $$(k^2-1)^2=k^4-2k^2+1,$$ which is smaller than $k^4-4k-4$ whenever $2k^2-1>4k+4$, and so the latter cannot be a square unless $2k^2-1leq 4k+4$.
– Servaes
Oct 6 '18 at 19:17
Sorry, I didn't understood how to finish using it
– Matheus Domingos
Oct 6 '18 at 19:24
|
show 1 more comment
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 Oct 6 '18 at 18:52
Dietrich Burde
77.7k64386
add a comment |
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 Oct 6 '18 at 18:54
greedoid
38k114794
Thanks, got this solution!
– Matheus Domingos
Oct 6 '18 at 19:25
add a comment |
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4 Answers
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4 Answers
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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 $b+1equiv0pmod{p}$, 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 $p=q+1$ which implies that $(p,q)=(3,2)$ which is not a solution. The equation above simplifies to
$$(ac-1)p=a+c,$$
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 Oct 6 '18 at 18:53
Servaes
22.4k33793
How do I finish this solution? Finding all the solution?
– Matheus Domingos
Oct 6 '18 at 19:05
@MatheusDomingos I completed the solution for you.
– Servaes
Oct 6 '18 at 19:05
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
Oct 6 '18 at 19:11
1
I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
– Servaes
Oct 6 '18 at 19:12
Understood, more elegant this way,thanks again
– Matheus Domingos
Oct 6 '18 at 19:14
add a comment |
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 $b+1equiv0pmod{p}$, 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 $p=q+1$ which implies that $(p,q)=(3,2)$ which is not a solution. The equation above simplifies to
$$(ac-1)p=a+c,$$
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 Oct 6 '18 at 18:53
Servaes
22.4k33793
How do I finish this solution? Finding all the solution?
– Matheus Domingos
Oct 6 '18 at 19:05
@MatheusDomingos I completed the solution for you.
– Servaes
Oct 6 '18 at 19:05
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
Oct 6 '18 at 19:11
1
I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
– Servaes
Oct 6 '18 at 19:12
Understood, more elegant this way,thanks again
– Matheus Domingos
Oct 6 '18 at 19:14
add a comment |
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 $b+1equiv0pmod{p}$, 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 $p=q+1$ which implies that $(p,q)=(3,2)$ which is not a solution. The equation above simplifies to
$$(ac-1)p=a+c,$$
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 Oct 6 '18 at 18:53
Servaes
22.4k33793
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 $b+1equiv0pmod{p}$, 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 $p=q+1$ which implies that $(p,q)=(3,2)$ which is not a solution. The equation above simplifies to
$$(ac-1)p=a+c,$$
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 Oct 6 '18 at 18:53
Servaes
22.4k33793
edited Oct 6 '18 at 19:13
answered Oct 6 '18 at 18:53
Servaes
22.4k33793
answered Oct 6 '18 at 18:53
Servaes
22.4k33793
answered Oct 6 '18 at 18:53
Servaes
22.4k33793
22.4k33793
How do I finish this solution? Finding all the solution?
– Matheus Domingos
Oct 6 '18 at 19:05
@MatheusDomingos I completed the solution for you.
– Servaes
Oct 6 '18 at 19:05
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
Oct 6 '18 at 19:11
1
I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
– Servaes
Oct 6 '18 at 19:12
Understood, more elegant this way,thanks again
– Matheus Domingos
Oct 6 '18 at 19:14
add a comment |
How do I finish this solution? Finding all the solution?
– Matheus Domingos
Oct 6 '18 at 19:05
@MatheusDomingos I completed the solution for you.
– Servaes
Oct 6 '18 at 19:05
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
Oct 6 '18 at 19:11
1
I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
– Servaes
Oct 6 '18 at 19:12
Understood, more elegant this way,thanks again
– Matheus Domingos
Oct 6 '18 at 19:14
How do I finish this solution? Finding all the solution?
– Matheus Domingos
Oct 6 '18 at 19:05
How do I finish this solution? Finding all the solution?
– Matheus Domingos
Oct 6 '18 at 19:05
@MatheusDomingos I completed the solution for you.
– Servaes
Oct 6 '18 at 19:05
@MatheusDomingos I completed the solution for you.
– Servaes
Oct 6 '18 at 19:05
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
Oct 6 '18 at 19:11
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
Oct 6 '18 at 19:11
1
1
I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
– Servaes
Oct 6 '18 at 19:12
I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step.
– Servaes
Oct 6 '18 at 19:12
Understood, more elegant this way,thanks again
– Matheus Domingos
Oct 6 '18 at 19:14
Understood, more elegant this way,thanks again
– Matheus Domingos
Oct 6 '18 at 19:14
add a comment |
$$p^3+p=q^2+q implies p|q^2+q implies p|q mathrm{or} p|(q+1).$$
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 Oct 6 '18 at 18:54
Carl Schildkraut
11.2k11441
Oh thanks, now I know how to finish it, thanks
– Matheus Domingos
Oct 6 '18 at 18:55
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
Oct 6 '18 at 19:12
@MatheusDomingos Note that $k^4-4k-4$ is very close to $(k^2)^2$. Specifically, for most positive $k$, it is between $(k^2-1)^2$ and $k^2$.
– Carl Schildkraut
Oct 6 '18 at 19:16
1
Note that $k^4=(k^2)^2$ is a square, and that the previous square is $$(k^2-1)^2=k^4-2k^2+1,$$ which is smaller than $k^4-4k-4$ whenever $2k^2-1>4k+4$, and so the latter cannot be a square unless $2k^2-1leq 4k+4$.
– Servaes
Oct 6 '18 at 19:17
Sorry, I didn't understood how to finish using it
– Matheus Domingos
Oct 6 '18 at 19:24
|
show 1 more comment
$$p^3+p=q^2+q implies p|q^2+q implies p|q mathrm{or} p|(q+1).$$
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 Oct 6 '18 at 18:54
Carl Schildkraut
11.2k11441
Oh thanks, now I know how to finish it, thanks
– Matheus Domingos
Oct 6 '18 at 18:55
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
Oct 6 '18 at 19:12
@MatheusDomingos Note that $k^4-4k-4$ is very close to $(k^2)^2$. Specifically, for most positive $k$, it is between $(k^2-1)^2$ and $k^2$.
– Carl Schildkraut
Oct 6 '18 at 19:16
1
Note that $k^4=(k^2)^2$ is a square, and that the previous square is $$(k^2-1)^2=k^4-2k^2+1,$$ which is smaller than $k^4-4k-4$ whenever $2k^2-1>4k+4$, and so the latter cannot be a square unless $2k^2-1leq 4k+4$.
– Servaes
Oct 6 '18 at 19:17
Sorry, I didn't understood how to finish using it
– Matheus Domingos
Oct 6 '18 at 19:24
|
show 1 more comment
$$p^3+p=q^2+q implies p|q^2+q implies p|q mathrm{or} p|(q+1).$$
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 Oct 6 '18 at 18:54
Carl Schildkraut
11.2k11441
$$p^3+p=q^2+q implies p|q^2+q implies p|q mathrm{or} p|(q+1).$$
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 Oct 6 '18 at 18:54
Carl Schildkraut
11.2k11441
edited Oct 6 '18 at 19:01
answered Oct 6 '18 at 18:54
Carl Schildkraut
11.2k11441
answered Oct 6 '18 at 18:54
Carl Schildkraut
11.2k11441
answered Oct 6 '18 at 18:54
Carl Schildkraut
11.2k11441
11.2k11441
Oh thanks, now I know how to finish it, thanks
– Matheus Domingos
Oct 6 '18 at 18:55
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
Oct 6 '18 at 19:12
@MatheusDomingos Note that $k^4-4k-4$ is very close to $(k^2)^2$. Specifically, for most positive $k$, it is between $(k^2-1)^2$ and $k^2$.
– Carl Schildkraut
Oct 6 '18 at 19:16
1
Note that $k^4=(k^2)^2$ is a square, and that the previous square is $$(k^2-1)^2=k^4-2k^2+1,$$ which is smaller than $k^4-4k-4$ whenever $2k^2-1>4k+4$, and so the latter cannot be a square unless $2k^2-1leq 4k+4$.
– Servaes
Oct 6 '18 at 19:17
Sorry, I didn't understood how to finish using it
– Matheus Domingos
Oct 6 '18 at 19:24
|
show 1 more comment
Oh thanks, now I know how to finish it, thanks
– Matheus Domingos
Oct 6 '18 at 18:55
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
Oct 6 '18 at 19:12
@MatheusDomingos Note that $k^4-4k-4$ is very close to $(k^2)^2$. Specifically, for most positive $k$, it is between $(k^2-1)^2$ and $k^2$.
– Carl Schildkraut
Oct 6 '18 at 19:16
1
Note that $k^4=(k^2)^2$ is a square, and that the previous square is $$(k^2-1)^2=k^4-2k^2+1,$$ which is smaller than $k^4-4k-4$ whenever $2k^2-1>4k+4$, and so the latter cannot be a square unless $2k^2-1leq 4k+4$.
– Servaes
Oct 6 '18 at 19:17
Sorry, I didn't understood how to finish using it
– Matheus Domingos
Oct 6 '18 at 19:24
Oh thanks, now I know how to finish it, thanks
– Matheus Domingos
Oct 6 '18 at 18:55
Oh thanks, now I know how to finish it, thanks
– Matheus Domingos
Oct 6 '18 at 18:55
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
Oct 6 '18 at 19:12
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
Oct 6 '18 at 19:12
@MatheusDomingos Note that $k^4-4k-4$ is very close to $(k^2)^2$. Specifically, for most positive $k$, it is between $(k^2-1)^2$ and $k^2$.
– Carl Schildkraut
Oct 6 '18 at 19:16
@MatheusDomingos Note that $k^4-4k-4$ is very close to $(k^2)^2$. Specifically, for most positive $k$, it is between $(k^2-1)^2$ and $k^2$.
– Carl Schildkraut
Oct 6 '18 at 19:16
1
1
Note that $k^4=(k^2)^2$ is a square, and that the previous square is $$(k^2-1)^2=k^4-2k^2+1,$$ which is smaller than $k^4-4k-4$ whenever $2k^2-1>4k+4$, and so the latter cannot be a square unless $2k^2-1leq 4k+4$.
– Servaes
Oct 6 '18 at 19:17
Note that $k^4=(k^2)^2$ is a square, and that the previous square is $$(k^2-1)^2=k^4-2k^2+1,$$ which is smaller than $k^4-4k-4$ whenever $2k^2-1>4k+4$, and so the latter cannot be a square unless $2k^2-1leq 4k+4$.
– Servaes
Oct 6 '18 at 19:17
Sorry, I didn't understood how to finish using it
– Matheus Domingos
Oct 6 '18 at 19:24
Sorry, I didn't understood how to finish using it
– Matheus Domingos
Oct 6 '18 at 19:24
|
show 1 more comment
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 Oct 6 '18 at 18:52
Dietrich Burde
77.7k64386
add a comment |
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 Oct 6 '18 at 18:52
Dietrich Burde
77.7k64386
add a comment |
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 Oct 6 '18 at 18:52
Dietrich Burde
77.7k64386
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 Oct 6 '18 at 18:52
Dietrich Burde
77.7k64386
answered Oct 6 '18 at 18:52
Dietrich Burde
77.7k64386
answered Oct 6 '18 at 18:52
Dietrich Burde
77.7k64386
answered Oct 6 '18 at 18:52
Dietrich Burde
77.7k64386
77.7k64386
add a comment |
add a comment |
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 Oct 6 '18 at 18:54
greedoid
38k114794
Thanks, got this solution!
– Matheus Domingos
Oct 6 '18 at 19:25
add a comment |
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 Oct 6 '18 at 18:54
greedoid
38k114794
Thanks, got this solution!
– Matheus Domingos
Oct 6 '18 at 19:25
add a comment |
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 Oct 6 '18 at 18:54
greedoid
38k114794
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 Oct 6 '18 at 18:54
greedoid
38k114794
edited Oct 6 '18 at 19:11
answered Oct 6 '18 at 18:54
greedoid
38k114794
answered Oct 6 '18 at 18:54
greedoid
38k114794
answered Oct 6 '18 at 18:54
greedoid
38k114794
38k114794
Thanks, got this solution!
– Matheus Domingos
Oct 6 '18 at 19:25
add a comment |
Thanks, got this solution!
– Matheus Domingos
Oct 6 '18 at 19:25
Thanks, got this solution!
– Matheus Domingos
Oct 6 '18 at 19:25
Thanks, got this solution!
– Matheus Domingos
Oct 6 '18 at 19:25
add a comment |
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