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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:

$p^2 = -1(\text{mod} \, q)$ 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

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Clearly $p\neq q$, and because $p$ and $q$ are prime and $$p(p^2+1)=p^3+p=q^2+q=q(q+1),$$ we must have $p\mid q+1$ and $q\mid p^2+1$. Write $$q+1=ap\qquad\text{ and }\qquad p^2+1=bq,$$ to find that $p^2-abp+b+1=0$. In particular $b+1\equiv0\pmod{p}$, 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 $p\geq2$ clearly we cannot have $a,c\geq2$. Hence $c=1$ and so $$p^2+1=bq=(cp-1)q=(p-1)q.$$ In particular $p-1\mid p^2+1$. As $p-1\mid p^2-1$ it follows that $p-1=2$, so $p=3$ and hence $q=5$.

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  • $\begingroup$ How do I finish this solution? Finding all the solution? $\endgroup$ – Matheus Domingos Oct 6 '18 at 19:05
  • $\begingroup$ @MatheusDomingos I completed the solution for you. $\endgroup$ – Inactive - avoiding CoC Oct 6 '18 at 19:05
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    $\begingroup$ Thank you, now I understood the limitation. I need to have a+c=>ac-1... What implies that (a-1)(c-1)=>2. Thanks $\endgroup$ – Matheus Domingos Oct 6 '18 at 19:11
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    $\begingroup$ I rearranged the argument a bit, as in fact $a>1$ follows very easily earlier on, simpifying the final step. $\endgroup$ – Inactive - avoiding CoC Oct 6 '18 at 19:12
  • $\begingroup$ Understood, more elegant this way,thanks again $\endgroup$ – Matheus Domingos Oct 6 '18 at 19:14
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$$p^3+p=q^2+q \implies p|q^2+q \implies p|q\ \mathrm{or}\ p|(q+1).$$

Obviously, $p\neq q$, so $p\nmid q$. 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?

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  • $\begingroup$ Oh thanks, now I know how to finish it, thanks $\endgroup$ – Matheus Domingos Oct 6 '18 at 18:55
  • $\begingroup$ 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? $\endgroup$ – Matheus Domingos Oct 6 '18 at 19:12
  • $\begingroup$ @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 $k^2$. $\endgroup$ – Carl Schildkraut Oct 6 '18 at 19:16
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    $\begingroup$ 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 $2k^2-1\leq 4k+4$. $\endgroup$ – Inactive - avoiding CoC Oct 6 '18 at 19:17
  • $\begingroup$ Sorry, I didn't understood how to finish using it $\endgroup$ – Matheus Domingos Oct 6 '18 at 19:24
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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?

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Remember that if $a,b$ are positive integers such that $a\mid b$ then $a\leq b$. I'll be using this frekvently here.

From $$p(p^2+1)= q(q+1)\implies p\mid q\;\;\;{\rm or}\;\;\;p\mid q+1$$

1. case $p\mid q$, then $q+1\mid p^2+1$. Write $q+1=s$ then we get $$ps\mid (p^2+1)(s-1) = p^2s-p^2+s-1\implies ps\mid p^2-s+1$$

Since $p^2+1\geq s$ we have 2 subcases:

1.1 case $p^2+1>s$, then $ps\leq p^2-s+1$ so $s(p+1)\leq p^2+1$, and thus $$s\leq {p^2+1\over p+1} <p\implies s\leq p-1$$

So $q+1\leq p-1 \leq q-1$ and thus no solution.

1.2 case $p^2+1=s$, then $q^2+1 = q+1$ and again no solution.

2. case $p\mid q+1$, then $q\mid p^2+1$. Then we get $$pq\mid (p^2+1)(q+1) = p^2q+p^2+q+1\implies pq\mid p^2+q+1$$

so we have $pq\leq p^2+q+1$ so $q \leq {p^2+1\over p-1} \leq p+2$ if $p\geq 3$. So if $p\geq 3$ and since $p\mid q$ that $q\in \{p,p+1,p+2\}$ which is easy to finish by hand.

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  • $\begingroup$ Thanks, got this solution! $\endgroup$ – Matheus Domingos Oct 6 '18 at 19:25

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