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Are there positive intgers $a,b,c$ such that both $2a(b^c+1),2a(b^c-1)$ are perfect powers?

My research

We have $2a(b^c+1)-2a(b^c-1)=4a$, so I looked at perfect powers which have a multiple of $4$ as a difference. I found this question which talks about differences of powers, and I found:

$5^3-11^2=4$

$47^2-13^3=12$

$312^2-46^3=8$

$2^{17}-362^2=28$

and a few more, but none of them seemed to be in our form.

I also found On the Diophantine equation $m^2 - p^k = 4z$, where $z \in \mathbb{N}$ and $p$ is a prime satisfying $p \equiv k \equiv 1 \pmod 4$ which is kind of similar to what are talking about. They say we don't have a solution to their problem in the special case due to Catalan's conjecture, so I guess we also don't have one for our problem, but couldn't prove it.

Kaira
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    Do you allow also $c=1$ ? In this case , plenty of solutions exist. For $a,b,c\le 500$ and $c\ge 2$ , the solutions are $(81,7,2)$ (giving the pair $(8100,7776)$) , $(144,7,2)$ (giving the pair $(14400,13824)$) , $(400,3,2)$ (giving the pair $(8000,6400)$) and $(500,3,2)$ (giving the pair $(10000,8000)$). – Peter Jul 15 '21 at 15:13
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    Two more solutions are $(1568,2,3)$ (giving the pair $(28224,21952)$) and $(3721,3,5)$ (giving the pair $(1815848,1800964)$). – Peter Jul 15 '21 at 15:18
  • @Peter Thank you. May I ask you how you found them? Is it just brute force? – Kaira Jul 15 '21 at 16:01
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    Yes, just brute force. – Peter Jul 15 '21 at 16:31
  • Let $a=2^{6r+4}\cdot 3^2$ with $r\ge0, b=7,c=2,$ then we get infinitely many pairs such that $2a(b^2+1)=(2^{3r+3}\cdot3\cdot5)^2$, $2a(b^2-1)=(2^{2r+3}\cdot3)^3.$ – Tomita Jul 16 '21 at 05:18

1 Answers1

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If $b$ is even then $b^c+1$ and $b^c-1$ are coprime, and so we have $$b^c+1=\prod p^{u_p}\qquad\text{ and }\qquad b^c-1=\prod q^{v_q},$$ where the $p$ and $q$ are all distinct prime numbers, and the $u_p$ and $v_q$ are distinct positive integers. To find a positive integer $a$ such that $2a(b^c+1)$ and $2a(b^c-1)$ are both perfect powers, it suffices to find nonnegative integers $s_p$ and $t_q$ such that $$u_p+s_p\equiv t_q\equiv0\pmod{m}\qquad\text{ and }\qquad v_q+t_q\equiv s_p\equiv0\pmod{n},$$ for some integers $m,n>1$, as then $a:=2^{mn-1}\prod p^{s_p}\prod q^{t_q}$ will make the two products $2a(b^c+1)$ and $2a(b^c-1)$ perfect $m$-th and $n$-th powers, respectively. By the Chinese remainder theorem, this can always be done by choosing $m$ coprime to the $v_q$, and $n$ coprime to the $u_p$, and $m$ and $n$ coprime to eachother.

For a concrete example, take $b=6$ and $c=3$ so that $$b^c+1=217=7\times31\qquad\text{ and }\qquad b^c-1=215=5\times43.$$ Then $u_7=u_{31}=1$ and $v_5=v_{43}=1$. We may simply take $m=2$ and $n=3$, for which the smallest $s_p$ and $t_q$ are $$s_7=3,\quad s_{31}=3,\quad t_5=2,\quad t_{43}=2,$$ corresponding to $$a=2^{2\times3-1}\times(7^3\times31^3)\times(5^2\times43^2)=15114928589600,$$ with \begin{eqnarray*} 2a(b^c+1)&=&80993080^2,\\ 2a(b^c-1)&=&186620^3. \end{eqnarray*} Easy exercise: Can you find an appropriate value for $a$ given $(b,c)=(2,21)$?

Harder exercise: Can you extend this construction to odd $b$?

Servaes
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  • Why is the assumption $m$ is coprime to $v_q$ and $n$ is coprime to $u_p$ necessary? I felt like $m$ being coprime to $n$ is enough to use CRT. – Kaira Jul 16 '21 at 02:14