11
$\begingroup$

Are there positive integers $A,B$ such that $A2^n+B$ is a perfect square for infinitely many $n$ ?

This is not my homework.

$\endgroup$
1
$\begingroup$

Suppose that there are such positive integers $A$, $B$. Write $c_m=A2^m+B$, where $m$ is a positive integer such that $c_m$ is a perfect square. By assumption, there are infinitely many such $m\ge 1$. I will make a further assumption: there is an $m_0$ such that $c_m$ is a perfect square for all $m\ge m_0$. Then construct a sequence $$ (x_m,y_m)=(2\sqrt{c_m}, \sqrt{A2^{m+2}+B }). $$ It satisfies $(x_m+y_m)(x_m-y_m)=3B$. Hence $x_m+y_m$ divides $3B$ for infinitely many $m\ge m_0$. But this is impossible because $|x_m+y_m|>3B$ for $m$ large enough,and $3B\neq 0$ by assumption. Hence there are no such positive integers $A$ and $B$.

Edit: Without the additional assumption, $y_m$ may not be an integer. Perhaps one can remove the additional assumtion, e.g., by taking such integer $k$ that with $c_m$ also $A2^{m+k}+B$ is a perfect square for infinitely many $m$, i.e., with $x_m=2^k\sqrt{a2^m+b}$, $y_m=\sqrt{a2^{m+2k}+b}$.

$\endgroup$
  • $\begingroup$ but y[m] may not be integer… $\endgroup$ – QiRenrui May 7 '14 at 9:49
  • $\begingroup$ but this k may not exist $\endgroup$ – QiRenrui May 7 '14 at 10:07
  • $\begingroup$ Hmm, even if for any such $m$ you can find a $k>1$ such that $A\cdot2^{m+2k} + B$ is a square, too (which is possible), your argument will merely give you that $(x_m+y_m)(x_m-y_m)$ divides $(2^{2k}-1)B$ which unfortunately is not bounded :( $\endgroup$ – jpvee May 7 '14 at 10:44
  • $\begingroup$ @jpvee yes, it would only work for fixed $k$. Anyway, the argument is not enough, but shows at least some result in this direction. $\endgroup$ – Dietrich Burde May 7 '14 at 10:54
  • $\begingroup$ @Dietrich Burde but i believe the right direction is pell equation… $\endgroup$ – QiRenrui May 7 '14 at 11:01
1
$\begingroup$

If one writes $n=n_0+3k$ for $n_0 \in \{ 0, 1, 2 \}$, then the existence of infinitely many $n$ for which $A 2^n+B$ is square would imply the existence of infinitely many integral points on (at least) one of the (genus $1$, generically) curves $y^2 = A 2 ^{n_0} x^3 + B$, contradicting Siegel's theorem.

$\endgroup$
  • $\begingroup$ but A*2^n[0]may not be perfect cube… $\endgroup$ – QiRenrui May 9 '14 at 8:30
  • $\begingroup$ That doesn't affect the genus. Just multiply the equation by $(A 2^{n_0})^2$ to get a new equation of the shape $Y^2=X^3+C$ with $C=(A 2^{n_0})^2 B$. $\endgroup$ – Mike Bennett May 9 '14 at 14:41
  • $\begingroup$ you are right,and could you prove it in elementary number theory?thanks $\endgroup$ – QiRenrui May 11 '14 at 9:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.