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

This is not my homework.

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2 Answers 2

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 $x_m=2^k\sqrt{a2^m+b}$, $y_m=\sqrt{a2^{m+2k}+b}$.

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but y[m] may not be integer… –  QiRenrui May 7 at 9:49
    
but this k may not exist –  QiRenrui May 7 at 10:07
    
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 :( –  jpvee May 7 at 10:44
    
@jpvee yes, it would only work for fixed $k$. Anyway, the argument is not enough, but shows at least some result in this direction. –  Dietrich Burde May 7 at 10:54
    
@Dietrich Burde but i believe the right direction is pell equation… –  QiRenrui May 7 at 11:01

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.

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but A*2^n[0]may not be perfect cube… –  QiRenrui May 9 at 8:30
    
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 $C=(A 2^{n_0})^2 B$. –  Mike Bennett May 9 at 14:41
    
you are right,and could you prove it in elementary number theory?thanks –  QiRenrui May 11 at 9:05

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