3
$\begingroup$

Let $X,Y \geq 1$. I am interested in the number of solutions of the following diophantine equations: $$S_1\colon \, \, x_1y_1^3 = x_2 y_2^3 $$ Let $N_1(X,Y) $ denote the number of solutions to $S_1$ with $1\leq x_i \leq X$ and $1\leq y_i \leq Y$. There are the obvious solutions $x_1=x_2$ and $y_1=y_2$ which contribute $XY$. The true order of magnitude should be $XY \log(XY)^A$ with some small power of log. I would like to get $A$ as small as possible.

Also consider the equation $S_2$ given by $$ x_1(y_1^3 - y_2^3) = x_2 (z_1^3 -z_2^3)$$ with $1\leq x_i\leq X$ and $1\leq y_i,z_i \leq Y$ with $y_1\neq y_2$ and $z_1\neq z_2$. Let $N_2(X,Y)$ denote the corresponding number of solutions. I expect something like $N_2(X,Y) \ll XY^2\log(XY)^B$, with $B$ again hopefully small.

Is there some clever way to get the exponent of the logarithm small?

$\endgroup$
1
  • $\begingroup$ One thing is certain. For the equation $a(x^3-y^3)=b(z^3-q^3)$ there are always solutions. $\endgroup$
    – individ
    May 3, 2018 at 15:23

1 Answer 1

7
$\begingroup$

For $S_1$ you can take $A=0$.

Setting $g=\gcd(x_1,x_2)$ and $v_i = x_i/g$, and $h=\gcd(y_1,y_2)$ and $w_i = y_i/h$, one obtains $v_1w_1^3 = v_2w_2^3$ with $\gcd(v_1,v_2)=\gcd(w_1,w_2)=1$; this implies that $v_1=w_2^3$ and $v_2=w_1^3$. Therefore all solutions to $S_1$ are of the form $(x_1,x_2,y_1,y_2) = (gw_2^3,gw_1^3,hw_1,hw_2)$ with $\gcd(w_1,w_2)=1$ and the four coordinates less than $X,X,Y,Y$, respectively.

Counting these solutions yields the sum \begin{align*} \sum_{g\le X} \sum_{h\le Y} \sum_{\substack{w_1,w_2 \le \min\{ (X/g)^{1/3},Y/h \} \\ \gcd(w_1,w_2)=1} } 1 &= \sum_{g\le X} \sum_{h \le Y(g/X)^{1/3}} \sum_{\substack{w_1,w_2 \le (X/g)^{1/3} \\ \gcd(w_1,w_2)=1}} 1 + \sum_{h\le Y} \sum_{g \le X(h/Y)^3} \sum_{\substack{w_1,w_2 \le Y/h \\ \gcd(w_1,w_2)=1}} 1 \\ &\le \sum_{g\le X} \sum_{h \le Y(g/X)^{1/3}} \frac{X^{2/3}}{g^{2/3}} + \sum_{h\le Y} \sum_{g \le X(h/Y)^3} \frac{Y^2}{h^2} \\ &\le \sum_{g\le X} Y \frac{X^{1/3}}{g^{1/3}} + \sum_{h\le Y} X\frac hY \ll YX^{1/3} X^{2/3} + \frac XY Y^2 \ll XY. \end{align*}

$\endgroup$

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.