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What is the smallest $\gamma_1,\gamma_2,\gamma_3>0$ such that given coprime $p,q=\Theta(\ell)$ and integer $t\geq3$ there are coprime $m,n=\Theta(\ell^{t-1})$ with $(mn,pq)=1$, $\alpha_i\in\Bbb Z$ and $m'\in\Bbb Z$ with $$\alpha_1p^{t-2}q^t+\gamma_1=m'm^2$$ $$\alpha_2p^{t-1}q^{t-1}+\gamma_2=m'mn$$ $$\alpha_3p^tq^{t-2}+\gamma_3=m'n^2$$ where $m,n$ where $0<m'<p^{t}q^t$ and $\alpha_i=O(\ell^{2t})$ holds?

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  • $\begingroup$ $\gcd(mn,pq) \ne 1$ is impossible, since it would be divisor of consecutive integers. $\endgroup$
    – joro
    Commented Feb 14, 2017 at 9:00

1 Answer 1

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There is simple infinite family of simple solutions with $m=n=1$.

The system is linear in $k_i$, giving the simple solutions:

$$ k_1 = (m^2*m' - 1)*p^{(-t + 2)}/q^t \\ k_2 = (m'*n^2 - 1)*q^{(-t + 2)}/p^t \\ k_3 = (m*m'*n - 1)*p^{(-t + 1)}*q^{(-t + 1)} $$

Solution is $m'=p^t q^t+1$.

For the edited question, solutions are $(m,n)=1+\mathbb{Z}p^tq^t,1+\mathbb{Z}p^tq^t$.

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  • $\begingroup$ So we cannot get $\Theta(\max(p^{t-1},q^{t-1}))$? $\endgroup$
    – Turbo
    Commented Feb 15, 2017 at 1:36
  • $\begingroup$ rephrased problem and count makes sense now. $\endgroup$
    – Turbo
    Commented Feb 15, 2017 at 2:39

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