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

Reduces to $$1\equiv m'm^2\bmod p^{t-2},\quad1\equiv m'mn\bmod p^{t-1},\quad1\equiv m'n^2\bmod p^{t}$$ $$1\equiv m'm^2\bmod q^{t},\quad1\equiv m'mn\bmod q^{t-1},\quad1\equiv m'n^2\bmod q^{t-2}$$

$\alpha_i=O(\ell^{2t})$ holds.

For simplification if $\alpha_1=\beta_1p^2$, $\alpha_2=\beta_2pq$ and $\alpha_3=\beta_3q^2$ then we are looking for $$\beta_1p^{t}q^t+1=m'm^2$$ $$\beta_2p^{t}q^{t}+1=m'mn$$ $$\beta_3p^tq^{t}+1=m'n^2$$ that is we are looking at special forms of integers in arithmetic progression $\beta p^tq^t+1$ where $\beta_i=O(\ell^{2t-2})$.

For any fixed $m'$ we have $O(\ell^{t-1})$ of $\beta_1$ and $O(\ell^{t-1})$ of $\beta_3$ produces square multiples of $m'$. $\beta_2$ sweeps $O(\ell^{2(t-1)})$ integers out of $\Theta(\ell^{4t-2})$ integers of which $\Theta(\ell^{2t-2})$ are integers of form $m'mn$ for some $m,n$ of size $\Theta(\ell^{t-1})$.

So expected intersection is $\Theta(\frac{\beta_1\beta_3mn}{\ell^{4t-2}})=\Theta(\frac{\ell^{4t-4}}{\ell^{4t-2}})=\Theta(\frac1{\ell^2})=o(1)$. So is there none?

What if I replace $1$ by variable $\gamma_i$ and seek the smallest $\gamma_i$ for which such a system produces solution? $$\beta_1p^{t}q^t+\gamma_1=m'm^2$$ $$\beta_2p^{t}q^{t}+\gamma_2=m'mn$$ $$\beta_3p^tq^{t}+\gamma_3=m'n^2$$

So is there some $\gamma_i$ of size $O(\ell^{\frac23+\epsilon})$ that will work?

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?

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

Reduces to $$1\equiv m'm^2\bmod p^{t-2},\quad1\equiv m'mn\bmod p^{t-1},\quad1\equiv m'n^2\bmod p^{t}$$ $$1\equiv m'm^2\bmod q^{t},\quad1\equiv m'mn\bmod q^{t-1},\quad1\equiv m'n^2\bmod q^{t-2}$$

$\alpha_i=O(\ell^{2t})$ holds.

For simplification if $\alpha_1=\beta_1p^2$, $\alpha_2=\beta_2pq$ and $\alpha_3=\beta_3q^2$ then we are looking for $$\beta_1p^{t}q^t+1=m'm^2$$ $$\beta_2p^{t}q^{t}+1=m'mn$$ $$\beta_3p^tq^{t}+1=m'n^2$$ that is we are looking at special forms of integers in arithmetic progression $\beta p^tq^t+1$ where $\beta_i=O(\ell^{2t-2})$.

For any fixed $m'$ we have $O(\ell^{t-1})$ of $\beta_1$ and $O(\ell^{t-1})$ of $\beta_3$ produces square multiples of $m'$. $\beta_2$ sweeps $O(\ell^{2(t-1)})$ integers out of $\Theta(\ell^{4t-2})$ integers of which $\Theta(\ell^{2t-2})$ are integers of form $m'mn$ for some $m,n$ of size $\Theta(\ell^{t-1})$.

So expected intersection is $\Theta(\frac{\beta_1\beta_3mn}{\ell^{4t-2}})=\Theta(\frac{\ell^{4t-4}}{\ell^{4t-2}})=\Theta(\frac1{\ell^2})=o(1)$. So is there none?

What if I replace $1$ by variable $\gamma_i$ and seek the smallest $\gamma_i$ for which such a system produces solution? $$\beta_1p^{t}q^t+\gamma_1=m'm^2$$ $$\beta_2p^{t}q^{t}+\gamma_2=m'mn$$ $$\beta_3p^tq^{t}+\gamma_3=m'n^2$$

So is there some $\gamma_i$ of size $O(\ell^{2+\epsilon})$ that will work?

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

Reduces to $$1\equiv m'm^2\bmod p^{t-2},\quad1\equiv m'mn\bmod p^{t-1},\quad1\equiv m'n^2\bmod p^{t}$$ $$1\equiv m'm^2\bmod q^{t},\quad1\equiv m'mn\bmod q^{t-1},\quad1\equiv m'n^2\bmod q^{t-2}$$

$\alpha_i=O(\ell^{2t})$ holds.

For simplification if $\alpha_1=\beta_1p^2$, $\alpha_2=\beta_2pq$ and $\alpha_3=\beta_3q^2$ then we are looking for $$\beta_1p^{t}q^t+1=m'm^2$$ $$\beta_2p^{t}q^{t}+1=m'mn$$ $$\beta_3p^tq^{t}+1=m'n^2$$ that is we are looking at special forms of integers in arithmetic progression $\beta p^tq^t+1$ where $\beta_i=O(\ell^{2t-2})$.

For any fixed $m'$ we have $O(\ell^{t-1})$ of $\beta_1$ and $O(\ell^{t-1})$ of $\beta_3$ produces square multiples of $m'$. $\beta_2$ sweeps $O(\ell^{2(t-1)})$ integers out of $\Theta(\ell^{4t-2})$ integers of which $\Theta(\ell^{2t-2})$ are integers of form $m'mn$ for some $m,n$ of size $\Theta(\ell^{t-1})$.

So expected intersection is $\Theta(\frac{\beta_1\beta_3mn}{\ell^{4t-2}})=\Theta(\frac{\ell^{4t-4}}{\ell^{4t-2}})=\Theta(\frac1{\ell^2})=o(1)$. So is there none?

What if I replace $1$ by variable $\gamma_i$ and seek the smallest $\gamma_i$ for which such a system produces solution? $$\beta_1p^{t}q^t+\gamma_1=m'm^2$$ $$\beta_2p^{t}q^{t}+\gamma_2=m'mn$$ $$\beta_3p^tq^{t}+\gamma_3=m'n^2$$

So is there some $\gamma_i$ of size $O(\ell^{\frac23+\epsilon})$ that will work?

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

Reduces to $$1\equiv m'm^2\bmod p^{t-2},\quad1\equiv m'mn\bmod p^{t-1},\quad1\equiv m'n^2\bmod p^{t}$$ $$1\equiv m'm^2\bmod q^{t},\quad1\equiv m'mn\bmod q^{t-1},\quad1\equiv m'n^2\bmod q^{t-2}$$

$\alpha_i=O(\ell^{2t})$ holds.

For simplification if $\alpha_1=\beta_1p^2$, $\alpha_2=\beta_2pq$ and $\alpha_3=\beta_3q^2$ then we are looking for $$\beta_1p^{t}q^t+1=m'm^2$$ $$\beta_2p^{t}q^{t}+1=m'mn$$ $$\beta_3p^tq^{t}+1=m'n^2$$ that is we are looking at special forms of integers in arithmetic progression $\beta p^tq^t+1$.

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

Reduces to $$1\equiv m'm^2\bmod p^{t-2},\quad1\equiv m'mn\bmod p^{t-1},\quad1\equiv m'n^2\bmod p^{t}$$ $$1\equiv m'm^2\bmod q^{t},\quad1\equiv m'mn\bmod q^{t-1},\quad1\equiv m'n^2\bmod q^{t-2}$$

$\alpha_i=O(\ell^{2t})$ holds.

For simplification if $\alpha_1=\beta_1p^2$, $\alpha_2=\beta_2pq$ and $\alpha_3=\beta_3q^2$ then we are looking for $$\beta_1p^{t}q^t+1=m'm^2$$ $$\beta_2p^{t}q^{t}+1=m'mn$$ $$\beta_3p^tq^{t}+1=m'n^2$$ that is we are looking at special forms of integers in arithmetic progression $\beta p^tq^t+1$ where $\beta_i=O(\ell^{2t-2})$.

For any fixed $m'$ we have $O(\ell^{t-1})$ of $\beta_1$ and $O(\ell^{t-1})$ of $\beta_3$ produces square multiples of $m'$. $\beta_2$ sweeps $O(\ell^{2(t-1)})$ integers out of $\Theta(\ell^{4t-2})$ integers of which $\Theta(\ell^{2t-2})$ are integers of form $m'mn$ for some $m,n$ of size $\Theta(\ell^{t-1})$.

So expected intersection is $\Theta(\frac{\beta_1\beta_3mn}{\ell^{4t-2}})=\Theta(\frac{\ell^{4t-4}}{\ell^{4t-2}})=\Theta(\frac1{\ell^2})=o(1)$. So is there none?

What if I replace $1$ by variable $\gamma_i$ and seek the smallest $\gamma_i$ for which such a system produces solution? $$\beta_1p^{t}q^t+\gamma_1=m'm^2$$ $$\beta_2p^{t}q^{t}+\gamma_2=m'mn$$ $$\beta_3p^tq^{t}+\gamma_3=m'n^2$$

So is there some $\gamma_i$ of size $O(\ell^{2+\epsilon})$ that will work?