Fix $(a,b)=1$, $a<b<2a$ and $a,b>n^{1/(2t)}$ and fix a prime $T\approx n^{\tau+\frac1k}$ where $\tau\geq1$ and $k=2(t-1)$. We can show using exponential sums there is an $m_{_T}$ such that $T/a^{2(t-1)}<m_{_T}<T-T/a^{2(t-1)}$ such that each of $c_{i,T}=m_{_T}a^{2(t-1)-i}b^i\bmod T$ is bound by $T^{\frac{2t-2+\kappa}{2t-1}}$ for some $\kappa=\frac{k(\tau-1)+2}{k\tau+1}>0$. Example if $t=2$ and $\tau=1$ we can get $\kappa=2/3$ and $T^{\frac{2t-2+\kappa}{2t-1}}=T^{8/9}$.

$\underline{\mbox{Problem} 1}$: Assume we have such $c_{0,T},...c_{k,T}$ that are proved to exist using exponential sums. Fix distinct primes $q_j>T^4>\max(c_{i,T})$ for $j\in\{0,\dots,k\}$. Are there integers $p_j$ such that $q_j/\min(c_{i,T})<p_j<q_j-q_j/\min(c_{i,T})$ such that each of $p_jc_{i,T}\bmod q_j$ at $i\neq j$ is bound in interval $[0,q_j^{r+\epsilon}]$ while $p_jc_{j,T}\bmod q_j$ is bound in $[k\cdot q_j^{r+\epsilon}+1,(k+1)\cdot q_j^{r+\epsilon}]$ for some $r\in(0,1)$?

Can we get $r<\frac{k^\alpha}{k^\alpha+1}$ for any fixed $\alpha\geq1$ in this case? If $c_{i,T}$s were uniform then $\alpha=1$ is possible.

Note that for every $\alpha_i\in\Bbb Z$ with $|\alpha_i|<n^\frac1{2t}$ and $(\alpha_1,\dots,\alpha_k)\neq(\underbrace{0,\dots,0}_{k\mbox{ times}})$ we have $$|\sum_{i=0}^k\alpha_ia^{2(t-1)-i}b^i|\leq(k+1)n^{\frac{2t-1}{2t}}$$ and if $(k+1)<n^{\frac1{2t}+\tau}$ then $$|\sum_{i=0}^k\alpha_ia^{2(t-1)-i}b^i|<T$$ holds. Moreover since $|\alpha_i|<\min(a,b)$ and $(\alpha_1,\dots,\alpha_k)\neq(\underbrace{0,\dots,0}_{k\mbox{ times}})$ $$0<|\sum_{i=0}^k\alpha_ia^{2(t-1)-i}b^i|<T$$ holds.

This implies $$\sum_{i=0}^k\alpha_i[m_{_T}a^{2(t-1)-i}b^i\bmod T]=\sum_{i=0}^k\alpha_ic_{i,T}\neq0\bmod T$$ as well. It means the discrepancy of the derived sequence $(c_{0,T},\dots,c_{k,T})\bmod T$ is at most $n^{-\frac1{2t}}$.

$\underline{\mbox{Problem} 2}$: Is it possible to get a prime $q>T^4$ such that discrepancy of $(c_{0,T},\dots,c_{k,T})\bmod q$ is at most $n^{-\frac1{t}}$? If so there is an $m'$ such that $$(m'c_{0,T},\dots,m'c_{k,T})\bmod q\in\underbrace{[0,n^{-\frac1{t}}q]\times\dots\times[0,n^{-\frac1{t}}q]}_{k+1\mbox{ times}}$$ holds?

Is it always true that if for a given sequence we can exclude a linear relation with coefficients $<n^\delta$ then we can always achieve $r=1-\delta$ for that sequence? Is there a proof or good reference? Is there a converse?