As we know for the most simple Diophantine equation $ax+by=1$ for positive $a,b$ if $1\le x\le b-1$ then $-(a-1)\le y\le -1$ but what if we have n variables and not just two , if equation be equals to 1 could we form smallest boundaries for each variable ?! this is a problem that I'm facing with it !
I was trying to proof that this equation for each $p_i(\text{primes}) \quad -(p_i-2)\le k_i\le(p_i-2)$ has answer $$\frac{P_n}{p_1}k_1+\frac{P_n}{p_2}k_2+\dots+\frac{P_n}{p_n}k_n=2 $$ which $$P_n=\prod_{i=0}^n{p_i} \qquad p_0=2 (primes)$$
In order to solve this equation we can symplify it to this $$\frac{P_n}{p_1p_2}(p_2k_1+p_1k_2)+\frac{P_n}{p_3}k_n+\dots+\frac{P_n}{p_n}k_n=2$$ $$\frac{P_n}{p_1p_2p_3}(p_3w_3+p_1p_2k_3)+\frac{P_n}{p_4}k_4+\dots+\frac{P_n}{p_n}k_n=2$$ $$.$$ $$.$$ $$.$$ $$\frac{P_n}{p_1p_2p_3\dots p_{n}}(p_nw_n+p_1p_2\dots p_{n-1}k_n)=2$$
which $w_j=p_{j-1}w_{j-1}+p_1\dots p_{j-2}k_{j-1} \qquad w_2=k_1$
so now we can solve $2(p_nw_n+p_1p_2\dots p_{n-1}k_n)=2$
and other w's
$$w_j=p_1\dots p_{j-1}m_j+w'_{j}w_{j+1}$$
$$k_j=-p_jm_j+k'_{j}w_{j+1}$$
but from here onwards I was not able to proof existing answer for given bound on each k's because it become complex .
It seems that if we change the boundaries of k like this
$$-(p_i-1)\le k_i\le(p_i-1)$$
the answer exists but I was not able to proof this either.
so what is minimum solutions for linear Diophantine equation in n variables or what are boundaries?
