Hamilton equations-Symplectic Euler method We know that $\dot{q} = \frac{\partial H}{\partial p}$ and $\dot{p} = -\frac{\partial H}{\partial q}$, and we also know the values $Q$ and $P$ respectively of $q$ and $p$ at a later time step $\Delta t$. How could we prove that the quantities
$$
\begin{align}
Q &= q + {\Delta}t\frac{\partial H}{\partial p}(q,p),\\
P &= p - {\Delta}t\frac{\partial H}{\partial q}(q,p)
\end{align}
$$
are not symplectic, while
$$
\begin{align}
Q &= q - {\Delta}t\frac{\partial H}{\partial p}(q,p),\\
P &= p + {\Delta}t\frac{\partial H}{\partial Q}(Q,p) 
\end{align}
$$
are symplectic?
Clarification:
The sets of equations define different numerical integrators: in the first case (qi+1,pi+1) directly in terms of (qi,pi), and in the second case qi+1 in terms of (qi,pi), and pi+1 in terms of (qi+1,pi).

See canonical transformations via generating functions.
You might also directly compute the preservation of the symplectic form,
\begin{align}
\sum _i dP_i∧dQ_i  &=\sum_i \left(dp_i-Δt \sum_j[H_{q_iq_j}(Q,p)dQ_j+H_{q_ip_j}(Q,p)dp_j]\right)∧ dQ_i
\\
&=\sum_i dp_i ∧ dQ_i-Δt \sum_i\sum_j[H_{q_iq_j}(Q,p)dQ_j+H_{q_ip_j}(Q,p)dp_j]∧ dQ_i
\\
&=\sum_i dp_i ∧ dQ_i-Δt \sum_jdp_j∧ \sum_i  H_{q_ip_j}(Q,p) dQ_i
\\
&=\sum_i dp_i ∧ \left[dQ_i-Δt   \sum_j  H_{p_iq_j}(Q,p) dQ_j\right]
\\&=\sum_i dp_i ∧  dq_i
\end{align}
Due to the symmetric coefficient matrix $\sum_{i,j}\sum_i\sum_j H_{q_iq_j}(Q,p)dQ_j∧ dQ_i=0$ and the same for the $p$ coordinates.
Note that except for a separable Hamiltionian like $H(q,p)=T(p)+V(q)$, the first equation
$$
Q=q+ΔtH_p(Q,p)
$$
would be implicit.

For the Hamiltonian compare
\begin{align}
H(q,p)&=H(Q-ΔtH_p,p)=H-ΔtH_q·H_p+O(Δt^2)
\\
\text{ and }
H(Q,P)&=H(Q,p-ΔtH_q)=H-ΔtH_p·H_q+O(Δt^2),
\end{align}
the arguments on the right all $(Q,p)$, to see that it stays largely constant. One can modify this to $\tilde H=H+\tfrac12ΔtH_q·H_p$ where the Taylor expansion to one order higher gives
\begin{align}
\tilde H(q,p)&=\tilde H(Q-ΔtH_p,p)=H-\tfrac12ΔtH_q·H_p-\tfrac12H_{qp}[H_p,H_q]+O(Δt^3)
\\
\text{ and }
\tilde H(Q,P)&=\tilde H(Q,p-ΔtH_q)=H-\tfrac12ΔtH_p·H_q-\tfrac12H_{pq}[H_q,H_p]+O(Δt^3).
\end{align}
This means that this modified energy functional has a global error of $O(Δt^2)$, the global $O(Δt)$ first-order error will mainly manifest as a time dilation.
