Is 0-1 integer linear programming with only equality constraints NP-Hard?
Consider the Maximum Independent Set problem ($\mathcal{NP}$-hard): given a graph $G=(V,E)$, find the maximum independent set in $G$, i.e., the subset of vertices $I \subseteq V$ such that every two vertices in $I$ are not adjacent $\forall v_1,v_2 \in I, (v_1,v_2) \notin E$.

This problem has a well known Integer programming formulation:
let $x_i$ be a binary variable for every $i \in V$ and equal one if $i \in I$.
\begin{align}
\max &amp; \sum\limits_{i=1}^n x_i \\
\text{subject to} &amp; \\
&amp; x_i + x_j \leq 1 \text{ for all } (i,j) \in E \\
&amp; x_i \in \{0,1\} \text{ for all } i \in V
\end{align}

If $|V| = n$, this problem has $n$ variables and $O(n^2)$ constraints.

To transform it to equality form (a.k.a. standard form) one might introduce slack variables for every constraint. Generally, if you have $Ax &lt;= b$ you can write $Ax + x_s = b$ where $x_s \geq 0$ and $x_s$ are called slack variables.

It appears that for Independent Set slack variables are actually binary!
Thus, you have the formulation
\begin{align}
\max &amp; \sum\limits_{i=1}^n x_i \\
\text{subject to} &amp; \\
&amp; x_i + x_j +y_{ij} = 1 \text{ for all } (i,j) \in E \\
&amp; x_i \in \{0,1\} \text{ for all } i \in V \\
&amp; y_{ij} \in \{0,1\} \text{ for all } i,j \in V
\end{align}

This formulation exactly satisfies yours 


  0-1 integer linear programming with only equality constraints (ILPEQ)


Now, you have a basic reduction argument. Suppose, you can solve ILPEQ fast, that means you can solve independent set fast, but it is known to be $\mathcal{NP}$-hard, so ILPEQ is also $\mathcal{NP}$-hard. The formal argument as well as checking that this is in fact a polynomial reduction I leave to the reader.