Find the total number of solutions to the following system of equations: \[ \begin{cases} a^2\plus{}bc\equiv a\pmod {37}\\ b(a\plus{}d)\equiv b\pmod {37}\\ c(a\plus{}d)\equiv c\pmod{37}\\ bc\plus{}d^2\equiv d\pmod{37}\\ ad\minus{}bc\equiv 1\pmod{37}\end{cases}\]
To find the total number of solutions to the given system of equations, we will analyze each equation step by step and use properties of modular arithmetic and field theory.

Given system of equations:
\[
\begin{cases}
a^2 + bc \equiv a \pmod{37} \\
b(a + d) \equiv b \pmod{37} \\
c(a + d) \equiv c \pmod{37} \\
bc + d^2 \equiv d \pmod{37} \\
ad - bc \equiv 1 \pmod{37}
\end{cases}
\]

1. **First Equation:**
   \[
   a^2 + bc \equiv a \pmod{37}
   \]
   Rearrange to:
   \[
   a^2 - a + bc \equiv 0 \pmod{37}
   \]
   This can be seen as a quadratic equation in \(a\):
   \[
   a^2 - a + bc = 0
   \]
   The roots of this quadratic equation are:
   \[
   a, d = \frac{1 \pm \sqrt{1 - 4bc}}{2}
   \]

2. **Second and Third Equations:**
   \[
   b(a + d) \equiv b \pmod{37}
   \]
   \[
   c(a + d) \equiv c \pmod{37}
   \]
   If \(b \neq 0\), then:
   \[
   a + d \equiv 1 \pmod{37}
   \]
   Similarly, if \(c \neq 0\), then:
   \[
   a + d \equiv 1 \pmod{37}
   \]
   Therefore, we have:
   \[
   a + d = 1
   \]

3. **Fourth Equation:**
   \[
   bc + d^2 \equiv d \pmod{37}
   \]
   Substitute \(d = 1 - a\):
   \[
   bc + (1 - a)^2 \equiv 1 - a \pmod{37}
   \]
   Simplify:
   \[
   bc + 1 - 2a + a^2 \equiv 1 - a \pmod{37}
   \]
   Using \(a^2 - a + bc = 0\):
   \[
   1 - 2a + a \equiv 0 \pmod{37}
   \]
   \[
   1 - a \equiv 0 \pmod{37}
   \]
   \[
   a \equiv 1 \pmod{37}
   \]
   Therefore, \(a = 1\) and \(d = 0\).

4. **Fifth Equation:**
   \[
   ad - bc \equiv 1 \pmod{37}
   \]
   Substitute \(a = 1\) and \(d = 0\):
   \[
   1 \cdot 0 - bc \equiv 1 \pmod{37}
   \]
   \[
   -bc \equiv 1 \pmod{37}
   \]
   \[
   bc \equiv -1 \pmod{37}
   \]
   Since \(bc \equiv -1 \pmod{37}\), \(b\) and \(c\) must be non-zero and their product must be \(-1\).

5. **Conclusion:**
   The only solution that satisfies all the equations is:
   \[
   a = 1, d = 0, b = 0, c = 0
   \]
   However, this contradicts \(bc \equiv -1 \pmod{37}\). Therefore, the only consistent solution is:
   \[
   a = 1, d = 1, b = 0, c = 0
   \]

The final answer is \(\boxed{1}\)