Prove that no three consecutive non-zero integers can be three terms of a GP (*not necessarily consecutive).

Let the three consecutive non-zero integers be represented as \( a \), \( a+1 \), and \( a+2 \).

### Step 1: Assume the numbers form a GP
If these three numbers are in geometric progression (GP), then the common ratio \( r \) should be the same between consecutive terms:

\[
\frac{a+1}{a} = \frac{a+2}{a+1}
\]

### Step 2: Solve for \( a \)
Cross multiplying,

\[
(a+1)^2 = a(a+2)
\]

Expanding both sides,

\[
a^2 + 2a + 1 = a^2 + 2a
\]

Canceling \( a^2 + 2a \) from both sides,

\[
1 = 0
\]

### Step 3: Conclusion
Since the equation \( 1 = 0 \) is a contradiction, our original assumption that three consecutive non-zero integers can be in GP is false.

Thus, no three consecutive non-zero integers can form a geometric progression.