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.