We want to determine whether the expression

$(x-1)(x-2)(x-3)$

can ever be a perfect square, except when it equals zero.

Step 1: Identifying When the Expression is Zero

First, the expression equals zero when any of its factors is zero, which happens at $x = 1, 2, 3$. Since we are looking for perfect squares other than zero, we ignore these values.

Step 2: Understanding the Expression

The given expression represents the product of three consecutive integers:

$(x-1), (x-2), (x-3)$

A key property of three consecutive integers is that their product is always divisible by $6$ (since among any three consecutive integers, one is divisible by 2, and one is divisible by 3).

Step 3: Analyzing the Square Condition

For the product to be a perfect square, every prime factor in its prime factorization must have an even exponent. However, the structure of three consecutive integers creates problems:

1. One of the terms is always divisible by 2, meaning at least one factor of 2 exists in the product.
2. One of the terms is always divisible by 3, meaning at least one factor of 3 exists.
3. Other prime factors come in an unpredictable way, depending on the value of $x$.

Since we are multiplying three consecutive numbers, the product contains both even and odd powers of prime numbers. This means there is usually at least one prime factor with an odd exponent, preventing the product from being a perfect square.

Step 4: Considering Special Cases

To further confirm, let's test small integer values:

• $x = 4 \Rightarrow (4-1)(4-2)(4-3) = 3 \times 2 \times 1 = 6$ (not a perfect square)
• $x = 5 \Rightarrow (5-1)(5-2)(5-3) = 4 \times 3 \times 2 = 24$ (not a perfect square)
• $x = 6 \Rightarrow (6-1)(6-2)(6-3) = 5 \times 4 \times 3 = 60$ (not a perfect square)

A perfect square must have all its prime factors raised to even powers. Since the product of three consecutive numbers often introduces primes with odd exponents, this prevents it from being a perfect square.

Step 5: Conclusion

Since no integer $x$ satisfies the equation

$(x-1)(x-2)(x-3) = k^2$

for some integer $k$ (other than the trivial case where the product is zero), we conclude that the expression cannot be a perfect square.

Would you like a deeper explanation or a different approach?

Related Questions:

1. What is the general rule for determining whether a product of consecutive integers is a perfect square?
2. Can the product of two consecutive integers ever be a perfect square?
3. How do you determine whether a given number is a perfect square using prime factorization?
4. What other mathematical expressions always fail to be perfect squares?
5. How does modular arithmetic help in proving this type of problem?

Tip:

When dealing with perfect squares, always check the parity of exponents in the prime factorization—this often reveals whether a number can be a square!