Is it possible to get a perfect square if you multiply three consecutive natural numbers?

(In reply to

An Alternative Methodology by K Sengupta)

We know that:

"The term "natural number" refers either to a member of the set of positive integers 1, 2, 3, ... (Sloane's A000027) or to the set of nonnegative integers 0, 1, 2, 3, ... (Sloane's A001477; e.g., Bourbaki 1968, Halmos 1974)"

Source: http://mathworld.wolfram.com/NaturalNumber.html

Alos, we note that:

"In mathematics, a natural number (also called counting number) can mean either an element of the set {1, 2, 3, ...} (the positive integers) or an element of the set {0, 1, 2, 3, ...} (the non-negative integers)."

Source: http://en.wikipedia.org/wiki/Natural_number

Having regard to the foregoing, it can be asserted that:

A natural number is:

(i) A positive integer

(ii) A nonnegative integer.

In consonance with the first definition, a comprehensive proof had been given in terms of my earlier post that it is not possible to get a perfect square if you multiply three consecutive natural numbers,

We observe that, in consonance with the second definition, we additionally need to check whether 0*1*2 is a perfect square or otherwise.

It readily seen that 0*1*2 = 1, and we note that:

"A square number, also called a perfect square, is a figurate number of the form , where is an integer. The square numbers for , 1, ... are 0, 1, 4, 9, 16, 25, 36, 49, ... (Sloane's A000290)."

Source: http://mathworld.wolfram.com/SquareNumber.html

Since 0 is now verified to be a perfect square, in consonance with the second definition, there is precisely one solution in conformity with all the given conditions.

Summarizing, we have:

(i) No solution exists when a "natural number" is considered to be a positive integer.

(ii) Precisely one solution exist for the given problem, whenever a "natural number" is deemed as a nonnegative integer.