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 Multiplying gets a square? (Posted on 2004-06-26)
Is it possible to get a perfect square if you multiply three consecutive natural numbers?

 See The Solution Submitted by Federico Kereki Rating: 3.0000 (5 votes)

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 Further Analysis Comment 25 of 25 |
(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)"

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)."

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)."

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.

 Posted by K Sengupta on 2009-01-31 12:33:39

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