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 = 0, 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.
Edited on May 25, 2022, 7:14 am
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