Mathematic rule about square numbers

Question

sorry for every possibile grammatical mistake I may make, anyway, I’d like to ask you if this rule has been “discovered” before: A square number is formed by the sum of the previous square number and the corresponsing odd number so:

25(5th square number)=16(previous square number)+9(5th odd number)

64(8th square number)=49(previous square number)+15(8th odd number)

4(2nd square number)=1(previous square number)+3(2nd odd number)

1(first square number)=1(because there’s no previous square number and 1 is the first odd number)

Eccetera eccetera. Has this ever been discovered?

This is very well known. See https://math.stackexchange.com/questions/1461779/sum-of-odd-numbers-make-squares, https://math.stackexchange.com/questions/136/why-are-the-differences-between-consecutive-squares-equal-to-the-sequence-of-odd, https://math.stackexchange.com/questions/639068/sum-of-odd-numbers-always-gives-a-perfect-square, etc., etc. — Hans Lundmark, Jan 18 '23 at 21:48
Good thing to observe - congratulations. That said, there are many duplicates of this question. Here's one: Direct Proof that $1 + 3 + 5 + \cdots+ (2n - 1) = n\cdot n$ — Ethan Bolker, Jan 18 '23 at 22:19

score 1 · Answer 1 · answered Jan 18 '23 at 21:25

1

Let $x^2$ be the xth square number. The xth odd number is just 2x - 1, and the previous square is (x - 1)^2. Thus, you're showing that:

$$x^2 = (x - 1)^2 + 2x - 1$$ $$x^2 = x^2 - 2x + 1 + 2x - 1$$ $$x^2 = x^2$$

answered Jan 18 '23 at 21:25

Math Man

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Mathematic rule about square numbers

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