Suppose we have two consecutive numbers a and b. The relation between them among other is a^2 - b^2 = a + b (a is the greatest of the two)

For example 48^2 - 47^2 = 48 + 47

Also if we have a^3 - b^3 = 3 * a * b +1 (a is the greatest of the two)

For example 48^3 - 47^3 = 3 * 48 * 47 + 1

## Comments

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TopNewestI'm not really sure we can go so far as to call this a Theorem, but it's a nice piece of quick trickery sometimes useful to find the difference between squares or to find a square of an awkward number. However, basically what you've written is just a re-write of (x+1)^2 = x^2 + 2x + 1. Since it's also true that (x+y) = x^2 + 2xy+y^2, this lets us do things like square numbers like 1011) very quickly, since we already know that 1000^2 = 1000000. – Daniel Hirschberg · 3 years, 9 months ago

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– Kon Tim · 3 years, 9 months ago

I was just joking for the word theorem and what this really is, is 30^2 - 29^2 = (29 + 1)^2 - 29^2 = 29^2 +2 * 29 * 1 + 1^2 - 29^2 = 2 * 29 + 1 = 29 + 28. I noticed that yesterday when i was doing some exercises and i thought it was interesting and decided to publish it.Log in to reply

– Daniel Hirschberg · 3 years, 9 months ago

Kon - I didn't mean to criticize, as I said, it's a good tool or trick to have around for certain situations like squaring numbers in your head.Log in to reply

– Kon Tim · 3 years, 9 months ago

I didn't thought you criticized me, as i said i can totally get that this is not a theorem !Log in to reply

Your "Theorem" explained: Let the consecutive numbers be n and (n+1) then (n+1)^2-n^2 = ((n+1)-n)

((n+1)+n) as a^2-b^2 = (a+b)(a-b) hence (n+1)^2-n^2 = 1*((n+1)+n) = (n+1) + n. Hence proved :) – Saurabh Dubey · 3 years, 9 months agoLog in to reply

– Kon Tim · 3 years, 9 months ago

I thought i wrote exactly the same in the comment 7 hours before you!!!!Log in to reply

– Saurabh Dubey · 3 years, 9 months ago

Oh Man!!! i was not being sarcastic or rude!!!!. There is a smiley at the end too!!! :) . Moreover the explanations are quite different.I'm sorry anyways.Log in to reply