"If n is an odd integer, climate it is the distinction of two perfect squares" how is the proof for this valid!?

I feel like I am acquisition crazy pills.

You are watching: Every odd integer is a difference of two squares

I'm in a data structures class. I'm right now studying because that an exam, i m sorry is tomorrow.

On one inquiry in our homework, we had to prove the explain in the title. Here is the supposed answer:

"Let n = 2k + 1, for part k. Then take 2 continually squares, (k+1)2 and also k2. The distinction is (k+1)2 - k2 = k2 + 2k + 1 - k2 = 2k + 1. Done!"

So the point that bothers me is the "consecutive" in the solution. Sure, this proof works for 32 and 42, yet it doesn't take right into account 32 and 52, walk it? due to the fact that the proof is only for continuous integers?

What am i missing?


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level 1
· 5y
You're not trying come prove that for any two squares, their distinction is odd. You're trying come prove that any type of odd number is the distinction of some pair the perfect squares. And also it transforms out the consecutive squares do work.

So, as an example, room there any kind of two square number whose distinction is 7? Sure, 32 and also 42. It doesn't matter that 52 - 32 is even--we simply seek a solitary example, and the evidence as composed gives an instance for every strange number: if n = 2k + 1, climate the squares k2 and (k+1)2 room an example.

See more: Fruit Doesn T Fall Far From The Tree In English, The Apple Doesn'T Fall Far From The Tree


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level 2
Op · 5y

ahhh that defines things. Give thanks to you!


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level 1
· 5y

You're right. It works for squares of continuous numbers, and also it has proven that it works just for continuous numbers. We don't understand (we actually might, but that's not included in the initial post) around nonconsecutive numbers n and also n+m whereby m>1.


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