Prime time
Show that any odd prime can be written as the difference of the squares of two integers.
Labels: mathschallenge
posted by Chris at
6:32 PM
Friday, September 25, 2009
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9 Comments:
a^2 - b^2 = (a+b)(a-b)
How can this be a prime?
4^2 - 3^2 = 7
i.e
16 - 9 = 7(odd prime number)
Hi Wiz. That's the question :)
Hi Vishav_011. Thanks, for the sample. Was that a teaser, or do you know the general result?
Good one, Vishav. I stupidly missed the obvious case where a = b+1. Then a^2-b^2 = 2b+1.
So, as b increases from 1, then all the odd numbers are covered, not just the primes.
I'll have to learn to stop and think a bit more and not rush in when I see 0 comments and try to get in first!
Hi Chris, I think this works. If b is a prime number, a is some number we to determine bellow, require
(a+b)^2-a^2=b, hence a=0.5(1-b). Because b is prime, it can't be even (why you say it's odd?), therefore 1-b is even and is divided by 2. Your integers are 0.5(1+b) and 0.5(1-b)
If b is not 2, of course. I have no questions to you any more, Chris.
Well done everyone. As Wiz pointed out, any odd number can be represented as the difference of two squares.
Quantense actually came up with the solution I was going to give. I had to say odd primes as 2 (the only even prime) cannot be represented that way.
Just cover it completely:
Let o be any odd number. Let a=(o+1)/2 and b=(o-1)/2.
Then a²-b² = (o²+2o+1)/4 - (o²-2o+1)/4 = 4o/4 = o.
What's more, if f is any number exactly divisible by 4, then:
Let a=(f+4)/4 and b=(f-4)/4 => f=a²-b².
In both cases I'm allowing 0 as a square. If you're not happy with that, that only gives a minor mod to the initial assertions.
Hi quantense, our posts crossed again.
Hi quantense. I've just deleted the Magnetism blog. I doubt that I could easily disagree with anything that Richard Feynman said.
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