That problem is too trivial.

by phiReKaLk6781, Oct 31, 2010, 9:36 PM

Try this much harder one.

Prove that every even prime is the sum of two odd numbers.

(Yes, in my math summer camp, this was given as the problem of the day on the very last day as a joke.)

EDIT BY BLOG OWNER: Fixed typo.
This post has been edited 2 times. Last edited by v_Enhance, Oct 31, 2010, 11:34 PM

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5 Comments

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Fail.

*every even

, not even even

by phiReKaLk6781, Oct 31, 2010, 9:37 PM

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Even even chang

Jk

lol which camp?

*That's pretty tricky.

Really long solution

Shorter Solution

by AwesomeToad, Oct 31, 2010, 10:11 PM

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You could use
return

instead of
print

.

by fries4guys, Nov 1, 2010, 2:09 PM

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lolfail. Thanks fries4guys

by AwesomeToad, Nov 1, 2010, 8:30 PM

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idk what I'm doing here and I don't know how to prove things properly but I'll give it a try

any even number can be represented as 2k, where k is an integer

we can write 2k as

2(k-1) + 2 = [2(k-1) + 1] + 1

we know that any number of the form 2m + 1 is an odd number and 1 is an odd number too

hence we can represent any even number as the sum of two odds

It is trivial to see that k can only be equal to 1 otherwise 2k wouldn't be prime

so we can represent 2 in the above form

[2(1-1) +1] +1 = 1 + 1 = 2


QED.

by Spinglasses, Apr 17, 2022, 6:21 AM

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