If k is the square of an even number, then k is a multiple of 4 (you should probably prove this).
so k = 4n for some .
2n+1 and 2n-1 are successive odd integers that sum to 4n = k.
a.) Prove the following theorem:
If k is the square of an integer that is even, then k is the sum of 2 successive odd integers.
b.) State the converse of the theorem. If it's true, prove it. If it's false, give a counterexample.
For a.), I'm not quite sure how to prove it, but I looked at what it means, that is, I know
2^2 = 1 + 3
4^2 = 7 + 9
6^2 = 17 + 19
8^2 = 31 + 33
etc.. I kind of see a pattern
For b.), the converse would be "If k is the sum of 2 successive odd integers, then k is the square of an even integer"
This is false. Example: 12 = 5 + 7, but 12 is not the square of an even integer.
Your right, my proof works for all multiples of 4. The first part of the proof should be to establish that all squares of even numbers are multiples of 4. I thought you should have a go at that part.Is what you said saying the same thing? It doesn't sound as though it is. Your thing is dealing with mod 4 stuff isn't it? Not sure how to prove this. Prove it by contrapositive maybe?
Hello, DiscreteW!
a) Prove the following theorem:
If is the square of an even integer,
then is the sum of two consecutive odd integers.
(a) An even integer has the form for some integer
2a)^2 \:=\:4a^2 \:=\:\underbrace{(2a^2-1)}_{\uparrow} + \underbrace{(2a^2 + 1)}_{\uparrow}" alt="\text{Then: }\;k \:=\2a)^2 \:=\:4a^2 \:=\:\underbrace{(2a^2-1)}_{\uparrow} + \underbrace{(2a^2 + 1)}_{\uparrow}" />
. . . . . . . . . . . . . . . two consecutive odd integers
.
Wow. That is so clear, and so obvious too. So if it's this obvious, how do I prove what you said? It looks "too obvious" to prove.
We know 2a^2 -1 is odd and 2a^2 + 1 is odd... hence it's what the theorem says (the addition of 2 odd numbers).
*EDIT*: Is that the proof? That looks like a direct proof to me. Can't get more clearer than that?