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Basic Proof help

 
 
lavaxy1
 
Reply Sat 25 Jan, 2014 04:42 pm
"m and n are integers such that n^2 + 1 = 2m" Prove that m is the sum of squares of two non-negative integers.

I'm having a lot of problems with this proof. I tried out a lot of tests and it comes out true, but i cant find a way to prove this works. I tried adding things/substituting numbers and rearranging, but I feel like there's an identity I'm missing.
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Kolyo
 
  1  
Reply Sat 25 Jan, 2014 05:42 pm
@lavaxy1,
One immediate insight I can give you is that n has to be odd. (Prove that to yourself.) Another insight is the n can be any positive odd number; another is that your choice of n completely determines the value of m.
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Kolyo
 
  1  
Reply Sat 25 Jan, 2014 05:48 pm
@lavaxy1,
Let's look at some low-n examples:

n=1 => m=1. 1 = 0^2 + 1^2
n=3 => m=5, 5 = 1^2 + 2^2
n=5 => m=13, 13 = 2^2 + 3^2

So you just have to prove two things:

(1) Whenever the equation holds, n=2k+1 for some non-negative integer k.
(2) If n=2k+1, then m= k^2 + (k+1)^2.
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markr
 
  1  
Reply Sat 25 Jan, 2014 07:57 pm
@lavaxy1,
As Kolyo pointed out, n is odd.
Let n=2p+1.
(2p+1)^2 + 1 = 2m
4p^2 + 4p + 1 + 1 = 2m
2p^2 + 2p + 1 = m
p^2 + (p^2 + 2p + 1) = m
p^2 + (p+1)^2 = m
m is the sum of two squares.
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