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 Integer Forms (Posted on 2012-01-07)
Let us denote F(n) = Σi= 1 to n i*(-1)i+1

For example, F(4) = 1-2+3-4, F(5) = 1-2+3-4+5, F(6) = 1-2+3-4+5-6... and, so on.

(1) Determine the general form of two positive integers x and y that satisfy:
F(x) + F(y) + F(x+y) = 2012

(2) Can you explain why no positive integer solution exists whenever F(x) + F(y) + F(x+y) = 2011?

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: Some Observations (partial solution) / The rest Comment 2 of 2 |
(In reply to Some Observations (partial solution) by Dustin)

I won't bother showing the proof of Dustin's assertion
When n is odd, F(n)=(n+1)/2
When n is even, F(n)=-n/2
Other that to mention that each case can easily be proven by induction.

In other words f(2a)= -a
and f(2a-1) = a

Now lets consider each of the 3 possibilities for x and y:

• 1) both are even x=2a, y=2b so x+y = 2a + 2b = 2(a+b)
• 2) both are odd x=2a-1, y=2b-1 so x+y = 2(a+b-1)
• 3) one of each x=2a, y=2b-1 (by symmetry it doesn't matter which) so x+y = 2(a+b) - 1
Case 1) F(x) + F(y) + F(x+y) = -a + -b + -(a+b) = -2(a+b)
In other words it is always negative if x and y are positive.

Case 2) F(x) + F(y) + F(x+y) = a + b + -(a+b-1) = 1
So you always get 1.

Case 3) F(x) + F(y) + F(x+y) = -a + b + (a+b) = 2b
In other words the even one doesn't matter but the odd one increases by one (from 2b-1 to 2b).

Since this 2b must be even, the sum of 2011 is not possible and the sum of 2012 requires 2b-1 to be 2011.

 Posted by Jer on 2012-01-07 20:29:52

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