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x = an integer one less than a square would be my guess.
I've proven that any solution must be an integer and that LHS ≥ RHS for any integer x, if that helps anyone else.
Why did the vector cross the road?
It wanted to be normal.
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If we define:
Then what we are asking is where in this function is there a fixed point. There are various theorems used to find fixed points, so I believe the next step would be to investigate the analytic properties of this noncontinuous function.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Without using fixed-point theorems, I can at least verify Mathsys assertion (but note that x = 0 is not a solution; x must be positive).
First, note that
for all integers . This is because for all integers , (easily proved).It remains to show that
for all integers . This means showing that for all integersor, since
for all integers ,i.e.
Well, unless Ive made a mistake somewhere, it is easy to see that the last line is true for all integers
,Hence all
, where n is any integer greater than or equal to 2, are solutions.Unfortunately we still need to show that there are no other solutions.
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Actually, we have that all
, where n is any positive integer, are solutions .So we have
and .Now suppose
, where , is a solution.Then we have
Then, in order for the solution to hold, we would need to have
This would mean
Now
and .If
, then .If
, then .In either case, we have a clear contradiction of
.So this proves that there are no integer solutions other than
.Last edited by JaneFairfax (2008-11-09 08:49:47)
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