Analytically determine all possible solutions of the equation :
[p/2]² + [p/3] = 498, whenever p is a positive integer.

Note: [x] is defined as the greatest integer less than or equal to x.

If we let
x=[p/2] (x integer)
y=[p/3] (y integer)
then we have
x^2 + y = 498

Now,
[p/2]=x
p/2 in [x, x+1)
p in [2x, 2x+2)
[p/3]=y
p/3 in [y, y+1)
p in [3y, 3y+3)

As p is in both [2x, 2x+2) and [3y, 3y+3) these must overlap.
So we have:
3y+3>2x AND 3y<2x+2
Which can be written as:
3y>2x-3 AND 3y<2x+2

We have:
x^2 + y = 498
y = 498 - x^2
3y = 1494 - 3(x^2)

So that:
1494 - 3(x^2) > 2x-3 ... AND ... 1494 - 3(x^2) < 2x+2
Which can be written as:
x^2 + (2/3)x - (1492/3) > 0
...AND...
x^2 + (2/x)x - 499 < 0

Considering x is integer, each inequality yields, respectively:
(x<=-23 OR x>=22)
...AND...
-22 <= x <= 22

Which yields only x=22. This means that:
p in [2x, 2x+2)
p in [44, 45)
p=44, the only solution.

Posted by Kurious on 2007-04-16 11:02:11